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 . The probability of B occurring is .
The probability of C occurring is , 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 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 .
So out of the four fractions, D is the largest. So our answer is
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 th and $ilities (note that the problem did not say a specific gender.) Therefore, 3 are of one gender and 1 will have the greatest probability of occurring.
Video Solution
https://youtu.be/3bF8BAvg0uY ~savannahsolver
See Also
2014 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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