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KarishmaB
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The probability that a family with 6 children has exactly 16 Aug 2022, 00:27
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Mugdho
KarishmaB
Bunuel
chetan2u

GMATinsight

why does it have to be a case of ordering? question didn’t say anything about order...it simply asked if there are 4 boys and 2 girls...

if we consider the cases, as-

6 Girls
5 girls and 1 boy
4 girls and 2 boys
3 girls and 3 boys
2 girls and 4 boys
1 girl and 5 boys

So there are 7 cases...and 4 boys can happen only in one way...

so probability : 1/7

what's wrong with this reasoning?

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Take a parallel example: You have a pair of dice. Is the probability of getting a sum of 2 on the throw the same as the probability of getting a sum of say, 6?
To get a sum of 2, you need both dice to show exactly 1. The probability of that is 1/6 * 1/6.

But to get a 6, one die could show 1 and the other 5 or one die could show 2 and the other 4 or both die could show 3 each. So probability of getting a sum of 6 is higher, right? It can happen in multiple ways.

Similarly, is the probability of getting all 6 girls same as the probability of getting 2 girls and 4 boys? No.
To get all 6 girls, you need to get a girl every time which means GGGGGG which gives (1/2)*(1/2)*(1/2)*(1/2)*(1/2)*(1/2).

But we can obtain 2 girls and 4 boys in many ways: GGBBBB, GBGBBB, BBGBBG, BBBBGG etc. So the probability of both is not the same. Then 1/7 doesn't work, right?
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The probability that a family with 6 children has exactly 16 Aug 2022, 01:27
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simply use the binomial formula
=nCr p(success)^r P(failure) ^ (n-r)

= 6C4 (1/2)^2 (1/2)^4

=15/64
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The probability that a family with 6 children has exactly 25 Aug 2022, 08:02
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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
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The probability that a family with 6 children has exactly 31 May 2024, 05:12
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­Sorry for sound quality hah. Set up one order of 4 boys and 2 girls (each has 1/2 probability). Multiply by the number of different orders:

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The probability that a family with 6 children has exactly 17 Aug 2025, 03:52
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can we also do (1/2)^6 x 6!/4!x2! ans comes out to be the same

here the chance of a child being boy or girl is 1/2 and since there are 6 we are looking for BBBBGG and thus (1/2)^6 but to avoid arrangement issues we can multiply by 6!/4!x2!
Bunuel
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.
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The probability that a family with 6 children has exactly 28 Aug 2025, 20:24
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This is an unpopular approach but I will disagree with your solutions, I would have marked E, a here is why:
The statement says a family with 6 children it could be 0B6G, 1B5G, ..., 6B0G, so 7 cases... nobody mentions anything about the order in which this family have the children... so the probability would be 1/7
GMATMadeeasy
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
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The probability that a family with 6 children has exactly 28 Aug 2025, 23:37
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PocusFocus
This is an unpopular approach but I will disagree with your solutions, I would have marked E, a here is why:
The statement says a family with 6 children it could be 0B6G, 1B5G, ..., 6B0G, so 7 cases... nobody mentions anything about the order in which this family have the children... so the probability would be 1/7

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You are misunderstanding the question. It asks about the probability of having 4 boys and 2 girls. Each of the correct solutions in the thread considers the order of BBBBGG. Please review the solutions again.
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