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Probability: chosen at least twice?

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by Alpha800 » Tue Aug 05, 2008 2:37 pm
18 children, including 9 girls and 9 boys, are watching a magician. The magician summons a total of 4 children, one at a time, up to the stage to assist. Once a child is finished assisting, he or she returns to the audience, and can potentially be chosen over again. What is the probability that a girl is chosen at least twice?
I can solve it for at least once:

1/2 * 1/2 * 1/2 * 1/2 = 1/16
1 - 1/16 = 15/16

but how do you solve for at least twice? or for that matter, how to solve for at least thrice? :o
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by parallel_chase » Tue Aug 05, 2008 3:56 pm
To simplify the question try and calculate the probability for choosing exactly 1 boy and subtract that from 1. That will give you the probability for at least 2 girls, at least 3 girls and at least 4 girls being chosen.

probability of exactly 1 boy being chosen = 4/16 = 1/4

1-1/4 = 3/4

is this the answer?
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by preetha_85 » Sat Aug 09, 2008 5:28 am
wats the OA ?
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by sudhir3127 » Sat Aug 09, 2008 6:29 am
My Answer is 11/16.

Atleast 2 girls for me would be 1- [ p(0)+ p(1) ]

chosing either a boy or girl is 1/2

hence it would be

1- [4C0 (1/2)^4 + 4c1 (1/2)^3(1/2) ]...using binomial theorem.

1= (1/16 + 4/16)

hence its 11/16

please let us know the OA
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by Sant » Sat Aug 09, 2008 7:18 am
I think sudhir is right. The ans should be 11/16.
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by parallel_chase » Sat Aug 09, 2008 1:01 pm
Sudhir is absolutely right. These are one of the silly mistakes that will stop me from getting a high score on quant.

Thanks Sudhir!!!!
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by Alpha800 » Sat Aug 09, 2008 3:13 pm
Hi everyone. Thank you all for posting.

Unfortunately, there's no official answer. I asked how to solve the problem because Princeton was only giving an example of solving for at least one. But I wanted to know how to solve for at least two or three.

Sorry for not replying earlier. The Olympics just started and I'm busy keeping up with all the Olympic coverage. :)
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by CITI29 » Sat Aug 09, 2008 8:16 pm
Hi Sudhir /Parallel chase

Can u pls explain me as how to get :
probability of exactly 1 boy being chosen = 4/16 = 1/4

Somehow I aways get stuck on such probability ques where probability of selecting either of the grp is equal. For eg- when a question asks that what is the prob of getting atleast 3 heads when a coin is tossed 5 times.!!!..I just dont seem to understand the logic here --since prob of getting either head or tail is same.

if u can pls explain me this concept I'll be really grateful....thanks
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by sudhir3127 » Sat Aug 09, 2008 8:30 pm
The probability of choosing either a boy or a girl anytime will always be 1/2

eaxctly 1 means

4C1 ( 1/2)^1(1/2)^3 .. iam using binomail theorem here the first 1/2 is for chosing exactly 1 boy and the next 1/2 is for 3 girls...

hence its 4/16 or 1/4

I use Binomial theorem to solve such problems as i find them error free.. but i am sure if u reslise that the total will always be 16 in this case . so choosing 1 boy out of 4 children is 4C1 .

hence 4/16.

hope it helps...
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by petra » Tue Feb 24, 2009 10:48 am
well done guys thank you
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