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

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==Problem==
 
==Problem==
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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
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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?
  
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<math> \textbf{(A) }</math> all <math>4</math> are boys <math>\qquad\textbf{(B) }</math> all <math>4</math> are girls <math>\qquad\textbf{(C) }</math> <math>2</math> are girls and <math>2</math> are boys <math>\qquad\textbf{(D) }</math> <math>3</math> are of one gender and <math>1</math> is of the other gender <math>\qquad\textbf{(E) }</math> all of these outcomes are equally likely  
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(A) all 4 are boys
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==Solution==
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(B) all 4 are girls
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We'll just start by breaking cases down. The probability of A occurring is <math>(\frac{1}{2})^4 = \frac{1}{16}</math>. The probability of B occurring is <math>(\frac{1}{2})^4 = \frac{1}{16}</math>. The probability of C occurring is <math>\frac{4!}{2!2!}*(\frac{1}{2})^4 = \frac{3}{8}</math>. Lastly, the probability of D occurring is <math>2*\frac{4!}{3!}*(\frac{1}{2})^4 = \frac{1}{2}</math>. So out of the four fractions, D is the largest. So our answer is D.
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(C) 2 are girls and 2 are boys
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(D) 3 are of one gender and 1 is of the other gender
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(E) all of these outcomes are equally likely
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==Solution 1==
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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>.
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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.  
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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>.
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So out of the four fractions, D is the largest. So our answer is <math>\boxed{\text{(D)}}.</math>
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==Video Solution (CREATIVE THINKING)==
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https://youtu.be/erCpR2wX-78
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~Education, the Study of Everything
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==Video Solution==
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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}}
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[[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 $\left(\frac{1}{2}\right)^4 = \frac{1}{16}$. 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 $\boxed{\text{(D)}}.$

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 (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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