Difference between revisions of "2014 AMC 8 Problems/Problem 18"

m (Problem)
m (Problem)
Line 2: Line 2:
 
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?
  
<math>\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 genre and 1 is not\qquad \textbf{(E) }They all have the same probability of happening</math>
+
<math>\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</math>
  
 
==Solution 1==
 
==Solution 1==

Revision as of 22:35, 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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png