2014 AMC 8 Problems/Problem 18: Difference between revisions

Latest revision as of 21:36, 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

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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+==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?
-<math>\textbf{(A) }\text{all 4 are boys}\\ \textbf{(B) }\text{all 4 are girls}\\ \textbf{(C) }\text{2 are girls and 2 are boys}\\ \...</math>
+<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
+==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) 3 are of one gender and 1 is of the other gender}}.</math>
+==Video Solution (CREATIVE THINKING)==
+https://youtu.be/erCpR2wX-78
+~Education, the Study of Everything
+==Video Solution==
+https://youtu.be/3bF8BAvg0uY ~savannahsolver
+==See Also==
+{{AMC8 box|year=2014|num-b=17|num-a=19}}
+[[Category:Introductory Probability Problems]]
+{{MAA Notice}}

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