## A team of 2 is to be selected from a group of 5. if the number of males in the group is n, then what's the value of n?

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### A team of 2 is to be selected from a group of 5. if the number of males in the group is n, then what's the value of n?

by BTGmoderatorDC » Mon Aug 23, 2021 6:40 pm

00:00

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## Global Stats

A team of 2 is to be selected from a group of 5. if the number of males in the group is n, then what's the value of n?

1. The number of males in the group is one more than the number of females
2. The probability the team consists of at least one male is 9/10

OA D

Source: e-GMAT

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### Re: A team of 2 is to be selected from a group of 5. if the number of males in the group is n, then what's the value of

by swerve » Tue Aug 24, 2021 5:49 am

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## Global Stats

BTGmoderatorDC wrote:
Mon Aug 23, 2021 6:40 pm
A team of 2 is to be selected from a group of 5. if the number of males in the group is n, then what's the value of n?

1. The number of males in the group is one more than the number of females
2. The probability the team consists of at least one male is 9/10

OA D

Source: e-GMAT
Let the number of males be n and the number of females be f then,
$$n+f=5$$

Statement 1.
If $$n=f+1$$ then $$f+1+f=5,$$ so $$f=2, n=3.$$ Sufficient $$\Large{\color{green}\checkmark}$$

Statement 2:
If n males then $$f=(5-n)$$ females.
Now the probability to select at least one male$$= 1-$$(prob to select all female)
$$\dfrac{1}{10}=\dfrac{(5-n)C2}{5C2}$$
$$\dfrac{1}{10}=\dfrac{(5-n)8\ast (5-n-1)}{5\ast 4}$$
$$(n-5)(n-4)=2$$
$$n2-9n+18=0$$
$$n=6$$ or $$3$$

But $$n=6$$ not possible (as $$n+f=5$$) hence $$n=3$$ and $$f=2.$$ Sufficient $$\Large{\color{green}\checkmark}$$

Therefore, D

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