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When 2 people are selected at random from a group of 4 femal

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by Max@Math Revolution » Thu Sep 27, 2018 11:34 pm
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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. 5/14
B. 7/14
C. 9/14
D. 11/14
E. 13/14
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by swerve » Fri Sep 28, 2018 8:39 am
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.

The probability of at least one female = 1 - (3/14) = 11/14.
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by Max@Math Revolution » Sun Sep 30, 2018 9:46 pm
=>
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 – { (4*3)/(1*2) } / { (8*7)/(1*2) } = 1 – (4*3)/(8*7) = 1 – 3/14 = 11/14

Therefore, the answer is D.
Answer: D
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by Scott@TargetTestPrep » Sat Oct 13, 2018 5:33 pm
Max@Math Revolution wrote:
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. 5/14
B. 7/14
C. 9/14
D. 11/14
E. 13/14
We can use the formula:

P(at least one female is selected) = 1 - P(no females are selected)

P(no females are selected) = 4/8 x 3/7 = 1/2 x 3/7 = 3/14

P(at least one female is selected) = 1 - 3/14 = 11/14.

Alternate Solution:

Without any restriction, there are 8C2 = 8!/(2!*6!) = 28 different ways to choose two people from a group of eight.

We use the following formula:

#ways to choose two people = #ways where at least one female is selected + #ways where no female is selected

Notice that since there are 4 men, there are 4C2 = 4!/(2!*2!) = 6 ways to choose two people where no female is selected. Thus, 28 - 6 = 22 of the choices include at least one female. Thus, the probability that at least one female is selected is 22/28 = 11/14.

Answer: D

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