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

(Problem)
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==Solution 2==
 
==Solution 2==
 
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 th and $ilities (note that the problem did not say a specific gender.) Therefore, 3 are of one gender and 1 will have the greatest probability of occurring.
 
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 th and $ilities (note that the problem did not say a specific gender.) Therefore, 3 are of one gender and 1 will have the greatest probability of occurring.
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==Video Solution==
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https://youtu.be/3bF8BAvg0uY ~savannahsolver
  
 
==See Also==
 
==See Also==

Revision as of 09:19, 27 April 2022

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

Solution 2

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 th and $ilities (note that the problem did not say a specific gender.) Therefore, 3 are of one gender and 1 will have the greatest probability of occurring.

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

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