Difference between revisions of "2014 AMC 8 Problems/Problem 18"
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| − | 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 | + | ==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? | ||
| + | |||
| + | (A) all 4 are boys | ||
| + | (B) all 4 are girls | ||
| + | (C) 2 are girls and 2 are boys | ||
| + | (D) 3 are of one gender and 1 is of the other gender | ||
| + | (E) all of these outcomes are equally likely | ||
| + | |||
| + | ==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>. | ||
| + | |||
| + | 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 slots 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 <math>\dbinom{4}{1}\cdot\left(\frac{1}{2}\right)^4 = \frac{1}{4}</math> because we need to choose 1 of the 4 slots 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>. | ||
| + | |||
| + | |||
| + | So out of the four fractions, D is the largest. So our answer is <math>\boxed{\text{(D)}}.</math> | ||
| + | |||
| + | ==Video Solution (CREATIVE THINKING)== | ||
| + | https://youtu.be/erCpR2wX-78 | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | |||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/3bF8BAvg0uY ~savannahsolver | ||
| − | |||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2014|num-b=17|num-a=19}} | {{AMC8 box|year=2014|num-b=17|num-a=19}} | ||
| + | |||
| + | [[Category:Introductory Probability Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 10:20, 2 July 2023
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?
(A) all 4 are boys (B) all 4 are girls (C) 2 are girls and 2 are boys (D) 3 are of one gender and 1 is of the other gender (E) all of these outcomes are equally 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 slots 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 slots 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 
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
Video Solution
https://youtu.be/3bF8BAvg0uY ~savannahsolver
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. 
