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Probability Made Easy!
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23 Sep 2015, 08:00
Probability on the GMAT The topics on the GMAT quantitative section are chosen because most test takers have some experience solving questions on these topics in high school. Subjects like algebra and geometry have given high school students white hairs and craned necks for generations (what? I was stretching, not copying off of her exam, honest!). Areas such as calculus and linear algebra are usually only covered in college, putting students at varying levels of exposure depending on which programs they chose to study. Therefore, topics covered on the GMAT tend to come from high school level math, in order to give everyone a fair starting point. However, the problem with high school is that a lot of the learning is based on memorization and repetition. While these two procedures are important in the learning process, they are building blocks, not end points. To that point, much of math in high school is based on memorizing and applying formulae (“formulas” is also an acceptable plural for formula, but I prefer the more apposite formulae). Mechanically applying the correct formula usually yields the correct answer, for example the probability of seeing a heads on two consecutive coin flips is P(Heads on first) + P(Heads on second) – P(Heads of both). Identifying the correct probabilities for each occurrence will yield you the correct answer, but understanding why the chosen formula works is more crucial to getting these questions right on the GMAT. Automatically applying formulae without understanding what is being asked is a typical GMAT trap given that the exam is about logic and reasoning. Consider the following probability question:
There is a 50% chance Jen will visit Chile this year, while there is a 25% chance that she will visit Madagascar this year. What is the probability that Jen will visit either Chile or Madagascar this year, but NOT both?A. 25% B. 50% C. 62.5% D. 63.5% E. 75% Solution: Putting aside her peculiar travel schedule for a moment, (visiting Western South America and South East Africa without stopping by the Falklands?) what is the probability that Jen travels to one of these locales but not the other? Dutifully applying our memorized probability formula: the probability of visiting Chile is 0.5, and the probability of visiting Madagascar is 0.25, we get 0.5 + 0.25 – (0.5*0.25) or 0.75 – 0.125 = 0.625. Or answer choice C. (As Admiral Ackbar tried to warn us in 1983: It’s a Trap!) But why doesn’t the formula yield the correct choice? The answer lies in the question stem. This particular question asks us the probability of visiting Chile or Madagascar, but not both. The formula gives us the probability of visiting either, implicitly allowing the choice of visiting both. The probability of visiting either will indeed be 62.5% (or 5/8), but this is the probability of visiting Chile, Madagascar, or both. Verifying the converse, the probability of visiting neither is P(¬Chile) and P(¬Madagascar), or 0.5 * 0.75 = 0.375 (or 3/8), confirming that our 0.625 is merely the probability of not staying home this year. The obvious question now is: Why doesn’t the formula work? Didn’t the formula already account for the possibility of both? How do I solve this question correctly? (Admittedly, these are three questions and not one). The key to answering all of them is the same, though. Let’s go through the logic of the formula P(C) + P(M) – P(C&M): The first argument allows for all possibilities of visiting Chile, regardless of what happens with Madagascar. The second argument allows for all possibilities of visiting Madagascar, regardless of what happens with Chile (or the Falklands) The third argument is the possibility of both occurring. The formula works because P(C) accounts for the both choice, and P(M) accounts for the both choice as well, indicating that this option has been double counted. In order to count it only once, we need to remove one instance of it. This is why the formula works and is popular; it addresses the inherent problem in probability, double counting certain situations. The same logic applies to Venn diagrams and other similar question types where counting the same argument twice (or thrice) can occur. In practice, this question could then be solved using the default formula of P(C) + P(M) – P(C&M) and then subtracting P(C&M) again… or simply P(C) + P(M) – 2*P(C&M). Simply put, the first two arguments in the formula account for the “both” possibility twice, so we must remove it twice to answer the question at hand. A somewhat similar yet more straight forward alternative is to go bottom up instead of top down. The probability of going only to Chile is P(C) * P(¬M) = 0.5 * 0.75 = 0.375. The probability of going only to Madagascar is P(M) * P(¬C) = 0.25 * 0.5 = 0.125. Adding those two probabilities together yields the correct answer of 0.5 or 50%. The revised formula also yields this result (0.5 + 0.25 – 2 (0.125) = 0.5), so we can feel confident in our answer choice and not any of the other tempting answer choices. The answer is (B). This question is discussed HERE. On the GMAT, knowing the formula is often a useful shortcut to get to the correct answer more efficiently than trying to write out all the possibilities or estimating values. However, not knowing the formula is not a guarantee of an incorrect answer, and knowing the formula is not necessarily a guarantee of a correct answer either. It is always important to grasp the underlying logic before blindly applying formulae to a problem. The GMAT is much more often an exam of how you think than what you know.
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Re: Probability Made Easy!
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23 Sep 2015, 08:06
When Does Order Matter? I have to admit that probability is confusing. The problem is not so much that students find it hard to understand as that teachers find it hard to explain. There are subtle points in a probability question that make all the difference in the world and it takes a ton of ingenuity to explain them in a manner that others understand. You either get the point right away, or you don’t. Here, I will try to explain a probability concept I have always found very difficult to explain in person so the fact that I am attempting to explain it in a post is making me queasy. Nevertheless, the concept is important and I think it deserves a post. Let me give you two questions: Question 1: First 15 positive integers are written on a board. If two numbers are selected one by one from the board at random (the numbers are not necessarily different), what is the probability that the sum of these numbers is odd?Question 2: There is a group of people consisting of 10 men and 6 women. Among these 16 people, there are 4 married couples (manwoman couples). If one man and one woman are selected at random, what is the probability that a married couple gets selected?Now, let me give you the solutions to these questions. Note that the two solutions are different. We will discuss the reasons behind the difference today. Solution 1:Numbers: 1, 2, 3, 4, …, 13, 14, 15 When will the sum of two of these numbers be odd? When one number is odd and the other is even. P(Sum is Odd) = P(First number is Odd)*P(Second number is even) + P(First number is Even)*P(Second number is Odd) P(Sum is Odd) = (8/15)*(7/15) + (7/15)*(8/15) = 112/225 Solution 2:P(Selecting a Married Man) = 4/10 P(Selecting the Wife of that Man) = 1/6 P(Married Couple is Selected) = (4/10)*(1/6) = 4/60 The question I come across here is this: Why is the second question not solved the way we solved the first question? After all, selecting two things together is the same as selecting them one after another (explained in your Combinatorics book) i.e. why don’t we solve the second question in this way: P(Selecting a Married Couple) = P(Selecting a Married Man)*P(Selecting the Wife of that Man) + P(Selecting a Married Woman)*P(Selecting the Husband of the Woman) = (4/10)*(1/6) + (4/6)*(1/10) Other than the fact that it gives the wrong answer, why can’t we solve it like this? Because the order doesn’t matter here. It doesn’t matter whether we pick the husband first or the wife first. The end result is the same. After we pick either one, the probability of picking the other one stays the same. The two selections have to be made from two different groups. They cannot be made from the same group (contrary to the first question). It doesn’t matter whether you catch hold of the man first or the woman first. In the first question, the probability of picking the correct second number depends on what you picked in the first selection. Hence we consider the order. I will explain this by trying to solve the first question the way we solved the second question: On first selection, we can pick any number so the probability is 1. The second selection depends on what you selected in the first pick. If you selected an odd number in the first pick, the probability of selecting an even number is 7/15. If you selected an even number in the first pick, the probability of selecting an odd number is 8/15. So what do you do? Do you use 7/15 or 8/15 with 1? You cannot say so you must take individual cases. Case 1: Select an odd number and then an even number: (8/15) * (7/15) Case 2: Select an even number and then an odd number: (7/15) * (8/15) The total cases considered here are 15*15 (select first number in 15 ways and select the second number in 15 ways since the second number can be the same as the first number). In 8*7 ways, you will select an odd number and then an even number. In 7*8 ways, you will select an even number and then an odd number. In both the cases, the sum will be odd. This gives us a probability of (56+56)/225 = 112/225 The total probability of 1 is obtained as follows: 1 = P(first number odd, second number even) + P(first number even, second number odd) + P(first number odd, second number odd) + P(first number even, second number even) = 56/225 + 56/225 + 64/225 + 49/225 = 1 We are only interested in the 56/225 + 56/225 part. In the second question, we need to select a couple. Here, it doesn’t matter whether you select the man first or the woman; the two member types are different and there is only one way in which you can select the corresponding partner. You cannot select two members of the same type e.g. two men or two women. Hence we don’t need to bother with calculating the different cases of selecting the man first or the woman first. Of course, even if we do it, we will get the correct answer. Let me show you the calculation. The total number of ways of selecting a man and a woman are ‘select a man in 10 ways’ and ‘a woman in 6 ways’. Then ‘select a woman in 6 ways’ and ‘then a man in 10 ways’ i.e. total 120 ways. To select a couple, you can select a man in 4 ways and the woman in 1 way. You can select a woman in 4 ways and the man in 1 way. So total 4 + 4 = 8 ways. Probability of selecting a couple = 8/120 = 4/60 (same as before). To sum it, the two questions are quite different.In the first question, you have two groups of numbers: Even Numbers and Odd Numbers You can select the two numbers from different groups or from the same group. Hence the total number of cases is 15*15. Also, you can select the same number again. In the second question, you have two groups of members: Men and Women You must select the two members from different groups. You cannot select two men or two women. Hence the total number of cases is only 10*6 (and not 16*15). You cannot select the same member again. In case of confusion, just use the combinations approach rather than probability. You will invariably get the correct answer.
