Difference between revisions of "2014 AMC 8 Problems/Problem 18"
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==Problem== | ==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 | + | 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? |
<math> \textbf{(A) }\text{all 4 are boys}\\ \textbf{(B) }\text{all 4 are girls}\\ \textbf{(C) }\text{2 are girls and 2 are boys}\\ \textbf{(D) }\text{3 are of one gender and 1 is of the other gender}\\ \textbf{(E) }\text{all of these outcomes are equally likely} </math> | <math> \textbf{(A) }\text{all 4 are boys}\\ \textbf{(B) }\text{all 4 are girls}\\ \textbf{(C) }\text{2 are girls and 2 are boys}\\ \textbf{(D) }\text{3 are of one gender and 1 is of the other gender}\\ \textbf{(E) }\text{all of these outcomes are equally likely} </math> | ||
− | ==Solution== | + | ==Solution 1== |
We'll just start by breaking cases down. The probability of A occurring is <math>\left(\frac{1}{2}\right)^4 = \frac{1}{16}</math>. The probability of B occurring is <math>\left(\frac{1}{2}\right)^4 = \frac{1}{16}</math>. | We'll just start by breaking cases down. The probability of A occurring is <math>\left(\frac{1}{2}\right)^4 = \frac{1}{16}</math>. The probability of B occurring is <math>\left(\frac{1}{2}\right)^4 = \frac{1}{16}</math>. | ||
− | The probability of C occurring is <math>\ | + | The probability of C occurring is <math>\dbinom{4}{2}\cdot \left(\frac{1}{2}\right)^4 = \frac{3}{8}</math>, because we need to choose 2 of the 4 children to be girls. |
− | So out of the four fractions, D is the largest. So our answer is <math>\boxed{\text{(D) | + | 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 <math>\dbinom{4}{1}\cdot\left(\frac{1}{2}\right)^4 = \frac{1}{4}</math> 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 <math>\frac{1}{4} \cdot 2 = \frac{1}{2}</math>. |
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+ | So out of the four fractions, D is the largest. So our answer is <math>\boxed{\text{(D)}}.</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 20:47, 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, D is the largest. So our answer is
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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