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

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

by BTGmoderatorLU » Fri May 19, 2023 11:16 am

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## Global Stats

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

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### Re: A couple decides to have $$4$$ children. If they succeed in having $$4$$ children and each child is equally likely t

by GMATJourneyman » Wed May 24, 2023 8:09 am
BTGmoderatorLU wrote:
Fri May 19, 2023 11:16 am
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.
Last edited by GMATJourneyman on Wed May 24, 2023 8:14 am, edited 1 time in total.

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### Re: A couple decides to have $$4$$ children. If they succeed in having $$4$$ children and each child is equally likely t

by Brent@GMATPrepNow » Fri May 26, 2023 5:52 pm
BTGmoderatorLU wrote:
Fri May 19, 2023 11:16 am
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