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Re: Probability Made Easy!
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23 Sep 2015, 08:10
At (the very) Least You Should Know This About Probability Ah, autumn. The busiest GMAT season of the year as application deadlines and backtoschool nostalgia fill the air, and that season always coincides with Major League Baseball’s pennant races and playoffs. And whether you’re a baseball fan or not, as an aspiring MBA you’ll find a fair amount of overlap between the two, as both the GMAT (and business) and baseball prominently feature the art of probability. Through that lens, let’s discuss one of the most helpful “tricks” to avoid some of the most timeconsuming types of problems on the GMAT, and we’ll lead with a problem: Whenever his favorite baseball team’s “closer” allows a hit, Sean becomes irate (just close out the game, Joe Nathan!). If the closer needs to get three outs to win the game, and each batter he will face has a .250 batting average (a 1/4 chance of getting a hit), what is the probability that he will give up at least one hit (assuming that there are no walks/errors/hitbatsmen)? And for those not consumed with baseball, this question essentially asks “if outcome A has a 25% chance of occurring in any one event, what is the probability that outcome A will happen at least once during three consecutive events?” Baseball makes for an excellent demonstration here, because if we take out the other “free base” situations, really only two things happen – a Hit or an Out. And since we need 3 Outs, we could have all kinds of sequences in which there is at least one hit: Hit, Out, Out, Out Out, Hit, Out, Out Out, Out, Hit, Out or episodes with multiple hits: Out, Hit, Hit, Out, Out Hit, Out, Hit, Out, Hit, Out or even Hit, Hit, Hit, Hit, Hit, Hit, Hit, Hit…(game called by mercy rule, Sean punches through his TV) The GMATrelevant point is this: when a problem asks you for the probability of “ at least one” of a certain event occurring, there are usually several ways that at least one could occur. But look at it this way: the ONLY way that you don’t get “ at least one” H is if all three Os come first. The opposite of “ at least one” is “none.” And there’s only one way to get “none” – it’s “Not Event A” then “Not Event A” then “Not Event A”… as many times as it takes to finish out the number of events. In other words, in this baseball analogy, if there’s a 25% chance of a hit then there’s a 75% chance of “not a hit” or “Out”, allowing us to set up the ONLY sequence in which there isn’t at least one hit: Out, Out, Out Which has a probability of: 3/4 * 3/4 * 3/4 Do the math, and you’ll find that there’s a 27/64 probability of “not at least one hit” and you can then know that the other 37/64 outcomes are “ at least one hit.” To the baseball fan, that means “take it easy on your closer – .250 is a pretty lackluster batting average and that even takes out the chance of walks and errors, and even with *that* there’s a betterthanlikely chance there will be baserunners in the 9th!” To the GMAT student, this example means that when you see a probability question that asks for the probability of “at least one” you should almost always try to calculate it by taking the probability of “none” (which is just one sequence and not several) and subtract that from 1. So your process is: 1) Recognize that the problem is asking for the probability of “at least one” of event A. 2) Find the probability for “not A” in any one event 3) Calculate the probability of getting “not A” in all outcomes by multiplying the “not A” probability as many times as there are outcomes 4) Subtract that total from 1 (and #5 – make sure the problem doesn’t involve any unique probabilitychanging events like “if outcome A doesn’t happen in the first try then the probability increases to X% for the second try” – that kind of language is rare but does complicate things) Probability factors into many autumn situations, so whether you’re a GMAT student or a baseball fan, if you know at least this one probability concept your autumn should be a lot less stressful!
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Re: Probability Made Easy!
