2014 AMC 8 Problems/Problem 18

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 slots 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 slots 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

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.

2014 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 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

Page

Toolbox

Search

2014 AMC 8 Problems/Problem 18

Contents

Problem

Solution 1

Solution 2

See Also