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 | |
| 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 | ||
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