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23 Sep 2015, 08:43
Separating the Couples Let’s take another tricky probability question today and employ two different methods to solve it. Question: Two couples and one single person occupy a row of five chairs at random. What is the probability that neither couple sits together (the husband and the wife should not occupy adjacent seats)?(A) 1/5 (B) 1/3 (C) 3/8 (D) 2/5 (E) 1/2 Solution: You can approach this GMAT problem in different ways. One way is a stepbystep case evaluation. Another is to go the reverse way: count all the arrangements in which at least one couple sits together and subtract that from the total arrangements possible. My method of choice is generally the second one. The only catch is that you have to remember to subtract from the total number of arrangements. What is the total number of arrangements (without any restrictions)? I hope you remember your basic counting principle and will agree that it should be 5! (Five people arranged in five seats). Now, let’s find out the number of favorable cases. We will discuss both the methods in detail. Method 1: Step – by – Step Case Evaluation Let’s say the two couples are {C1h, C1w} and {C2h, C2w} and the single person is S. Case 1: S takes the first/last chair If S takes the first chair, any one of the remaining people can take the chair next to him (4 ways). Say, C1h takes this spot. The next place cannot be occupied by C1w but either of C2h and C2w can occupy it (2 ways). Say, C2h occupies it. The fourth chair can be occupied by only one of the remaining 2 people since C2w cannot take it now (1 way). The last chair has only one person remaining for it. Total number of acceptable arrangements in which S takes the first chair = 4*2*1 = 8 The case would be exactly the same if S took the fifth seat. Think of it this way: The chairs have seat numbers 15. Now the numbers have reversed, 1 switched with 5, 2 switched with 4 and 3 is as it is. Now S is sitting on seat number 5 and we have exactly 8 more arrangements possible. Total number of acceptable arrangements in which S takes the first or fifth chair = 8*2 = 16 Case 2: S takes the second/fourth chair If S takes the second seat, any of the remaining four people could sit next to S on either side. However, we need to ensure that both people sitting on either side of S are not a couple i.e. C1h, S, C1w should not occupy the first, second and third seats respectively because then C2h and C2w are left and only 2 adjacent seats are vacant. But they cannot take adjacent seats. This means that there are 4 ways in which the first seat can be occupied — i.e., anyone can take it but there are only 2 ways in which the third seat can be occupied since the person taking the third seat must be from the other couple — i.e., if C1h takes the first seat, only C2h or C2w could take the third one. Now we have 2 people and 2 seats leftover. Fourth seat can be occupied in only one way since if C2h takes the third seat, C2w cannot take it i.e. one of the remaining two people cannot take it. Thereafter, one person and one seat are leftover so the fifth seat can be occupied in one way. Total number of acceptable arrangements in which S takes the second chair = 4*2*1*1 = 8 The case would be exactly the same if S took the fourth seat. Total number of acceptable arrangements in which S takes the second or fourth chair = 8*2 = 16 Case 3: S takes the middle seat i.e. the third seat If S takes the third seat, there are two seats on his left and two on his right. We have to ensure that a couple doesn’t sit on one side and the other side would automatically be couplefree. Any one of the four people can occupy the first seat (say C1h takes it). The second seat can be taken by one person from the other couple i.e. by C2h or C2w so it can be occupied in only 2 ways. Now we have two people leftover and two seats. Either one of them could take the fourth seat so it can be occupied in 2 ways. The fifth seat can be occupied in one way. Total number of acceptable arrangements in which S takes the third chair = 4*2*2*1 = 16 Total number of favorable arrangements = 16 + 16 + 16 = 48 Total number of arrangements = 120 Probability that neither couple sits together = 48/120 = 2/5 Answer (D) This question is discussed HERE. Method 2: The logic I use here is the one we use to solve SETS questions. It needs a little bit of thought but minimum case evaluations. There are two couples. We don’t want either couple to sit together. Let’s go the reverse way – let’s make at least one of them sit together. We can then subtract this number from the total arrangements to get the number of arrangements in which neither couple sits together. Would you agree that it is easy to find the number of arrangements in which both couples sit together? It is. We will work on it in a minute. Let’s think ahead now. How about ‘finding the number of ways in which one couple sits together?’ Sure we can easily find it but it will include those cases in which both couples are sitting together too. But we would have already found the number of ways in which both couples sit together. We just subtract ‘both couples together’ number from ‘one couple together’ number and get the number of arrangements in which ONLY one couple sits together. Think of SETS here. Let’s do this now. Number of arrangements in which both couples sit together: Let’s say the two couples are {C1h, C1w} and {C2h, C2w} and the single person is S. There are three groups/individuals. They can be arranged in 3! ways. But in each couple, husband and wife can be arranged in 2 ways (husband and wife can switch places) Hence, number of arrangements such that both couples are together = 3!*2*2 = 24 Number of arrangements such that C1h and C1w are together: C1 acts as one group. We can arrange 4 people/groups in 4! ways. C1h and C1w can be arranged in 2 ways (husband and wife can switch places). Number of arrangements in which C1h and C1w are together = 4! * 2 = 48 But this 48 includes the number of arrangements in which C2h and C2w are also sitting together. Therefore, number of arrangements such that ONLY C1h and C1w sit together = 48 – 24 = 24 Similarly, number of arrangements such that ONLY C2h and C2w sit together = 24 Number of arrangements in which at least one couple sits together = 24 + 24 + 24 = 72 Number of arrangements in which neither couple sits together = 120 – 72 = 48 Probability that neither couple sits together = 48/120 = 2/5 I believe that the second method is much faster and easier. Nevertheless, it’s good to know and understand both.
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Re: Probability Made Easy!
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23 Sep 2015, 08:14
Using Symmetry in Probability on the GMAT We know that Combinatorics and Probability are tricky topics. It is easy to misinterpret questions of these topics and get the incorrect answer – which, unfortunately, we often find in the options, giving us a false sense of accomplishment. In many questions, we need to account for different cases one by one but we don’t really see such questions on the GMAT since we have limited time. Also, we don’t tire of repeating this again and again – GMAT questions are more reasoning based than calculation intensive. Usually, there will be an intellectual method to solve every GMAT question – a method that will help you solve it in seconds. We have discussed using symmetry in Combinatorics before. It can be used in many questions though most people don’t realize that. In our ongoing endeavor to expose you to intellectual methods, here we present how most people tackle a question and how you can tackle it instead to be in the top 1%ile. Question: Let S be the set of permutations of the sequence 2, 3, 4, 5, 6 for which the first term is not 2. A permutation is chosen randomly from S. The probability that the second term is 5 is given by a/b (in lowest terms). What is a+b?(A) 5 (B) 6 (C) 11 (D) 16 (E) 19 Solution:Most Common Solution: What are the permutations of sequence S? They are the different ways in which we can arrange the elements of S. For example, 3, 2, 4, 5, 6 or 4, 2, 3, 6, 5 or 6, 3, 4, 5, 2 etc In how many different ways can we make the sequence? The first element can be chosen in 4 ways – one of 3, 4, 5 and 6. (You are given that 2 cannot be the first element). The second element can be chosen in 4 ways (2 and the leftover 3 numbers). The third element can be chosen in 3 ways. The fourth element can be chosen in 2 ways. And finally there will be only 1 element left for the last spot. Number of ways of making set S = 4*4*3*2*1 = 96 In how many of these sets will 5 be in the second spot? If 5 is reserved for the second spot, there are only 3 ways of filling the first spot (3 or 4 or 6). The second spot has to be taken by 5. The third element will be chosen in 3 ways (ignoring 5 and the first spot) The fourth element can be chosen in 2 ways. And finally there will be only 1 element left for the last spot. Number of favorable cases = 3*1*3*2*1 = 18 Required Probability = Favorable Cases/Total Cases = 18/96 = 3/16 = a/b a+b = 3 + 16 = 19 Answer (E) This question is discussed HERE. Intellectual Approach:Use a bit of logic of symmetry to solve this question without any calculations. Set S would include all such sequences as 3, 2, 4, 5, 6 or 4, 2, 3, 6, 5 or 6, 3, 4, 5, 2 etc – starting with 3, with 4, with 5 or with 6 with equal probability. By symmetry, note that 1/4th of them will start with 5 – which we need to ignore – so we are left with the rest of the 3/4th sequences. Now, in these 3/4th sequences which start with either 3 or 4 or 6, 5 could occupy any one of the 4 positions – second, third, fourth or fifth with equal probability. So we need 1/4th of these sequences i.e. only those sequences in which 5 is in the second spot. Probability that 5 is the second element of the sequence = (3/4)*(1/4) = 3/16 Therefore, a+b = 3+16 = 19 Answer (E)
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Re: Probability Made Easy!
