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Check Ian's and Staurt's response for the same question here

http://www.beatthegmat.com/semi-hard-probability-question-t8877.html

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First the answer: 5C3(1/2)^3(1/2) = 5/16

The concept here is that we could have YYYNN, YYNYN, NNYYY and we have to basically count all such events and divide by all possible events such as YYYYY, NYYYY etc. Where N - no rain, Y = rain. As you can see this can get really messy, hence some creative math minds of yore developed the Binomial method, you might want to lok it up in detail but here is what I understand of it:
If the probability of an event occuring is "p" and this is independent across tries, then if an activity is carried out "n" times, the probability of exactly "r" successes is given by:
nCr(p)^r(1-p)^r.

Some clarifications - "independent across tries" what is the probability of getting an Ace of spades from a deck = 1/52, if this is repeated again with replacement of the card picked what is the probability = 1/52 BUT if card is not replaced what is the probability = 1/51 (assuming the ace has not been already picked). Thus in the former case (when card replaced) the probability is independent, whereas in the latter they are not.

Some places where this can be applied - coin flips, rainy days, days when a temp will cross a certain number, dice roll, card picking.

Hope this is useful

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To All

Shortest possible way:

3 days can be chosen from 5 days in 5C3=10days
Chance of rain =50/100=1/2=no chance of rain

Total Probability = 10*(1/2)^5=5/16

Hi, i just thought this might be helpful:
I looked at this problem as a coin flip problem (i am more confident with that type), so we can assume the question is: what is the probability that coin will fall on heads 3 times in 5 flips.

5C3/2^5=5/16

Am i doing right?