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655-705 (Hard)|   Probability|                              
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Bunuel
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A couple decides to have 4 children. If they succeed in having 4 child 15 Jan 2024, 00:55
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Thib33600
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By fundamental counting principle,
Total No of out comes: 2^4 = 16

Total Desired outcomes: No of ways to arrange BBGG by MISSISSIPPI rule= 4!/(2!*2!) = 6

Probability is 6/16 = 3/8

Could please explain that MISSISSIPPI rule?
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THEORY

Permutations of \(n\) things, of which \(P_1\) are alike of one kind, \(P_2\) are alike of a second kind, \(P_3\) are alike of a third kind, ... \(P_r\) are alike of \(r_{th}\) kind such that: \(P_1+P_2+P_3+...+P_r=n\), is:

    \(\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}\)

For example, the number of permutations of the letters of the word "gmatclub" is \(8!\), as there are 8 DISTINCT letters in this word.

The number of permutations of the letters of the word "google" is \(\frac{6!}{2!2!}\), as there are 6 letters, with "g" and "o" each appearing twice.

The number of permutations of 9 balls, out of which 4 are red, 3 green, and 2 blue, would be 9!/4!3!2!.

The number of permutations of the letters of the word "ILLUSION" is \(\frac{8!}{2!2!}\), as there are 8 letters, with "L" and "I" each appearing twice.

21. Combinatorics/Counting Methods



For more:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread

Hope it helps.
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DanTheGMATMan
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A couple decides to have 4 children. If they succeed in having 4 child 09 Nov 2024, 17:33
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Very standard probability problem. Define one order, put in the probabilities, and then multiply by the number of orders it could have been in:

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wickedvikram
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A couple decides to have 4 children. If they succeed in having 4 child 08 Apr 2026, 08:15
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The probability of either a B/G 1/2 each time, thus = (1/2)^4
Number of combination possible = 4C2
Probability of exactly 2B&2G = 4C2*(1/2)^4

Bunuel
A couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and 2 boys?

(A) 3/8
(B) 1/4
(C) 3/16
(D) 1/8
(E) 1/16
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A couple decides to have 4 children. If they succeed in having 4 child 19 Jun 2026, 03:12
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To anyone still confused, this is as good as a problem that involves tossing a coin 4 times & asking the probability of getting exactly 2 Heads & 2 Tails, with each toss having the probability of 1/2

Therefore, sample space = 2*2*2*2 (i.e. each toss has 2 possible outcomes & since 4 tosses in total) = 16 total outcomes

Required: HHTT (i.e. same as BBGG), 4 letter word can be arrange in 4! ways, but since repetition, to get rid of 2 times H & 2 times T, divided by 2!*2!

i.e. favorable outcomes = 4!/(2!*2!) = 6

Therefore required probability of exactly HHTT (or BBGG) = 6/16 i.e. 3/8

Please leave a kudos if this helps.

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A couple decides to have 4 children. If they succeed in having 4 child 30 Jun 2026, 07:16
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Here's my video solution:
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