Probability Question from GMAT

[nikhil.goel]

7. 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

What is the answer . I think E.

Please suggest

[esledge]

The wording of your subject made me nervous: questions from official GMAT exams are definitely verboten! However, I checked and this is one of our Challenge Problems.

The answer is (C). If you have a specific question about how to do this one, feel free to post a follow-up.

I do have a suggestion for how to logically eliminate (E) as an answer. The chance of rain or sun is equal, so you can exploit the symmetry:

The chance of 5 rainy days is equal to the chance of 5 sunny days. [chance(5R) = chance(5S)]

Likewise, chance(4R1S)=chance(4S1R) and chance(3R2S)=chance(3S2R).

Summing, the chance(a majority of rainy days) = chance(a majority of sunny days). The total probability is 1, so the chance of rain on the majority of days (either 3R, 4R, or 5R) must be 1/2. The chance of rain on exactly 3 days must be less than 1/2.

[Guest]

Probability for RRRSS = 1/32
and its the same for all combinations of 3R and 2S. Now we need to find how many combinations:
We have three slots and 5 dates to pick (July 4 through July 8) = 5C3 = 10
Answer = 10*1/32 = 5/16

[GMAT Soon!]

as stated above... any three rainy days rrrss, rsrsr, etc. gives (1/2)^5 or 1/32 chance of this scenario. Since both rain or shine are equal probability, 1/2, any combination has a 1/32 shot of happening.

So... how many different ways can you make a group of 3 from 5 days. Just like how many teams of 3 could you pick from 5 players, this is 5!/(3!*2!) or 10.

Multiply the original 1/32 possibility of any scenario times your 10 ways to pick "teams" of 3 days and you get 10/32 or 5/16

[JonathanSchneider]

nicely done, both of you