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Updated on: 10 Feb 2012, 12:42
Originally posted by GMATMadeeasy on 08 Jan 2010, 17:56.
Last edited by Bunuel on 10 Feb 2012, 12:42, edited 1 time in total.
Last edited by Bunuel on 10 Feb 2012, 12:42, edited 1 time in total.
Edited the question and added the OA
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C
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The probability that a family with 6 children has exactly four boys is:
A. 1/3
B. 1/64
C. 15/64
D. 3/8
E. none of the above
A. 1/3
B. 1/64
C. 15/64
D. 3/8
E. none of the above
08 Jan 2010, 23:35
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The probability that a family with 6 children has exactly four boys is:
A. 1/3
B. 1/64
C. 15/64
D. 3/8
E. none of the above
NOTE: If the probability of a certain event is \(p\) (\(\frac{1}{2}\) in our case), then the probability of it occurring \(k\) times (4 times in our case) in \(n\)-time (6 in our case) sequence is:
\(P = C^k_n*p^k*(1-p)^{n-k}\)
\(P = C^k_n*p^k*(1-p)^{n-k}= C^4_6*(\frac{1}{2})^4*(1-\frac{1}{2})^{6-4}= C^4_6*(\frac{1}{2})^4*(\frac{1}{2})^{2}=\frac{6!}{4!2!}*\frac{1}{2^6}=\frac{15}{64}\)
Consider this:
We are looking for the case BBBBGG, probability of each B or G is \(\frac{1}{2}\), hence \(\frac{1}{2^6}\). But BBBBGG can occur (these letters can be arranged) in \(\frac{6!}{4!2!}\) # of ways. So, the total probability would be \(\frac{6!}{4!2!}*\frac{1}{2^6}=\frac{15}{64}\).
Answer: C.
A. 1/3
B. 1/64
C. 15/64
D. 3/8
E. none of the above
NOTE: If the probability of a certain event is \(p\) (\(\frac{1}{2}\) in our case), then the probability of it occurring \(k\) times (4 times in our case) in \(n\)-time (6 in our case) sequence is:
\(P = C^k_n*p^k*(1-p)^{n-k}\)
\(P = C^k_n*p^k*(1-p)^{n-k}= C^4_6*(\frac{1}{2})^4*(1-\frac{1}{2})^{6-4}= C^4_6*(\frac{1}{2})^4*(\frac{1}{2})^{2}=\frac{6!}{4!2!}*\frac{1}{2^6}=\frac{15}{64}\)
Consider this:
We are looking for the case BBBBGG, probability of each B or G is \(\frac{1}{2}\), hence \(\frac{1}{2^6}\). But BBBBGG can occur (these letters can be arranged) in \(\frac{6!}{4!2!}\) # of ways. So, the total probability would be \(\frac{6!}{4!2!}*\frac{1}{2^6}=\frac{15}{64}\).
Answer: C.
09 Jan 2010, 05:18
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I think this way:
For each child we've two options i.e., that child can either be a boy or a girl.Hence we have 2^6 = 64 total outcomes.
Now out of 6 children any 4 of them can boys and this can be done in 6C4 = 15 ways.
Now here I don't think that ordering does matter in this case.
So the probability comes out to be 15/64.
P.S. Please explain me if ordering really matters in this case as my understanding is quite naive in these problems.
For each child we've two options i.e., that child can either be a boy or a girl.Hence we have 2^6 = 64 total outcomes.
Now out of 6 children any 4 of them can boys and this can be done in 6C4 = 15 ways.
Now here I don't think that ordering does matter in this case.
So the probability comes out to be 15/64.
P.S. Please explain me if ordering really matters in this case as my understanding is quite naive in these problems.
