# 2014 AMC 8 Problems/Problem 18

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## Problem

Four children were born at City Hospital yesterday. Assume each child is equally likely to be a boy or a girl. Which of the following outcomes is most likely?

$\textbf{(A) }\text{all 4 are boys}\\ \textbf{(B) }\text{all 4 are girls}\\ \textbf{(C) }\text{2 are girls and 2 are boys}\\ \textbf{(D) }\text{3 are of one gender and 1 is of the other gender}\\ \textbf{(E) }\text{all of these outcomes are equally likely}$

## Solution 1

We'll just start by breaking cases down. The probability of A occurring is $\left(\frac{1}{2}\right)^4 = \frac{1}{16}$. The probability of B occurring is $\left(\frac{1}{2}\right)^4 = \frac{1}{16}$.

The probability of C occurring is $\dbinom{4}{2}\cdot \left(\frac{1}{2}\right)^4 = \frac{3}{8}$, because we need to choose 2 of the 4 children to be girls.

For D, there are two possible cases, 3 girls and 1 boy or 3 boys and 1 girl. The probability of the first case is $\dbinom{4}{1}\cdot\left(\frac{1}{2}\right)^4 = \frac{1}{4}$ because we need to choose 1 of the 4 children to be a boy. However, the second case has the same probability because we are choosing 1 of the 4 children to be a girl, so the total probability is $\frac{1}{4} \cdot 2 = \frac{1}{2}$.

So out of the four fractions, D is the largest. So our answer is $\boxed{\text{(D)}}.$

## Solution 2 -SweetMango77

We can also find out how many total cases there are for one solution. This will work, because before simplifying, the denominators of the fraction will be the same. Both $\text{A}$ and $\text{B}$ have ${4 \choose 0}=1$ possibility. $\text{C}$ will have ${4 \choose 2}=6$ possibilities. $\text{D}$ has $2\cdot{4 \choose 1}=8$ possibilities (note that the problem did not say a specific gender.) Therefore, $\boxed{\left(\text{D}\right)\text{ 3 are of one gender and 1 is of the other gender}}$ will have the greatest probability of occurring.

## See Also

 2014 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 17 Followed byProblem 19 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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