2014 AMC 8

Complete problem set with solutions and individual problem pages

Problem 18 Hard

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 are boys

  • B.

    All are girls

  • C.

    2 are boys and 2 are girls

  • D.

    3 are the same gender and 1 is not

  • E.

    They all have the same probability of happening

Answer:D

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) 3 are of one gender and 1 is of the other gender}}.

Related Problems
Problem 16 2014 AMC 8
Problem 17 2014 AMC 8
Problem 19 2014 AMC 8
Problem 20 2014 AMC 8
Think Academy Powered by ChatGPT