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23 Sep 2015, 08:17
Probability with Conditions! Let’s look at the concept of conditional probability in detail today. (As if the probability questions weren’t tricky enough!) But since I like to discuss advanced concepts in this blog (in addition to alternative approaches and very important fundamentals), it would not be fair on my part to end the probability discussion without a quick review of conditional probability. Let me start by tossing a question at you. Question 1: Alex tosses a coin four times. On two of the tosses (we don’t know which two), he gets ‘Heads’. What is the probability that he gets ‘Tails’ on other two tosses?Solution: Wait a minute! Isn’t it something like the Binomial Probability questions we saw last week? It is but notice that it is also a conditional probability question. You are given that on at least 2 tosses, he got ‘Heads’. Under this condition, you want to find the probability that he got 2 tails i.e. he got 2 heads and 2 tails on his 4 tosses. Conditional Probability is calculated as given below: P(A given B) = P(A)/P(B) Here, we are trying to find the probability that event A happens given that event B happens. To understand this formula, think of it this way: Say there are a total of 100 cases and event B takes place in 10 cases. Also, event A takes place in 5 of the 10 cases in which event B takes place (A is a more restricted event under event B). Let’s say we know that event B has taken place. This means that one of the 10 cases has occurred. The probability that A has taken place is 5/10 = 1/2 and not 5/100. I hope this makes sense to you. Let me take an example to make this clearer. GMAT score can take one of 61 values (200/210/220 … 780/790/800). So there are a total of 61 cases. What is the probability that I will score above 700 on GMAT? (well, it should be 100% because otherwise I should not be writing blog posts on GMAT but let’s assume that all the scores are equally likely) There are 10 possible scores above 700 (710/720/730 … 800). Probability of a score above 700 = 10/61. That is our simple probability that we have been working on till date. Now, consider this: You know that I scored above 600. How much exactly, you do not know! What will you say is the probability that I scored above 700? (again assuming that all the scores are equally likely) I did score above 600. Now, what is the probability that I scored above 700? There are 20 possible scores above 600 (610/620/630 … 800). Any of them could have been my score. What is the probability that I actually scored above 700? It is 10/20. The event that I scored more than 700 is event A. It is more restrictive than event B i.e. the event that I scored more than 600. Given that event B took place i.e. I scored above 600, the probability that event A took place i.e. I scored above 700 is P(Score above 700)/P(Score above 600). This is conditional probability. I hope you see the difference between probability and conditional probability. Let’s go back to the original question now. We want to find this probability: P(‘2 Heads and 2 Tails’ given ‘At least 2 Heads’) = P(2 Heads and 2 Tails)/P(At least 2 Heads) We can easily find P(2 Heads and 2 Tails) and P(At least 2 Heads) since we are comfortable with the concepts of binomial probability! (right?) P(2 Heads and 2 Tails) = (1/2)*(1/2)*(1/2)*(1/2) * 4!/(2!*2!) = 3/8 You multiply by 4!/(2!*2!) because out of the four tosses, any 2 could be heads and the other two would be tails. So you have to account for all arrangements: HHTT, HTHT, TTHH etc P(Atleast 2 Heads) = P(2 Heads and 2 Tails) + P(3 Heads, 1 Tails) + P(4 Heads) Let me remind you here that we can also find P(Atleast 2 Heads) in the reverse way like this: P(Atleast 2 Heads) = 1 – [P(4 Tails) + P(3 Tails, 1 Heads)] Let me show you the calculations involved in both the methods. P(2 Heads and 2 Tails) = 3/8 (calculated above) P(3 Heads, 1 Tails) = (1/2)*(1/2)*(1/2)*(1/2) * 4!/3! = 1/4 We multiply by 4!/3! to account for all arrangements e.g. HHHT, HHTH etc P(4 Heads) = (1/2)*(1/2)*(1/2)*(1/2) = 1/16 P(Atleast 2 Heads) = 3/8 + 1/4 + 1/16 = 11/16 OR P(4 Tails) = (1/2)*(1/2)*(1/2)*(1/2) = 1/16 P(3 Tails, 1 Heads) = (1/2)*(1/2)*(1/2)*(1/2) * 4!/3! = 1/4 P(Atleast 2 Heads) = 1 – (1/16 + 1/4) = 11/16 As expected, the value of P(Atleast 2 Heads) is the same using either method. P(‘2 Heads and 2 Tails’ given ‘At least 2 Heads’) = P(2 Heads and 2 Tails)/P(At least 2 Heads) = (3/8)/(11/16) = 6/11 Notice here that you can ignore all the (1/2)s since in every case, you get (1/2)*(1/2)*(1/2)*(1/2) because Heads and Tails have equal probability. You can simply solve this question using this method: No of arrangements with 2 Heads and 2 Tails = 4!/(2!*2!) = 6 No of arrangements with 3 Heads and 1 Tails = 4!/3! = 4 No of arrangements with 4 Heads = 4!/4! = 1 No of arrangements with at least 2 Heads = 6 + 4 + 1 = 11 P(‘2 Heads and 2 Tails’ given ‘At least 2 Heads’) = 6/11 Out of the total number of arrangements of ‘At least 2 Heads’ (which is 11), only 6 are such that you get 2 Heads and 2 Tails. Mind you, you cannot do that if the probabilities differ. Look at the question given below: Question 2: Alex has five children. He has at least two girls (you do not know which two of his five children are girls). What is the probability that he has at least two boys too? (The probability of having a boy is 0.4 while the probability of having a girl is 0.6)Think about what you are going to do here. We will look at the solution of this question in the next post.
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Re: Probability Made Easy!
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23 Sep 2015, 08:36
Pairs Probability (And How You Can Use It to Win Super Bowl Bets) People are generally pretty bad at pairs probability. Here’s an example – if you were to bet a friend on “will this year’s Super Bowl champion repeat as next year’s Super Bowl champion?”, your friend might see the *random* odds as 1/64 (since the GMAT only deals in random probability, we’ll take actual talent, coaching, contract status, draft position out of the equation!). That’s because, in order for the 49ers, say, to repeat, they’ll have to win this year’s championship (a 1/2 chance) and then next year’s championship (and they’re 1 team out of 32). But this is wrong – your bet doesn’t ask for the probability of one *particular* team winning both Super Bowls, but rather the probability of “this year’s champion” (whichever team wins) doing it again next year. This year’s probability does not matter! Someone will win, and so you’re only concerned with that team’s (whatever it is – and there’s a 100% probability that there will be a winner) probability of repeating. Whatever that team is will have a 1/32 chance (again, just keeping it random) of repeating. This is a concept that does get tested on the GMAT, and when it does there’s always a trap answer. Consider the question: On three consecutive flips of a coin, what is the probability that all three produce the same result?(A) 1/16 (B) 1/8 (C) 1/4 (D) 3/8 (E) 1/2 Solution: The trap answer here is 1/8 – you might look at this as a 1/2 probability on the first flip, then a 1/2 on the second, and a 1/2 on the third for a 1/8 probability, but remember – in this case the result of the first flip doesn’t have to be one or the other. Your job is just to match whatever the first result was on the next two. If the first was heads, then you need heads next (a 1/2 chance) and heads again (a 1/2 chance). And if it were tails, then you need tails (1/2) then tails (1/2). But because “any match will do” and you don’t care that it’s a specific match – the question doesn’t specify all heads or all tails, just “all of one of them” – your probability doubles because you’re not concerned about the result of the first event, you’re only concerned about matching whatever that result was. This question is discussed HERE. So for probability questions that ask about pairs or matches, remember: 1) Check whether you need a *specific* pair/match or not. 2) If you don’t need a specific pair, but “any pair will do,” then the probability of the first result is 100% – something will happen. 3) If you need to guess, keep in mind that if it’s an unspecified pair/match, it’s almost certain that one of the trap answers will be a smaller number than the correct answer (in the above case, 1/8 is a trap and 1/4 is correct), so you can confidently rule out the smallest number and use number properties to try to eliminate another 12 answers.
