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

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

Solution

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 $\frac{4!}{2!2!}\cdot (\frac{1}{2})^4 = \frac{3}{8}$ . Lastly, the probability of D occurring is $2\cdot \frac{4!}{3!}\cdot\left(\frac{1}{2}\right)^4 = \frac{1}{2}$ . So out of the four fractions, D is the largest. So our answer is $\boxed{\text{(D) 3 of one gender and 1 of the other}}.$

2014 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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All AJHSME/AMC 8 Problems and Solutions

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2014 AMC 8 Problems/Problem 18

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