Difference between revisions of "2014 AMC 8 Problems/Problem 18"
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− | So out of the four fractions, | + | So out of the four fractions, C is the largest. So our answer is <math>\boxed{\text{(C)}}.</math> |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2014|num-b=17|num-a=19}} | {{AMC8 box|year=2014|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:30, 9 November 2019
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, C is the largest. So our answer is
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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