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23 Sep 2015, 08:39
99 Problems But Probability Ain't One Some of the GMAT’s hardest Problem Solving problems can be made exponentially easier by keeping a famous JayZ lyric in the back of your mind. When you hear the phrase: If you’re having girl problems, I feel bad for you son? What immediately springs to mind? I got 99 problems but a b**** ain’t one. Now, what’s the GMAT genius in Hova’s lyric? He didn’t tell you what his problems WERE, he just told you what they WEREN’T. Explaining 99 problems would take way more than the two minutes you’d have for a quant problem or the ~3 minutes that Jay wants to spend on a track. And, like JayZ, you want to be Mr. One Take on GMAT problems, doing things the efficient way and getting to the answer much more quickly. So heed his advice when you see a problem like: Solange takes four roundhouse swings at her brotherinlaw. If she is just as likely to connect on any one punch as she is to not connect on that punch, what is the probability that she connects on at least one punch? Now, there are plenty of sequences in which she can connect: Hit, Miss, Hit, Miss Miss, Miss, Miss, Hit Hit, Hit, Hit, Hit (ouch!) etc. Trying to list out all the different ways in which she can land a punch is almost as timeconsuming as listing all of one’s 99 problems. But think of it this way – which of the sequences available “ain’t one”; which ways does she NOT land a punch. There’s only one: Miss, Miss, Miss, Miss And so if we’re calculating the probability among the 16 total sequences (each of two things can happen at each of four points, so the total number of sequences is 2^4 = 16), then if one doesn’t work the other 15 must work. So the probability is 15/16. And the “formula” to use on this essentially derives straight from JayZ’s lyrics about what “ain’t one”: For complementary events (when the probability of A + the probability of B = 100%), the probability of A = (1 – “not A”). And most strategically, this can be used as: The probability of “At least one” = (1 – probability of “none”)So if you’re calculating the probability of an outcome that has many different paths, see if it’s a cleaner calculation to determine the number of paths that “ain’t one” of your desired outcomes, and then just subtract those from one. Note that this ideology doesn’t just extend to probability. In many problems, calculating all the outcomes that “are” desired is a whole lot harder than calculating the outcomes that “ain’t one” of the desired. Consider this problem from this week’s GMATT Mondays session: Matt is touring a nation in which coins are issued in two amounts, 2¢ and 5¢, which are made of iron and copper, respectively. If Matt has ten iron coins and ten copper coins, how many different sums from 1¢ to 70¢ can he make with a combination of his coins?A) 66 B) 67 C) 68 D) 69 E) 70 Here look at the answer choices – they’re all very, very high numbers for the range (170) in question. So if your goal is to try to come up with all the possible coin combinations that work, you’ll be there a while. But what about the combinations that “ain’t one” of the possibilities? Since the maximum is 70, if you find the combinations that don’t work you’re doing this much more efficiently…and the answer choices tell you that at maximum only four won’t work so your job just became a lot easier. With 2 and 5 cent coins as your options, you can’t get to 1 and you can’t get to 3, so those are two “ain’t one” possibilities. And then “100% minus… comes back into play” – Notice too that 70¢ is the maximum possible sum (that would use all the coins), so 70¢ – 1¢, or 69¢, and 70¢ – 3¢, or 67¢ are impossible too. So the answer is 66, but the takeaway is bigger: when calculating all the possibilities looks to be far too timeconsuming, you often have the opportunity to calculate the possibilities that “ain’t one.” You’ve got a lot of problems to tackle on test day; hopefully this strategy allows you to make one question much less of one. This question is discussed HERE.
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23 Sep 2015, 08:50
The Intricacies of Probability Now that we have laid the groundwork for permutations and combinations, probability will be a piece of cake. We just need to build up on what we have already learned. The single most important concept in probability is the following: The probability of an event A is calculated as P(A) = No. of outcomes when A occurs/Total no. of outcomes. In this post, we will just extend the combinatorics concepts and apply them to probability. Let me explain how we will do it using some examples. Example 1: Six friends live in the city of Monrovia. There are four natural attractions around Monrovia – a waterfall, a safari, a lake and some caves. The friends decide to take a vacation together at one of these attractions. To select the attraction, each one of them votes for one of the attractions. What is the probability that each one of them votes for the safari?Solution: Here, A, the event for which we want to find the probability is ‘all six friends vote for the safari’ P(A) = No of ways in which all six can vote for the safari/Total no. of ways in which they can vote. What is the no. of ways in which all six vote for the safari? Only one way. They all vote for the safari! What is the no. of ways in which the friends can vote? Say, the friends are A, B, C, D, E and F. A can vote in 4 ways. B can vote in 4 ways. C can vote in 4 ways and so on… Total no of ways in which the 6 friends can vote = 4*4*4*4*4*4 = 4^6 (Using our old friend, the basic counting principle). We discussed this concept in our post on Unfair Distributions. Therefore, P(A) = \(\frac{1}{(4^6)}\) Finding this probability involved the use of the concepts we have already learned in combinatorics. I hope you see that it is quite simple and straight forward. Let’s tweak this example a little to make it slightly complicated. Example 2: Six friends live in the city of Monrovia. There are four natural attractions around Monrovia – a waterfall, a safari, a lake and some caves. The friends decide to take a vacation together at one of these attractions. To select the attraction, each one of them votes for one of the attractions. What is the probability that each one of them votes for the same attraction?Solution: Here, A, the event for which we want to find the probability is ‘all six friends vote for the same attraction’. We don’t have a specific attraction given to us. So the selected attraction could be any one of the given four. P(A) = No of ways in which all six can vote for the same attraction/Total no. of ways in which they can vote. What is the no. of ways in which all six vote for the same attraction? They could all vote for the waterfall or for the safari or for the lake or for the caves. All of them can vote for the same attraction in 4 ways. What is the no. of ways in which the friends can vote? As we saw in question no. 1, total no of ways in which the 6 friends can vote = 4*4*4*4*4*4 = 4^6 Therefore, P(A) = \(\frac{4}{(4^6)} = \frac{1}{(4^5)}\) Now, let’s make the question even trickier. Example 3: Six friends live in the city of Monrovia. There are four natural attractions around Monrovia – a waterfall, a safari, a lake and some caves. The friends decide to take a vacation together at one of these attractions. To select the attraction, each one of them votes for one of the attractions. What is the probability that each attraction gets at least one vote?Solution: Here, A, the event for which we want to find the probability is ‘each attraction gets at least one vote’. P(A) = No of ways in which each attraction gets at least one vote /Total no. of ways in which the friends can vote. Each attraction should get at least one vote. 6 votes can be divided among 4 attractions in the following ways: (1, 1, 1, 3) and (1, 1, 2, 2) Case 1: (1, 1, 1, 3) First, we select the attraction that will get 3 votes in 4 ways (= 4C1) Now, we can select the 3 people who will vote for this attraction in 6*5*4/3! = 20 ways (= 6C3 ) The other 3 votes will be distributed among the other 3 attractions in 3! = 6 ways The 6 people could vote for the 4 attractions in this case in 4*20*6 = 480 ways Case 2: (1, 1, 2, 2) Let’s select the two attractions that will get 2 votes each in 4*3/2! = 6 ways (= 4C2). Say we select caves and waterfall. Now, we can select the 2 people who will vote for one of the selected attractions in 6*5/2! = 15 ways (= 6C2) We can select the other 2 people who will vote for the other selected attraction in 4*3/2! = 6 ways (= 4C2) The other 2 votes will be distributed among the other 2 attractions in 2! = 2 ways The 6 people could vote for the 4 attractions in this case in 6*15*6*2 = 1080 ways Total number of ways in which 6 votes can be distributed among 4 attractions such that each attraction gets at least one vote = 480 + 1080 = 1560 ways As we saw in the questions above, the total no. of ways in which the friends can vote = 4^6 Therefore, P(A) = \(\frac{1560}{(4^6)}\) I hope you see that probability is just an extension of combinatorics. Some important concepts in Probability e.g. Independent events, mutually exclusive events, dependent events etc are discussed in detail in your Combinatorics and Probability book. Go through that theory before next Monday. We will discuss some tricky questions related to those concepts next week.
