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Probability Made Easy!

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Re: Probability Made Easy! [#permalink]

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New post 16 Dec 2017, 12:22
Erjan_S wrote:
Bunuel wrote:
Braving the Binomial Probability

BY Karishma, VERITAS PREP


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_S

when 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! [#permalink]

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New post 05 Feb 2018, 07:40
Bunuel wrote:
When Does Order Matter?

BY Karishma, VERITAS PREP


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 (man-woman 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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Spiritual Yoda

Re: Probability Made Easy!   [#permalink] 05 Feb 2018, 07:40

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