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Let me try to elaborate Each day, we have 2 possibilities, rain or not rain, therefore the total possibility of 5 days is 2^5=32 Then, we pick 3 rainy days randomly from 5 days, 5C3 = 10 10/32=5/16

Binomial distribution formula: C(n,k) * p^k * (1-p)^(n-k)

Given that the probability of Rain happening is p (=1/2) and not happening is 1-p (=1-1/2=1/2), => Probability of Rain happening k times (=3) in n repeated tests (=5) = C(5,3) * (1/2)^3 * (1-1/2)^(5-3) = C(5,3) * (1/2)^5 = 10/32 = 5/16
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If the probability of rain on any given day in City X is 50 percent, what is the probability that it rains on exactly 3 days in a 5-day period?

(A) 8/125 (B) 2/25 (C) 5/16 (D) 8/25 (E) 3/4

Soln: Let Y represent raining on a day and N represent Not raining on a day So for it to rain 3 out of 5 days we have YYYNN. Since the order in which it rains in 5 days matters hence we find the total number of possibilities But since Y has a happened 3 times and N has happened 2 times we cannot distinguish between them and hence divide by the repetitions . Thus we need to divide by (3! * 2!) => 5!/(2! * 3!) => 10 ways

Now since on each day it can rain or it cannot rain. Thus total outcomes is => 2 * 2 * 2 * 2 * 2 => 32

So probability that it rains on exactly 3 days is = 10/32 = 5/16

I'm confused by the answers, only one answer actually has the 50% probability factored in. How come all the other answers just seemed to mention how many ways to choose 3 days in 5 days?

Seems like 50% is a special case, because when you do 5c3, the 3 days could be 3 days of rain or even 3 days of no rain. They all end up being the same probability because of the 50%.

I'm confused by the answers, only one answer actually has the 50% probability factored in. How come all the other answers just seemed to mention how many ways to choose 3 days in 5 days?

Seems like 50% is a special case, because when you do 5c3, the 3 days could be 3 days of rain or even 3 days of no rain. They all end up being the same probability because of the 50%.

how would this be solved with 20%? 80%?

Eg: For the 20% possibility of rain it will be (1/5)^3 * (4/5)^2* (5C3)

If the probability of rain on any given day in City X is 50 percent, what is the probability that it rains on exactly 3 days in a 5-day period?

(A) 8/125 (B) 2/25 (C) 5/16 (D) 8/25 (E) 3/4

Could you tell me why it is a combination and not a permutation?

N = 3C5 x 1/8 x 1/4

why do i have to divide by 2! ?

The probability of rain each day is 1/2 and the probability of no rain is also 1/2. \(C^3_5=10\) represent ways to choose on which 3 days out of 5 there will be a rain, so \(P=C^3_5*(\frac{1}{2})^3*(\frac{1}{2})^2=\frac{10}{32}=\frac{5}{16}\).

Answer: C.

Or think about it this way: we want the probability of the following event: RRRNN, where R represent rain day and N represents no-rain day. Now, each R and each N have individual probability of 1/2, so \((\frac{1}{2})^5\).

But the case of RRRNN can occur in many ways: RRRNN, RRNRRN, RNRRN, NRRRN, ... basically it will be equla to # of arrangements (permutations) of 5 letters RRRNN out of which there are 3 identical R's and 2 identical N's. That # of arrangements is \(\frac{5!}{3!2!}\), (notice that it's the same as \(C^3_5\)). So, finally \(P=\frac{5!}{3!2!}*(\frac{1}{2})^5=\frac{5}{16}\).

I still dont understand why we use the combination formula and not the permutation one.

Is it because we do not differentiate between one rainy day and the other? are they all the same for us?

Are most Gmat questions at this level?

Yes, we do not differentiate between RRR and NN: for example RRRNN means that it was raining on the first three days and we have no reason to differentiate between them. Also notice that most GMAT combination/probability questions are fairly straightforward and can be solved in several ways. This problem is also dealing with a simple concept explained in the links I sited above, so I'd recommend to brush your fundamentals on combinations before you move to the problems.
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17 Jun 2015, 00:40

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Re: If the probability of rain on any given day in City X is 50 [#permalink]

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