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23 Sep 2015, 20:54
KS15 wrote: Bunuel wrote: This is beautiful. Thanks Bunuelmuch needed. Can we have one for Inequalities, Permutation and Combinations, Statistics and Standard Deviation? We have similar topic on SD: statisticsmadeeasyallinonetopic203966.htmlOther topics are coming soon.
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Calculating the Probability of Intersecting Events We know our basic probability formulas (for two events), which are very similar to the formulas for sets: P(A or B) = P(A) + P(B) – P(A and B)P(A) is the probability that event A will occur. P(B) is the probability that event B will occur. P(A or B) gives us the union; i.e. the probability that at least one of the two events will occur. P(A and B) gives us the intersection; i.e. the probability that both events will occur. Now, how do you find the value of P(A and B)? The value of P(A and B) depends on the relation between event A and event B. Let’s discuss three cases: 1) A and B are independent eventsIf A and B are independent events such as “the teacher will give math homework,” and “the temperature will exceed 30 degrees celsius,” the probability that both will occur is the product of their individual probabilities. Say, P(A) = P(the teacher will give math homework) = 0.4 P(B) = P(the temperature will exceed 30 degrees celsius) = 0.3 P(A and B will occur) = 0.4 * 0.3 = 0.12 2) A and B are mutually exclusive eventsIf A and B are mutually exclusive events, this means they are events that cannot take place at the same time, such as “flipping a coin and getting heads” and “flipping a coin and getting tails.” You cannot get both heads and tails at the same time when you flip a coin. Similarly, “It will rain today” and “It will not rain today” are mutually exclusive events – only one of the two will happen. In these cases, P(A and B will occur) = 0 3) A and B are related in some other wayEvents A and B could be related but not in either of the two ways discussed above – “The stock market will rise by 100 points” and “Stock S will rise by 10 points” could be two related events, but are not independent or mutually exclusive. Here, the probability that both occur would need to be given to you. What we can find here is the range in which this probability must lie. Maximum value of P(A and B):The maximum value of P(A and B) is the lower of the two probabilities, P(A) and P(B). Say P(A) = 0.4 and P(B) = 0.7 The maximum probability of intersection can be 0.4 because P(A) = 0.4. If probability of one event is 0.4, probability of both occurring can certainly not be more than 0.4. Minimum value of P(A and B):To find the minimum value of P(A and B), consider that any probability cannot exceed 1, so the maximum P(A or B) is 1. Remember, P(A or B) = P(A) + P(B) – P(A and B) 1 = 0.4 + 0.7 – P(A and B) P(A and B) = 0.1 (at least) Therefore, the actual value of P(A and B) will lie somewhere between 0.1 and 0.4 (both inclusive). Now let’s take a look at a GMAT question using these fundamentals: There is a 10% chance that Tigers will not win at all during the whole season. There is a 20% chance that Federer will not play at all in the whole season. What is the greatest possible probability that the Tigers will win and Federer will play during the season?(A) 55% (B) 60% (C) 70% (D) 72% (E) 80% Let’s review what we are given. P(Tigers will not win at all) = 0.1 P(Tigers will win) = 1 – 0.1 = 0.9 P(Federer will not play at all) = 0.2 P(Federer will play) = 1 – 0.2 = 0.8 Do we know the relation between the two events “Tigers will win” (A) and “Federer will play” (B)? No. They are not mutually exclusive and we do not know whether they are independent. If they are independent, then the P(A and B) = 0.9 * 0.8 = 0.72 If the relation between the two events is unknown, then the maximum value of P(A and B) will be 0.8 because P(B), the lesser of the two given probabilities, is 0.8. Since 0.8, or 80%, is the greater value, the greatest possibility that the Tigers will win and Federer will play during the season is 80%. Therefore, our answer is E. This question is discussed HERE.
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23 Sep 2015, 08:28
Probability with Conditions Part II In the last post I left you with a conditional probability question. Let’s look at its solution now. This will be my last post on GMAT Combinatorics and Probability (for a while at least) until and unless you want me to take up a particular concept/question related to this topic. Next week, we will start a new topic. Back to question at hand: Question 2: Alex has five children. He has at least two girls (you do not know which two of her five children are girls). What is the probability that he has at least two boys too? (The probability of having a boy is 0.4 while the probability of having a girl is 0.6)Solution:We want to find this probability: P(‘At least 2 Boys and at least 2 Girls’ given ‘At least 2 Girls’) = P(At least 2 Boys and at least 2 Girls)/P(At least 2 Girls) Let’s try and find P(At least 2 Boys and at least 2 Girls) and P(At least 2 Girls) ‘At least 2 Boys and at least 2 Girls’ can be obtained in two ways: ‘3 Boys and 2 Girls’ or ‘2 Boys and 3 Girls’ P(At least 2 Boys and at least 2 Girls) = P(3 Boys and 2 Girls) + P(2 Boys and 3 Girls) \(P(2 \ Boys \ and \ 3 \ Girls) = 0.4*0.4*0.6*0.6*0.6 * \frac{5!}{(2!*3!)} = (0.4)^2 * (0.6)^3 * 10\) You multiply by 5!/(2!*3!) because out of the five children, any 2 could be boys and the other three would be girls. So you have to account for all arrangements: BBGGG, BGBGG, GGBGB etc \(P(3 \ Boys \ and \ 2 \ Girls) = 0.4*0.4*0.4*0.6*0.6 * \frac{5!}{(3!*2!)} = (0.4)^3 * (0.6)^2 * 10\) \(P(At \ least \ 2 \ Boys \ and \ at \ least \ 2 \ Girls) = [(0.4)^2 * (0.6)^3 * 10] + [(0.4)^3 * (0.6)^2 * 10] = (0.4)^2 * (0.6)^2 *10 (0.6 + 0.4) = (1.6)(0.36)\) Now that we have P(At least 2 Boys and at least 2 Girls), let’s focus on getting P(At least 2 Girls). Again, as we saw last week, there are 2 ways of arriving at P(At least 2 Girls). P(At least 2 Girls) = P(2 Girls and 3 Boys) + P(3 Girls and 2 Boys) + P(4 Girls + 1 Boy) + P(5 Girls) OR P(At least 2 Girls) = 1 – P(5 Boys) – P(1 Girl and 4 Boys) Let me show you the calculations involved in both the methods. Method 1: P(At least 2 Girls) = P(2 Girls and 3 Boys) + P(3 Girls and 2 Boys) + P(4 Girls + 1 Boy) + P(5 Girls) \(P(2 \ Girls \ and \ 3 \ Boys) = (0.4)^3 * (0.6)^2 * 10\) (from above) \(P(3 \ Girls \ and \ 2 \ Boys) = (0.4)^2 * (0.6)^3 * 10\) (from above) \(P(4 \ Girls \ + \ 1 \ Boy) = (0.4)*(0.6) *(0.6)*(0.6)*(0.6)*5!