A couple decides to have \(4\) children. If they succeed in having \(4\) children and each child is equally likely to be

[BTGmoderatorLU]

Source: Official Guide

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 \(2\) girls and \(2\) boys?

A. \(3/8\)
B. \(1/4\)
C. \(3/16\)
D. \(1/8\)
E. \(1/16\)

The OA is A

[GMATJourneyman]

Source: Official Guide

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 \(2\) girls and \(2\) boys?

A. \(3/8\)
B. \(1/4\)
C. \(3/16\)
D. \(1/8\)
E. \(1/16\)

The OA is A

The total number of outcomes when having 4 children, each of which can be a boy or a girl, is \(2^4 = 16\).

Among these outcomes, we are interested in the ones with exactly 2 boys and 2 girls. The number of such outcomes can be obtained using the combination formula \({{n}\choose{k}} = \frac{n!}{k!(n-k)!}\), where \(n\) is the total number of trials (children, in this case), \(k\) is the number of successful outcomes we are interested in (girls or boys), and \(!\) denotes factorial.

So the number of ways to have 2 boys and 2 girls out of 4 children is \({{4}\choose{2}} = \frac{4!}{2!(4-2)!} = 6\).

Thus, the probability is \(\frac{6}{16} = \frac{3}{8}\).

So, the answer is A. \(\frac{3}{8}\)

Not sure that's the most efficient approach, but that's what I did.

[Brent@GMATPrepNow]

Source: Official Guide

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 \(2\) girls and \(2\) boys?

A. \(3/8\)
B. \(1/4\)
C. \(3/16\)
D. \(1/8\)
E. \(1/16\)

The OA is A

We can solve this question using counting methods.
P(exactly 2 girls and 2 boys) = (number of 4-baby outcomes with exactly 2 girls and 2 boys)/(TOTAL number of 4-baby outcomes)

As always, we'll begin with the denominator.

TOTAL number of 4-baby arrangements
There are 2 ways to have the first baby (boy or girl)
There are 2 ways to have the second baby (boy or girl)
There are 2 ways to have the third baby (boy or girl)
There are 2 ways to have the fourth baby (boy or girl)
By the Fundamental Counting Principle (FCP), the total number of 4-baby arrangements = (2)(2)(2)(2) = 16


Number of 4-baby outcomes with exactly 2 girls and 2 boys
This portion of the question boils down to "In how many different ways can we arrange 2 G's and 2 B's (where each G represents a girl, and each B represents a boy)?"

----------ASIDE-------------------------
When we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this:

If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....]

So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows:
There are 11 letters in total
There are 4 identical I's
There are 4 identical S's
There are 2 identical P's
So, the total number of possible arrangements = 11!/[(4!)(4!)(2!)]
------------BACK TO THE QUESTION---------------------------

Our goal is to arrange the letters G, G, B, and B
There are 4 letters in total
There are 2 identical G's
There are 2 identical B's
So, the total number of possible arrangements = 4!/[(2!)(2!)] = 6


So.....
P(exactly 2 girls and 2 boys) = 6/16 = 3/8

Answer: A