2014 AMC 8 Problems/Problem 18
Contents
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?
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 children 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 children 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
The possibilities are listed out in the fourth row of Pascal's triangle, with the leftmost being the possibility of all boys and the rightmost
being the possibility of all girls. Since the fourth row of Pascal's Triangle goes
and
are all the possibilities of two children from each gender, there are a total of
possibilities of three children from one gender and one from the other. Since there are a total of
total possibilities for the gender of the children,
has the highest probability.
See Also
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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