/4! = (0.4) * (0.6)^4 * 5\) \(P(5 \ Girls) = (0.6)*(0.6)*(0.6)*(0.6)*(0.6) = (0.6)^5\) \(P(At \ least \ 2 \ Girls) = [(0.4)^3 * (0.6)^2 * 10] + [(0.4)^2 * (0.6)^3 * 10] + [(0.4) * (0.6)^4 * 5] + [(0.6)^5]\) Method 2: \(P(At \ least \ 2 \ Girls) = 1 – P(5 \ Boys) – P(1 \ Girl \ and \ 4 \ Boys)\) \(P(5 \ Boys) = (0.4)* (0.4)* (0.4)* (0.4)* (0.4) = (0.4)^5\) \(P(1 \ Girl \ and \ 4 \ Boys) = (0.6)* (0.4)*(0.4)*(0.4)*(0.4)*\frac{5!}{4!} = (0.6)*(0.4)^4 * 5\) \(P(At \ least \ 2 \ Girls) = 1 – [(0.4)^5] – [(0.6)*(0.4)^4 * 5]\) The values in bold are the same even if they don’t look same. (Trust me, I checked on my financial calculator!) \(P(‘At \ least \ 2 \ Boys \ and \ at \ least \ 2 \ Girls’ \ given \ ‘At \ least \ 2 Girls’) = P(At \ least \ 2 \ Boys \ and \ at \ least \ 2 Girls)/P(At \ least \ 2 \ Girls)\) \(P(‘At \ least \ 2 \ Boys \ and \ at \ least \ 2 \ Girls’ \ given \ ‘At \ least \ 2 Girls’) = \frac{(1.6)(0.36)}{[1 – (0.4)^5 – (0.6)*(0.4)^4 * 5]}\) Even though the solution looks complicated, I hope you see that the approach is quite logical and straight forward.
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23 Sep 2015, 08:47
Braving the Binomial Probability I would like to take up a couple of questions on binomial probability today. The concepts of the topic have been covered in detail in the book so I am assuming that you know how to solve questions such as “What is the probability of getting at least 3 heads on 5 tosses of a coin?” etc. Therefore, let’s work on a couple of questions which use the binomial probability with a twist. Question 1: Martin and Joey are playing a coin game in which each player tosses a fair coin alternately. The player who gets a ‘Heads’ first wins. The maximum number of tosses allowed in a single game for any player is 6. What is the probability that the person who tosses first will win the game?Solution:Probability of getting ‘Heads’ on a single toss = 1/2 Probability of getting ‘Tails’ on a single toss = 1/2 The person who starts the game can win the game if one of the following scenarios plays out: The first person tosses the coin and gets a ‘Heads’ right away. The first person wins! The first person tosses the coin and gets a ‘Tails’. The second person gets ‘Tails’ too. The first person tosses again and gets a ‘Heads’. The first person wins! The first person tosses the coin and gets a ‘Tails’. The second person gets ‘Tails’ too. The first person tosses again and gets a ‘Tails’ again. The second person gets ‘Tails’ again too. Finally, the first person tosses and this time, gets a ‘Heads’. The first person wins! and so on… In the worst case, the first person will have to toss 6 times to get a ‘Heads’. He and the second person would end up getting ‘Tails’ on five previous tosses. Probability that the first person tosses the coin and gets a ‘Heads’ right away = 1/2 Probability that the first person tosses the coin and gets a ‘Tails’ (1/2), the second person gets ‘Tails’ (1/2) and then the first person gets a ‘Heads’(1/2) = (1/2)*(1/2)*(1/2) = (1/2)^3 Probability that the first person tosses the coin and gets a ‘Tails’ (1/2), the second person gets ‘Tails’ (1/2) , the first person tosses again and gets a ‘Tails’ again (1/2), the second person gets ‘Tails’ again (1/2) and finally, the first person tosses and this time, gets a ‘Heads’ (1/2) = (1/2)^5 and so on… Probability that the first person will have to toss 6 times to get a ‘Heads’ = (1/2)^11 To get the probability of the first person winning, we just need to add all these probabilities now. Probability that the first person will win = (1/2) + (1/2)^3 + (1/2)^5 + (1/2)^7 + (1/2)^9 + (1/2)^11 On the same lines, can you find the probability that the person who tosses second wins? I hope you understand that it is very similar to what we have already discussed. The person who tosses second will win if one of the following happens: The person who tosses first gets ‘Tails’ and then the person who tosses second gets ‘Heads’. The person who tosses first gets ‘Tails’, the person who tosses second gets ‘Tails’, the person who tosses first gets ‘Tails’ again and the second person then gets ‘Heads’. and so on… Probability that the second person will win = \((\frac{1}{2})^2 + (\frac{1}{2})^4 + (\frac{1}{2})^6 + (\frac{1}{2})^8 + (\frac{1}{2})^10 + (\frac{1}{2})^{12}\) I hope you see that the question is quite straight forward. Now, let’s take a question very similar to one from a GMAT Prep test. Question 2: For one toss of a certain coin, the probability that the outcome is heads is 0.7. If the coin is tossed 6 times, what is the probability that the outcome will be tails at least 5 times?Solution: This question is very similar to the questions we saw in the Probability book. The only difference is that we are not tossing a fair coin. The probability of getting heads is 0.7 not 0.5. So the probability of getting tails must be 0.3 since the total probability has to add up to 1. The only acceptable cases are those in which we get ‘tails’ on all 6 tosses or we get tails on exactly 5 of the 6 tosses. P(Tails on all 6 tosses) = \((0.3)*(0.3)*(0.3)*(0.3)*(0.3)*(0.3) = (0.3)^6\) P(Tails on exactly 5 tosses and Heads on one toss) = \((0.3)^5*(0.7)*6\) We multiply by 6 because 5 tails and 1 heads can be obtained in 6 different ways: HTTTTT, THTTTT, TTHTTT, TTTHTT, TTTTHT, TTTTTH Probability that the outcome will be tails at least 5 times = Probability that the outcome will be tails 5 times + Probability that the outcome will be tails 6 times Probability that the outcome will be tails at least 5 times = \((0.3)^6 + (0.3)^5*(0.7)*6\) Again, the question is straight forward. It just has a little twist which sometimes throws people off during the test. It is these little things that differentiate a medium level question from a high level question.
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23 Sep 2015, 18:45
Bunuel wrote: This is beautiful. Thanks Bunuelmuch needed. Can we have one for Inequalities, Permutation and Combinations, Statistics and Standard Deviation?



