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? | 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== | ==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>\dbinom{4}{2}\cdot \left(\frac{1}{2}\right)^4 = \frac{3}{8}</math>, because we need to choose 2 of the 4 | + | 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 | + | 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> | So out of the four fractions, D is the largest. So our answer is <math>\boxed{\text{(D)}}.</math> | ||
− | ==Solution | + | ==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== |
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 |
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