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

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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>
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So out of the four fractions, D is the largest. So our answer is <math>\boxed{\text{(D)}}.</math>
  
==Solution 2==
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==Solution 2 -SweetMango77==
 
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We can also find out how many total cases there are for one solution. This will work, because before simplifying, the denominators of the fraction will be the same. Both <math>\text{A}</math> and <math>\text{B}</math> have <math>{4 \choose 0}=1</math> possibility. <math>\text{C}</math> will have <math>{4 \choose 2}=6</math> possibilities. <math>\text{D}</math> has <math>2\cdot{4 \choose 1}=8</math> possibilities (note that the problem did not say a specific gender.) Therefore, <math>\boxed{\left(\text{D}\right)\text{ 3 are of one gender and 1 is of the other gender}}</math> will have the greatest probability of occurring.
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}}
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[[Category:Introductory Probability Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 08:40, 5 November 2020

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 $\boxed{\text{(D)}}.$

Solution 2 -SweetMango77

We can also find out how many total cases there are for one solution. This will work, because before simplifying, the denominators of the fraction will be the same. Both $\text{A}$ and $\text{B}$ have ${4 \choose 0}=1$ possibility. $\text{C}$ will have ${4 \choose 2}=6$ possibilities. $\text{D}$ has $2\cdot{4 \choose 1}=8$ possibilities (note that the problem did not say a specific gender.) Therefore, $\boxed{\left(\text{D}\right)\text{ 3 are of one gender and 1 is of the other gender}}$ will have the greatest probability of occurring.

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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