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A company is to hire two employees. They have prepared a final list of eight candidates, all of whom are equally qualified. Of these eight candidates, five are women. If the company decides to select two persons randomly from these eight candidates, what is the probability that:
i. Both candidates are women.
ii. The second candidate is a woman.
iii. The first candidate is a woman given that second one is a woman.

Respuesta :

Answer:

i) Probability that both candidates employed are women = 5/14

ii) Probability that the second candidate is a woman = 5/8

iii) Probability that the first candidate is a woman given that second one is a woman = 4/5

Step-by-step explanation:

Let the probability that a man is employed be P(M) = 3/8

Probability that a woman is employed P(W) = 5/8

a) Probability that both candidates employed are women = (5/8) × (4/7) = 5/14

b) Probability that the second candidate is a woman = (probability that first candidate is a man and second candidate is a woman) + (probability that first candidate is a woman & second candidate is a woman)

= (3/8)(5/7) + (5/8)(4/7) = (15/56) + (20/56) = 35/56 = 5/8

c) Probability that the first candidate is a woman given that second one is a woman

Given that the second candidate was a women, means that the first candidate-women was selected among other four women.

Probability = (4/8)/(5/8) = 4/5

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