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Probability question

by ch0719 » Sun Oct 26, 2008 4:10 pm
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?

1/32
2/25
5/16
8/25
3/4
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by canuckclint » Sun Oct 26, 2008 5:01 pm
Let me edit my answer:

So you are looking for 3 rainy days out of 5.

rrrnn. This can be arranged (5 c 2) = 10 ways.

2 ^ 5 = 32 total possible combinations.

10/32 = 5/16
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by cramya » Sun Oct 26, 2008 8:39 pm
2 ^ 5 = 32 total possible combinations.


can u eloborate how u get the total combinations? I am missing something here. Thanks!
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Re: Probability question

by parallel_chase » Sun Oct 26, 2008 11:07 pm
ch0719 wrote: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?

1/32
2/25
5/16
8/25
3/4
probability of rain on any given day = 50/100 = 1/2
probability of not raining on any given day = 1-1/2 = 1/2

no. of days = 4, 5, 6, 7, 8 = 5 days

exactly 3 days from July 4 - July 8

total combinations = 5C3 = 10

3 days raining * 2 days not raining = (1/2)^5

total probability = 10*(1/2)^5 = 10/2^5 = 5/16

Hope this helps.
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by ch0719 » Sun Oct 26, 2008 11:07 pm
thanks, I over analyzed this question... :)
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by Anaira Mitch » Thu Jan 05, 2017 6:57 pm
The period from July 4 to July 8, inclusive, contains 8 - 4 + 1 = 5 days, so we can rephrase the question as "What is the probability of having exactly 3 rainy days out of 5?"
Since there are 2 possible outcomes for each day (R = rain or S = shine) and 5 days total, there are 2 x 2 x 2 x 2 x 2 = 32 possible scenarios for the 5 day period (RRRSS, RSRSS, SSRRR, etc...) To find the probability of having exactly three rainy days out of five, we must find the total number of scenarios containing exactly 3 R's and 2 S's, that is the number of possible RRRSS anagrams:
= 5! / 2!3! = (5 x 4)/2 x 1 = 10
The probability then of having exactly 3 rainy days out of five is 10/32 or 5/16.
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by Jay@ManhattanReview » Thu Jan 05, 2017 9:54 pm
ch0719 wrote: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?

1/32
2/25
5/16
8/25
3/4
We are given that the probability of rain on a given day = 1/2.

There are 5 days during July 4 through July 8, inclusive.

We wish that it rains on any of the three days and does not rain on the two days.

Probability that it does not rain on a given day = 1 - 1/2 = 1/2.

Say, it rains on the first three days and does not rain in the last days.

Probability that it rains on the first three days and does not rain in the last two days (RRRNN) = (1/2)*(1/2)*(1/2)*(1/2)*(1/2) = (1/2)^5

Since the order of rainy days is not a constraint, i.e., it can rain on any three of the given five days, we have a combination problem.

Ways of choosing 3 days out of 5 days = 5C3 = 5C2 = (5.4)/(1.2) = 10 ways; [nCr = nC(n-r)]

Probability that it rains on any three days = 10*(1/2)^5 = 5/(2^4) = 5/16.

Answer: C

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by [email protected] » Fri Jan 06, 2017 10:20 am
Hi All,

The most efficient way to answer this question is in understanding how the Combination Formula applies to the math involved. However, even if you don't fully understand that concept, you can still get to the correct answer with a bit of 'brute force.'

To start, you have to understand that since each day has two equally-likely outcomes ('rain' or 'not rain'), then the various arrangements of 5 days of weather = (2)(2)(2)(2)(2) = 32. Thus, the probability of having EXACTLY 3 rain days must be some fraction out of 32...

From this point, we just have to figure out how many options consistent of just 3 rain days. We can 'map' those out rather easily...

RRRNN

RRNRN
RRNNR

RNRRN
RNRNR
RNNRR

NRRRN
NRRNR
NRNRR
NNRRR

10 total options out of 32. Reducing that fraction gives us... 10/32 = 5/16

Final Answer: C

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Rich
Contact Rich at [email protected]
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by Jeff@TargetTestPrep » Mon Jan 09, 2017 9:47 am
ch0719 wrote: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?

1/32
2/25
5/16
8/25
3/4
The number of days from July 4 to July 8, inclusive, is 5 days. Thus, we need to determine the probability of having 3 rainy days within a 5-day period. We are given that the probability of a rainy day is ½, and thus the probability of no rain is also 1/2. We need to determine the probability of 3 rainy days within a 5-day period. We can assume the first 3 days are rainy (R) and the last 2 days are not rainy (N). Thus,

P(R-R-R-N-N) = 1/2 x 1/2 x 1/2 x 1/2 x 1/2 = 1/32

However, we need to determine in how many ways it can rain 3 out of 5 days. That number will be equivalent to how many ways we can organize the letters R-R-R-N-N.

We use the indistinguishable permutations formula to determine the number of ways to arrange R-R-R-N-N: 5!/(3! x 2!) = 10 ways

Each of these 10 ways has the same probability of occurring. Thus, the total probability is:

10(1/32) = 10/32 = 5/16

Answer: C

Jeffrey Miller
Head of GMAT Instruction
[email protected]

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