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When 2 people are selected at random from a group of 4 females and 4 m

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Math Revolution GMAT Instructor
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When 2 people are selected at random from a group of 4 females and 4 m  [#permalink]

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New post 27 Sep 2018, 23:42
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A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

77% (00:57) correct 23% (01:56) wrong based on 48 sessions

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[Math Revolution GMAT math practice question]

When \(2\) people are selected at random from a group of \(4\) females and \(4\) males, what is the probability that at least one female is selected?

\(A. \frac{5}{14}\)
\(B. \frac{7}{14}\)
\(C. \frac{9}{14}\)
\(D. \frac{11}{14}\)
\(E. \frac{13}{14}\)

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When 2 people are selected at random from a group of 4 females and 4 m  [#permalink]

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New post 27 Sep 2018, 23:51
Total possibilities 8c2 = (8*7)/2 = 28

Now since we are interested in probability at least one is female

We can find the probability that all 2 are males 4c2 = (4*3)/2 = 6

Now 6/28 = 3/14

Probability of at least one female = 1 - (3/14) = 11/14

Answer choice D

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Re: When 2 people are selected at random from a group of 4 females and 4 m  [#permalink]

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New post 27 Sep 2018, 23:54
MathRevolution wrote:
[Math Revolution GMAT math practice question]

When \(2\) people are selected at random from a group of \(4\) females and \(4\) males, what is the probability that at least one female is selected?

\(A. \frac{5}{14}\)
\(B. \frac{7}{14}\)
\(C. \frac{9}{14}\)
\(D. \frac{11}{14}\)
\(E. \frac{13}{14}\)


Probability of 1 Female in selected team = 1- Probability of NO Female in selected team

Probability of NO Female in selected team = 4C2/8C2 = 6/28 = 3/14

i.e. Probability of Atleast 1 female = 1-(3/14) = 11/14

Answer: Option D


ALTERNATIVE

Probability of atleast 1 female in team = Probability of exactly 1 female + probability of exactly two female = [(4C1*4C1)+(4C2)] / 8C2 = (16+6)/28 = 22/28 = 11/14

Answer: Option D
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Re: When 2 people are selected at random from a group of 4 females and 4 m  [#permalink]

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New post 30 Sep 2018, 21:58
=>

The probability that at least one female is selected is equal to 1 minus the probability that two males are selected.
There are 4C2 ways of selecting 2 males from the four males and 8C2 ways of selecting 2 people from the group of 8. Therefore, the probability of selecting two males from the group is 4C2/8C2, and the probability of selecting at least one female is
1 – 4C2/8C2\(= 1 – { \frac{(4*3)}{(1*2) } / { (8*7)/(1*2)} } = 1 – \frac{(4*3)}{(8*7)} = 1 – \frac{3}{14} = \frac{11}{14}\)

Therefore, the answer is D.
Answer: D
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Re: When 2 people are selected at random from a group of 4 females and 4 m &nbs [#permalink] 30 Sep 2018, 21:58
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When 2 people are selected at random from a group of 4 females and 4 m

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