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
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<math>\textbf{(A) }</math> All are boys <math>\qquad\textbf{(B) }</math> All are girls <math>\qquad\textbf{(C) }</math> 2 are boys and 2 are girls <math>\qquad\textbf{(D) }</math> 3 are the same gender and 1 is not <math>\qquad \textbf{(E) }</math> They all have the same probability of happening
(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==
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So out of the four fractions, D is the largest. So our answer is <math>\boxed{\text{(D)}3 are of one gender and 1 is of the other gender}.</math>
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So out of the four fractions, D is the largest. So our answer is <math>\boxed{\text{(D) 3 are of one gender and 1 is of the other gender}}.</math>
  
 
==Video Solution (CREATIVE THINKING)==
 
==Video Solution (CREATIVE THINKING)==

Latest revision as of 22:36, 11 August 2025

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?

$\textbf{(A) }$ All are boys $\qquad\textbf{(B) }$ All are girls $\qquad\textbf{(C) }$ 2 are boys and 2 are girls $\qquad\textbf{(D) }$ 3 are the same gender and 1 is not $\qquad \textbf{(E) }$ They all have the same probability of happening

Solution 1

We'll just start by breaking cases down. The probability of A occurring is $\left(\frac{1}{2}\right)^4 = \frac{1}{16}$. The probability of B occurring is $\left(\frac{1}{2}\right)^4 = \frac{1}{16}$.

The probability of C occurring is $\dbinom{4}{2}\cdot \left(\frac{1}{2}\right)^4 = \frac{3}{8}$, 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 $\dbinom{4}{1}\cdot\left(\frac{1}{2}\right)^4 = \frac{1}{4}$ 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 $\frac{1}{4} \cdot 2 = \frac{1}{2}$.


So out of the four fractions, D is the largest. So our answer is $\boxed{\text{(D) 3 are of one gender and 1 is of the other gender}}.$

Video Solution (CREATIVE THINKING)

https://youtu.be/erCpR2wX-78

~Education, the Study of Everything


Video Solution

https://youtu.be/3bF8BAvg0uY ~savannahsolver

See Also

2014 AMC 8 (ProblemsAnswer KeyResources)
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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