Life insurance salesman meets his prospective customers separately. At each meeting, he makes the same effort...

Life insurance salesman meets his prospective customers separately. At each meeting, he makes the same effort to persuade the prospective client to be insured. The vendor's understanding after a long term in the insurance business is that the probability of persuading the customer to be insured (probability of success) is 0,1. What is the probability in 4 trials to insure one of them?

Expert Solution

Binomial distribution: P(X) = nCx p^x q^n-x

P(a customer buys an insurance), p = 0.1

q = 1 - 0.1 = 0.9

Number of trials, n = 4

P(exactly one is insured) = 4C1 x 0.1 x 0.9³

= 0.2916

orchestra answered 2 years ago

Life insurance salesman meets his prospective customers separately. At each meeting, he makes the same effort...

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Life insurance salesman meets his prospective customers separately. At each meeting, he makes the same effort...

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