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17 Jun 2016, 09:51
Bunuel wrote: The Intricacies of Probability Example 3: Six friends live in the city of Monrovia. There are four natural attractions around Monrovia – a waterfall, a safari, a lake and some caves. The friends decide to take a vacation together at one of these attractions. To select the attraction, each one of them votes for one of the attractions. What is the probability that each attraction gets at least one vote?Solution: Here, A, the event for which we want to find the probability is ‘each attraction gets at least one vote’. P(A) = No of ways in which each attraction gets at least one vote /Total no. of ways in which the friends can vote. Each attraction should get at least one vote. 6 votes can be divided among 4 attractions in the following ways: (1, 1, 1, 3) and (1, 1, 2, 2) Case 1: (1, 1, 1, 3) First, we select the attraction that will get 3 votes in 4 ways (= 4C1) Now, we can select the 3 people who will vote for this attraction in 6*5*4/3! = 20 ways (= 6C3 ) The other 3 votes will be distributed among the other 3 attractions in 3! = 6 ways The 6 people could vote for the 4 attractions in this case in 4*20*6 = 480 ways Case 2: (1, 1, 2, 2) Let’s select the two attractions that will get 2 votes each in 4*3/2! = 6 ways (= 4C2). Say we select caves and waterfall. Now, we can select the 2 people who will vote for one of the selected attractions in 6*5/2! = 15 ways (= 6C2) We can select the other 2 people who will vote for the other selected attraction in 4*3/2! = 6 ways (= 4C2) The other 2 votes will be distributed among the other 2 attractions in 2! = 2 ways The 6 people could vote for the 4 attractions in this case in 6*15*6*2 = 1080 ways Total number of ways in which 6 votes can be distributed among 4 attractions such that each attraction gets at least one vote = 480 + 1080 = 1560 ways As we saw in the questions above, the total no. of ways in which the friends can vote = 4^6 Therefore, P(A) = \(\frac{1560}{(4^6)}\) Why isn't it possible to solve the question with this approach?  # of ways that every attraction has 1 vote: Selection of 4 from 6 and the order does matter: \(\frac{6!}{2!}= 360\)  now there are 2 votes left which have to be distributed over the 4 attractions: (comparable with the distribution of 2 balls over 4 urns and the balls are not distinguishable): \(\frac{(n+k1)!}{(k! * (n1)!)} = \frac{(4+21)!}{(2! * (41)!)} = \frac{5!}{(2!*3!)}=10\)  total # of possibilities: 360 * 10 = 3600 Therefore, \(P(A) = \frac{3600}{(4^6)}\) Thanks!



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Probability Made Easy!
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26 Jul 2017, 04:09
Hi, are you able to please explain the conditional probability question from understanding rather than simply formulas, or, more precisely, see exactly where my logic is failing. If I am thinking (what I believe to be) logically, and I am told that I flip a coin 4 times, atleast two of those tosses were heads, what is the probability of tossing two heads and two tails. I know that I can fix two heads into my list of all possible combinations. From here I simply think, ok, I'll stick the two heads into my initial two tosses. I can get a possible 4 solutions from here, HHHH, HHHT, HHTH and HHTT. So I see the probability from this point forward looks like it is 1/4. Now, I think that those two heads which I fixed could be in other positions, but using the symmetry which you taught me, regardless of where I am fixing my heads, my odds of two heads and two tails remains at 1/4. How is my logic incorrect.
I suppose further to this, with the equation P(A given B) = P(A)/P(B). How does this work in conjunction with the equation I learnt previously whereby P(A given B) = P(A & B) / P(B). The latter is more logical to me considering I would have thought there would be scenarios where P(A) is actually greater than P(B), making your answer with the first formula >1.
Thanks for your help, love your work!



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07 Dec 2017, 10:07
Bunuel wrote: When Does Order Matter? Of course, even if we do it, we will get the correct answer. Let me show you the calculation. The total number of ways of selecting a man and a woman are ‘select a man in 10 ways’ and ‘a woman in 6 ways’. Then ‘select a woman in 6 ways’ and ‘then a man in 10 ways’ i.e. total 120 ways. To select a couple, you can select a man in 4 ways and the woman in 1 way. You can select a woman in 4 ways and the man in 1 way. So total 4 + 4 = 8 ways. Probability of selecting a couple = 8/120 = 4/60 (same as before). Hey Bunuel, I did understand that the second questions cant be 16*15 ways, but still, my calculation would be: 4/60 + 4/60 = 8/60 (and not 8/120), can you please help my understand way am i wrong? Thanks!



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07 Dec 2017, 10:23
omavsp wrote: Bunuel wrote: When Does Order Matter? Of course, even if we do it, we will get the correct answer. Let me show you the calculation. The total number of ways of selecting a man and a woman are ‘select a man in 10 ways’ and ‘a woman in 6 ways’. Then ‘select a woman in 6 ways’ and ‘then a man in 10 ways’ i.e. total 120 ways. To select a couple, you can select a man in 4 ways and the woman in 1 way. You can select a woman in 4 ways and the man in 1 way. So total 4 + 4 = 8 ways. Probability of selecting a couple = 8/120 = 4/60 (same as before). Hey Bunuel, I did understand that the second questions cant be 16*15 ways, but still, my calculation would be: 4/60 + 4/60 = 8/60 (and not 8/120), can you please help my understand way am i wrong? Thanks! The way Karishma is dong this is (man)(woman)/(total) + (woman)(man)/(total) = 4/120 + 4/120 = 8/120.
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Probability Made Easy!
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Updated on: 16 Dec 2017, 21:32
Bunuel wrote: Braving the Binomial Probability Question 2: For one toss of a certain coin, the probability that the outcome is heads is 0.7. If the coin is tossed 6 times, what is the probability that the outcome will be tails at least 5 times?Solution: This question is very similar to the questions we saw in the Probability book. The only difference is that we are not tossing a fair coin. The probability of getting heads is 0.7 not 0.5. So the probability of getting tails must be 0.3 since the total probability has to add up to 1. The only acceptable cases are those in which we get ‘tails’ on all 6 tosses or we get tails on exactly 5 of the 6 tosses. P(Tails on all 6 tosses) = \((0.3)*(0.3)*(0.3)*(0.3)*(0.3)*(0.3) = (0.3)^6\) [color=#ffff00]P(Tails on exactly 5 tosses and Heads on one toss) = \((0.3)^5*(0.7)*6\)
We multiply by 6 because 5 tails and 1 heads can be obtained in 6 different ways: HTTTTT, THTTTT, TTHTTT, TTTHTT, TTTTHT, TTTTTH
Probability that the outcome will be tails at least 5 times = Probability that the outcome will be tails 5 times + Probability that the outcome will be tails 6 times
Probability that the outcome will be tails at least 5 times = \((0.3)^6 + (0.3)^5*(0.7)*6\)[/color]I am confused by this multiplication by 6  why when probability that the outcome will be tails 6 times we simply (0.3)^6 and when probability that the outcome will be tails 5 times  we do multiply (0.3)^5*(0.7) by 6? Why tails on all 6 tosses is (0.3)^6 without any multiplication by 6! and Tails on exactly 5 tosses and heads on one toss needs to be multliplied by 6?
Originally posted by Erjan_S on 16 Dec 2017, 11:10.
Last edited by Erjan_S on 16 Dec 2017, 21:32, edited 1 time in total.




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