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If the probability of rain on any given day in Chicago

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If the probability of rain on any given day in Chicago [#permalink] New post 23 Jan 2008, 14:11
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If the probability of rain on any given day in Chicago during the summer is 50%, independent of what happens on any other day, what is the probability of having exactly 3 rainy days from July 4 through July 8, inclusive?
(A) 1/32
(B) 2/25
(C) 5/16
(D) 8/25
(E) 3/4
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Re: PS - Probability [#permalink] New post 23 Jan 2008, 14:25
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C is the answer.
C(5,3)/2^5 = (5!/(3!*2!))/2^5 = 10/32=5/16
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Re: PS - Probability [#permalink] New post 23 Jan 2008, 14:39
maratikus wrote:
C is the answer.
C(5,3)/2^5 = (5!/(3!*2!))/2^5 = 10/32=5/16


don't get it. can you explain further please
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Re: PS - Probability [#permalink] New post 23 Jan 2008, 14:48
buffdaddy wrote:
maratikus wrote:
C is the answer.
C(5,3)/2^5 = (5!/(3!*2!))/2^5 = 10/32=5/16


don't get it. can you explain further please


read about binomial distribution:

http://en.wikipedia.org/wiki/Binomial_distribution
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Re: PS - Probability [#permalink] New post 23 Jan 2008, 15:02
Here, there are 5 days from July 4 to July 8.
The prob of rains is given as 50%, i.e: 1/2.

There are 10 possible ways to pick 3 days (when it rains) from 5 days,
i.e 5C3 = 5!/3!2! = 10.

So the probability for any one of the ten combinations would be:
1/2*1/2*1/2*1/2*1/2 = 1/32

So the prb for all 10 combinations are:
1/32 * 10 = 5/16.

Ans: C :-D
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Re: PS - Probability [#permalink] New post 23 Jan 2008, 15:06
It is the 'or' rule used here, i.e: It can rain on 4th, 5th and 7th or 4th, 6th and 8th etc.
So you would be adding 1/32 ten times.
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Re: PS - Probability [#permalink] New post 23 Jan 2008, 15:11
Same logic as in this one:

A man can hit a target once in 4 shots. If he fires 4 shots in succession, what is the probability that he will hit his target?

A. 1
B. 1/256
C. 81/256
D. 175/256
E. 144/256
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Re: PS - BINOMIAL DISTRIBUTION [#permalink] New post 23 Jan 2008, 23:46
The binomial distribution applies when:

1: The number of observations n is fixed.
2: Each observation is independent.
3: Each observation represents one of two outcomes ("success" or "failure").
4: The probability of "success" p is the same for each outcome.


If these conditions are met, then X has a binomial distribution with parameters n and p, abbreviated B(n,p).

nCx *(p)^x*(1-p)^(n-x)

n= 5
x=3
p=1/2


Or if you want to go by logic,

No. of ways in which it can rain on 3 days out of 5 is

5C3 = 10ways

10 * ( 1/2)^3 * (1-1/2)^2 = 10/32
Re: PS - BINOMIAL DISTRIBUTION   [#permalink] 23 Jan 2008, 23:46
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