Question

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?

Answer 1

Given that,

Probability of success, p = 0.1

Probability of failure, q = 1 - p = 1 - 0.1 = 0.9

Number of total outcomes n = 4

Number of favorable outcomes x = 1

By using the formula for binomial distribution,

P = $nCx\ *\ p^x * q^{n-x}$

P = $4C1\ *\ 0.1^1\ *\ 0.9^{4-1}$

P = 0.2916

Hence, the probability that in 4 trials of meeting 1 client gets insured is 0.2916