Difference between revisions of "2014 AMC 8 Problems/Problem 18"
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<math> \textbf{(A) }</math> all <math>4</math> are boys <math>\qquad\textbf{(B) }</math> all <math>4</math> are girls <math>\qquad\textbf{(C) }</math> <math>2</math> are girls and <math>2</math> are boys <math>\qquad\textbf{(D) }</math> <math>3</math> are of one gender and <math>1</math> is of the other gender <math>\qquad\textbf{(E) }</math> all of these outcomes are equally likely | <math> \textbf{(A) }</math> all <math>4</math> are boys <math>\qquad\textbf{(B) }</math> all <math>4</math> are girls <math>\qquad\textbf{(C) }</math> <math>2</math> are girls and <math>2</math> are boys <math>\qquad\textbf{(D) }</math> <math>3</math> are of one gender and <math>1</math> is of the other gender <math>\qquad\textbf{(E) }</math> all of these outcomes are equally likely | ||
==Solution== | ==Solution== | ||
− | 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>\frac{4!}{2!2!}\cdot \left(\frac{1}{2}\right)^4 = \frac{3}{8}</math>. Lastly, the probability of D occurring is <math>2\cdot \frac{4!}{3!}\cdot\left(\frac{1}{2}\right)^4 = \frac{1}{2}</math>. So out of the four fractions, D is the largest. So our answer is <math>\boxed{\text{(D) 3 of one gender and 1 of the other}}.</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>. |
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+ | The probability of C occurring is <math>\frac{4!}{2!2!}\cdot \left(\frac{1}{2}\right)^4 = \frac{3}{8}</math>. Lastly, the probability of D occurring is <math>2\cdot \frac{4!}{3!}\cdot\left(\frac{1}{2}\right)^4 = \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) 3 of one gender and 1 of the other}}.</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 13:01, 28 November 2014
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
all are boys all are girls are girls and are boys are of one gender and is of the other gender all of these outcomes are equally likely
Solution
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 . Lastly, the probability of D occurring 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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