Difference between revisions of "2014 AMC 8 Problems/Problem 18"
(→Solution) |
(→Solution) |
||
| Line 4: | Line 4: | ||
<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 (\frac{1}{2})^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>. 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> |
==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 18:04, 27 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 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
