An insurance agent has appointments with four prospective clients tomorrow. From past experience the agent knows that the probability of making a sale on any appointment is 1 out of 5. Using the rules of probability, what is the likelihood that the agent will sell a policy to 3 of the 4 prospective clients?
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A. B. C. D. E.D
This is a binomial probability. The probability of getting r successes out of n trials where the probability of success each trial is p and probability of failure each trial is q (where q = 1-p) is given by: n!(p^r)[q^(n-r)]/r!(n-r)!. Here n = 4, r = 3,p = 0.20 and q = 0.80. Therefore we have 4!(0.2^3)(0.8^1)/3!1! = 0.026.