## If a child is randomly selected from Columbus elementary school, what is the probability that the child will be a boy?

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### If a child is randomly selected from Columbus elementary school, what is the probability that the child will be a boy?

by AAPL » Fri Apr 28, 2023 1:38 pm

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Magoosh

If a child is randomly selected from Columbus elementary school, what is the probability that the child will be a boy?

1) If $$25$$ boys are removed from the school, the probability of selecting a boy will be $$0.75$$

2) There are $$35$$ more boys than there are girls

OA C

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### Re: If a child is randomly selected from Columbus elementary school, what is the probability that the child will be a bo

by Brent@GMATPrepNow » Mon May 15, 2023 6:13 am

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## Global Stats

AAPL wrote:
Fri Apr 28, 2023 1:38 pm
Magoosh

If a child is randomly selected from Columbus elementary school, what is the probability that the child will be a boy?

1) If $$25$$ boys are removed from the school, the probability of selecting a boy will be $$0.75$$

2) There are $$35$$ more boys than there are girls

OA C
Target question: What is the probability that the child will be a boy?
This is a good candidate for rephrasing the target question.
Let G = # of girls in the school
Let B = # of boys in the school
So, G + B = total number of children in the school
So, P(selected child is a boy) = B/(G + B)
REPHRASED target question: What is the value of B/(G + B)?

The video posted below has tips on rephrasing the target question

Statement 1: If 25 boys are removed from the school, the probability of selecting a boy will be 0.75.
So, the number of boys = B - 25, and the total number of children = G + (B - 25)
We can write: (B - 25)/(G + B - 25) = 3/4
Since we have a linear equation with TWO variables, there's no way to solve this equation for B and G. So, statement 1 is NOT SUFFICIENT

If you're not convinced, consider these two CONFLICTING cases:
Case a: B = 28 and G = 1. After 25 boys leave, there are 3 boys and 1 girl. So, P(boy) = 3/4 = 0.75, which satisfies statement 1. In this case, the answer to the REPHRASED target question is B/(G + B) = 28/(1 + 28) = 28/29
Case b: B = 31 and G = 2. After 25 boys leave, there are 6 boys and 2 girls. So, P(boy) = 6/8 = 0.75, which satisfies statement 1. In this case, the answer to the REPHRASED target question is B/(G + B) = 31/(2 + 31) = 31/33
Since we cannot answer the REPHRASED target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: There are 35 more boys than there are girls.
There are several CONFLICTING cases that satisfy statement 2. Here are two:
Case a: B = 36 and G = 1. In this case, the answer to the REPHRASED target question is B/(G + B) = 36/(1 + 36) = 36/37
Case b: B = 37 and G = 2. In this case, the answer to the REPHRASED target question is B/(G + B) = 37/(2 + 37) = 37/39
Since we cannot answer the REPHRASED target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
From statement 1, we can write: (B - 25)/(G + B - 25) = 3/4
Cross multiply to get: 3(G + B - 25) = 4(B - 25)
Expand: 3G + 3B - 75 = 4B - 100
Rearrange to get: 3G - B = - 25

From statement 2, we can write: B = G + 35

At this point, we have two different linear equations with two variables. So, we COULD solve the system for B and G, which means we COULD answer the REPHRASED target question with certainty.
So, the combined statements are SUFFICIENT