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from a group of 8 men and 4 women, a team of 4 will be formed by random selection. what is the probability that the team will consist of 4 men

User Smdvlpr
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Explanation:

remember, a probability is always the ratio of

desired cases / totally possible cases

it starts with all the combinations of 4 out of 12 (= 8 + 4) choices

C(12, 4) = 12! / (4! × (12-4)!) = 12! / (4! × 8!) =

= 12×11×10×9 / 24 = 11×5×9 = 495

these are our totally possible outcomes.

the desired outcomes with only 4 men in the selected group out of 12 people is the same amount as the possible selections of 4 men out of the available 8 men :

C(8, 4) = 8! / (4! × (8-4)!) = 8! / (4! × 4!) =

= 8×7×6×5 / 24 = 2×7×5 = 70

any other combination of C(12, 4) must contain at least 1 woman.

so the probability of getting 4 men out of the random pull is

70/495 = 0.1414141414...

we could get that also by saying the probability to get a man on the first pull is (12 people in total, 8 "desired" period with the right gender)

8/12 = 2/3 = 0.666666666...

now, we have 7 men out of 11 total people for the second pull. the probability here is

7/11 = 0.636363636...

then we have 6 men out of 10 total people for the third pull. the probability here is

6/10 = 3/5 = 0.6

and lastly, we have 5 men out of total 9 people for the fourth pull. the probability is

5/9 = 0.555555555...

the probability to get 4 men out of the random pulling is the combination of these 4 individual probabilities :

2/3 × 7/11 × 3/5 × 5/9 = 210 / 1485 = 70/495 =

= 0.1414141414...

it is confirmed.

User Laurendoss
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