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tricky one

by mjjking » Wed Feb 25, 2009 8:58 am
In Rwanda, the chance for rain on any given day is 50%. What is the probability that it rains on 4 out of 7 consecutive days in Rwanda?

4/7
3/7
35/128
4/28
28/135
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Re: tricky one

by x2suresh » Wed Feb 25, 2009 9:28 am
mjjking wrote:In Rwanda, the chance for rain on any given day is 50%. What is the probability that it rains on 4 out of 7 consecutive days in Rwanda?

4/7
3/7
35/128
4/28
28/135
7C4 *(1/2)^4*(1/2)^3 = 35/128

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Re: tricky one

by Stuart@KaplanGMAT » Wed Feb 25, 2009 1:16 pm
mjjking wrote:In Rwanda, the chance for rain on any given day is 50%. What is the probability that it rains on 4 out of 7 consecutive days in Rwanda?

4/7
3/7
35/128
4/28
28/135
Any question which involves a 50/50 probability is really a coin flip question in disguise. The question could have been written as:

"If you flip a fair coin 7 times, what's the chance of getting exactly 4 heads?"

For coin flip questions, we can use the following formula:

Prob of getting k results out of n flips = nCk/2^n

Here, n = 7 and k = 4, so:

7C4/2^7

(7!/4!3!)/128

(7*6*5/3*2)/128

35/128

* * *

As an aside, if we understand coin flip questions we can get the right answer with 100% certainty in under 5 seconds.

With a 50/50 probability, the denominator will ALWAYS be a power of 2, i.e. 2, 4, 8, 16, 32, ....

In this particular question, only choice (c) has a denominator that's a power of 2, therefore it MUST be the correct answer to the question.
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Re: tricky one

by Mr2Bits » Wed Feb 25, 2009 4:51 pm
Stuart Kovinsky wrote:
mjjking wrote:In Rwanda, the chance for rain on any given day is 50%. What is the probability that it rains on 4 out of 7 consecutive days in Rwanda?

4/7
3/7
35/128
4/28
28/135
Any question which involves a 50/50 probability is really a coin flip question in disguise. The question could have been written as:

"If you flip a fair coin 7 times, what's the chance of getting exactly 4 heads?"

For coin flip questions, we can use the following formula:

Prob of getting k results out of n flips = nCk/2^n

Here, n = 7 and k = 4, so:

7C4/2^7

(7!/4!3!)/128

(7*6*5/3*2)/128

35/128

* * *

As an aside, if we understand coin flip questions we can get the right answer with 100% certainty in under 5 seconds.

With a 50/50 probability, the denominator will ALWAYS be a power of 2, i.e. 2, 4, 8, 16, 32, ....

In this particular question, only choice (c) has a denominator that's a power of 2, therefore it MUST be the correct answer to the question.
Thanks for the formula and trick, I find this to be very useful. The only way your statement about being a power of two is incorrect is if the answers are reduced and you don't see it. In this case, no reduction could be done.
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by Brent@GMATPrepNow » Wed Feb 25, 2009 6:01 pm
Thanks for the formula and trick, I find this to be very useful. The only way your statement about being a power of two is incorrect is if the answers are reduced and you don't see it. In this case, no reduction could be done.
Even if the answer can be reduced, the deniminator will always be a power of 2.
Consider the probability k/(2^n).
If we expand this we get k/(2x2x2x2...x2x2)
We can only reduce this fraction if k is divisible by some power of 2, in which case we will lose some of the denominator's 2's in the reduction. Even so, we can see that only 2's will remain in the denominator.
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