Question

40 points for 2 questions.

Please help me :)

3 persons A, B, C independently fire at a target. What is the probability that
a. Exactly one of them hits the target,
b. At least one of them hits the target?
Given: Probability of hitting the target. P(A) = 2/5, P(B) = 1/4, P(C) = 1/2.

2. Company provides to their clients the following products: accidental insurance and
health insurance.
If 55% of male clients opt for accidental insurance and 65% of female clients opt for
health insurance, then what is the probability that health insurance contract is chosen if
60% of the company’s clients are females?

Answer 1

1. a. 9/20 = 0.45; b. 31/40 = 0.775
2. 0.57 = 57%

Explanation:

You want to know various probabilities that A, B, and C will hit a target, given their individual probabilities of success are 0.4, 0.25, and 0.5, respectively. And, you want to know the probability a health insurance contract will be chosen if 45% of males and 65% of females favor it, and the clients are 60% female.

1. Target

a. Exactly one

The probability that exactly one person will hit the target is ...

P(AB'C') +P(A'BC') +P(A'B'C) = (0.4·0.75·0.5) +(0.6·0.25·0.5) +(0.6·0.75·0.5)

= 0.5(0.4·0.75 +0.6(0.25 +0.75)) = 0.5(0.3 +0.6) = 0.45

The probability that exactly one person hits the target is 9/20 = 0.45.

b. At least one

The probability that at least one hits the target is the complement of the probability that none do:

P(n≥1) = 1 -P(n=0) = 1 -P(A'B'C') = 1 -(0.6·0.75·0.5) = 0.775

The probability that at least one hits the target is 31/40 = 0.775.

2. Insurance

The probability that health insurance is chosen is the sum of the probabilities that the males will choose it and that the females will choose it, adjusted by the proportions of males and females.

P(health) = P(males&health)·%(males) +P(females&health)·%(females)

= (1 -0.55)(1-0.60) +(0.65)(0.60) = (0.45)(0.40) +(0.65)(0.60) = 0.57

The probability that clients will choose health insurance is 0.57 = 57%.

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