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

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<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}\\ \textbf{(D) }\text{3 are of one gender and 1 is of the other gender}\\ \textbf{(E) }\text{all of these outcomes are equally likely} </math>
 
<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}\\ \textbf{(D) }\text{3 are of one gender and 1 is of the other gender}\\ \textbf{(E) }\text{all of these outcomes are equally likely} </math>
  
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==Solution==
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==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>.
  
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So out of the four fractions, D is the largest. So our answer is <math>\boxed{\text{(D) 3 of one gender and 1 of the other}}.</math>
 
So out of the four fractions, D is the largest. So our answer is <math>\boxed{\text{(D) 3 of one gender and 1 of the other}}.</math>
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==Solution 2==
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The possibilities are listed out in the fourth row of Pascal's triangle, with the leftmost <math>1</math> being the possibility of all boys and the rightmost <math>1</math> being the possibility of all girls. Since the fourth row of Pascal's Triangle goes <cmath>1, 4, 6, 4, 1</cmath> and <math>6</math> are all the possibilities of two children from each gender, there are a total of <math>8</math> possibilities of three children from one gender and one from the other. Since there are a total of <math>2^4 = 16</math> total possibilities for the gender of the children, <math>\boxed{\text{D}}</math> has the highest probability.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2014|num-b=17|num-a=19}}
 
{{AMC8 box|year=2014|num-b=17|num-a=19}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:29, 3 January 2017

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

$\textbf{(A) }\text{all 4 are boys}\\ \textbf{(B) }\text{all 4 are girls}\\ \textbf{(C) }\text{2 are girls and 2 are boys}\\ \textbf{(D) }\text{3 are of one gender and 1 is of the other gender}\\ \textbf{(E) }\text{all of these outcomes are equally likely}$

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 children 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 children 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 $\boxed{\text{(D) 3 of one gender and 1 of the other}}.$

Solution 2

The possibilities are listed out in the fourth row of Pascal's triangle, with the leftmost $1$ being the possibility of all boys and the rightmost $1$ being the possibility of all girls. Since the fourth row of Pascal's Triangle goes \[1, 4, 6, 4, 1\] and $6$ are all the possibilities of two children from each gender, there are a total of $8$ possibilities of three children from one gender and one from the other. Since there are a total of $2^4 = 16$ total possibilities for the gender of the children, $\boxed{\text{D}}$ has the highest probability.

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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