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?

(A) all 4 are boys (B) all 4 are girls (C) 2 are girls and 2 are boys (D) 3 are of one gender and 1 is of the other gender (E) 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 $\boxed{\text{(D)}}.$

Video Solution (CREATIVE THINKING)

https://youtu.be/erCpR2wX-78

~Education, the Study of Everything

Video Solution

https://youtu.be/3bF8BAvg0uY ~savannahsolver

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

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

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2014 AMC 8 Problems/Problem 18

Contents

Problem

Solution 1

Video Solution (CREATIVE THINKING)

Video Solution

See Also