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Re: Probability Made Easy!
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16 Dec 2017, 11:22
Erjan_S wrote: 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) = \((0.3)*(0.3)*(0.3)*(0.3)*(0.3)*(0.3) = (0.3)^6\) 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 = \((0.3)^6 + (0.3)^5*(0.7)*6\)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? Hi Erjan_Swhen you are sure that only tails will occur, then the situation will look like TTTTTT i.e only 1 possibility. Hence probability of tails on all 6 counts will be (0.3)^6 but when 5 tails and 1 heads has to occur, then this situation throws up additional possibilities  HTTTTT, the probability of this possibility is 0.7*(0.3)^5 THTTTT, the probability of this possibility is 0.7*(0.3)^5 TTHTTT, the probability of this possibility is 0.7*(0.3)^5 TTTHTT, the probability of this possibility is 0.7*(0.3)^5 TTTTHT, the probability of this possibility is 0.7*(0.3)^5 TTTTTH, the probability of this possibility is 0.7*(0.3)^5 Hence total possibility in this case will be sum of all the above possibilities, which simply means you multiply 0.7*(0.3)^5 by 6



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Re: Probability Made Easy!
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05 Feb 2018, 06:40
Bunuel wrote: 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. In question 1, why is it said 'the numbers are not necessarily different'? Isn't it ambiguous to say it? Wouldn't the probabilty change if some of the numbers repeat?
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Probability Made Easy!
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18 Jan 2019, 01:48
VeritasKarishmaRegarding this problem: 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)Since it is already given that there are at least 2 girls, cant the answer be: P(at least two boys GIVEN at least 2 girls)= P(2Boys,3Girls)+P(3 Boys,2Girls) P(2b,3g)=0.4*0.4+0.6*0.6*0.6]*(5!/(2!*3!)) P(3b,2g)=0.4*0.4*0.4+0.6*0.6]*(5!/(2!*3!)) My reasoning: Since its already conditioned that there are 2 girls atleast, so we donot include the arrangements of (4Boys,1 Girl) and (5 Boys) Please put ur valuable feedback Thanks in advance



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Re: Probability Made Easy!
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18 Jan 2019, 03:35
Debashis Roy wrote: VeritasKarishmaRegarding this problem: 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)Since it is already given that there are at least 2 girls, cant the answer be: P(at least two boys GIVEN at least 2 girls)= P(2Boys,3Girls)+P(3 Boys,2Girls) P(2b,3g)=0.4*0.4+0.6*0.6*0.6]*(5!/(2!*3!)) P(3b,2g)=0.4*0.4*0.4+0.6*0.6]*(5!/(2!*3!)) My reasoning: Since its already conditioned that there are 2 girls atleast, so we donot include the arrangements of (4Boys,1 Girl) and (5 Boys) Please put ur valuable feedback Thanks in advance Yes, but Probability = Favourable cases/Total cases Favourable cases are (2 boys, 3 girls) and (3 boys, 2 girls)  Cases in which there are at least 2 boys too along with 2 girls. Total cases are (5 girls), (1 boy, 4 girls), (2 boys, 3 girls) and (3 boys, 2 girls)  these are all cases in which there are at least 2 girls.
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Probability Made Easy!
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23 Jan 2019, 04:59
Quote: 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?
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 Hey Bunuel VeritasKarishma chetan2uWhy cant i solve the above using the normal combinations method? Total outcomes = \(\frac{15*14}{2}\) What does this mean?  "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." Please elaborate! Super confused



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Re: Probability Made Easy!
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23 Jan 2019, 05:21
blitzkriegxX wrote: Quote: 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?
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 Hey Bunuel VeritasKarishma chetan2uWhy cant i solve the above using the normal combinations method? Total outcomes = \(\frac{15*14}{2}\) What does this mean?  "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." Please elaborate! Super confused By using combinations, you are choosing two numbers from 15 in a way that the two number are DIFFERENT, but we have been told that they need not be different, so after choosing 1 number from 15, we have all 15 again to choose the second on efrom. Two statements.. (1) choose two numbers from 15  15C2 (1) choose two numbers one by one and the two numbers can be same  15*15
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Re: Probability Made Easy!
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23 Jan 2019, 05:27
chetan2u wrote: blitzkriegxX wrote: Quote: 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?
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 Hey Bunuel VeritasKarishma chetan2uWhy cant i solve the above using the normal combinations method? Total outcomes = \(\frac{15*14}{2}\) What does this mean?  "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." Please elaborate! Super confused By using combinations, you are choosing two numbers from 15 in a way that the two number are DIFFERENT, but we have been told that they need not be different, so after choosing 1 number from 15, we have all 15 again to choose the second on efrom. Two statements.. (1) choose two numbers from 15  15C2 (1) choose two numbers one by one and the two numbers can be same  15*15 Thank you chetan2u! I dont know how i missed that part! It is like choosing "with replacement" if i am not wrong! I lost half my hair scratching my head on this! Thank you again for the super quick response! (you saved me from going bald)



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Re: Probability Made Easy!
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23 Oct 2019, 01:51
Bunuel wrote: 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) BunuelVeritasKarishmaCould someone please elaborate the symmetry approach? I didn't get it at all.



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Re: Probability Made Easy!
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24 Oct 2019, 19:46
Bunuel wrote: 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. How can this be solved with combinatorics approach? Please help!




Re: Probability Made Easy!
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