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## Help on this probability problem

This topic has 1 expert reply and 4 member replies
kellogs4toniee Just gettin' started!
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Help on this probability problem Wed Jan 18, 2012 4:40 pm
Elapsed Time: 00:00
• Lap #[LAPCOUNT] ([LAPTIME])
If the probability of rain on any given day in City X is 50 percent, what is the probability that it rains on exactly 3 days in a 5-day period?

() 8/25
() 2/25
() 5/16
() 8/25
() 3/4

Please explain in detail how would you do this, and what concepts I should re-familiarize myself with for similar problems in the future.

Thank you!

Tony

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neelgandham Community Manager
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Wed Jan 18, 2012 5:20 pm
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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mj78ind Really wants to Beat The GMAT!
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Thu Jan 19, 2012 8:22 am
kellogs4toniee wrote:
If the probability of rain on any given day in City X is 50 percent, what is the probability that it rains on exactly 3 days in a 5-day period?

() 8/25
() 2/25
() 5/16
() 8/25
() 3/4

Please explain in detail how would you do this, and what concepts I should re-familiarize myself with for similar problems in the future.

Thank you!

Tony
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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ArunangsuSahu GMAT Destroyer!
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Thu Jan 19, 2012 8:51 am
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

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Thu Jan 19, 2012 9:13 am
I posted a solution (and a link to a trickier version of this problem) here:

http://www.beatthegmat.com/rain-prob-t88396.html

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ziko Just gettin' started!
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Wed Mar 14, 2012 2:15 am
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?

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