Final answer:
The likelihood that an insurance agent will sell policies to exactly 3 out of the 4 clients, given a 1 in 5 success rate per appointment, is 0.0256, corresponding to option B.
Step-by-step explanation:
The question asks for the likelihood that the agent will sell insurance policies to 3 out of the 4 prospective clients. The probability of making a sale to any client is 1 out of 5, which implies a probability of not making a sale is 4 out of 5. The problem is a case of a binomial distribution where the number of successes 'k' is 3, the number of trials 'n' is 4, and the probability of success 'p' is 0.20 (1/5).
Using the binomial probability formula: P(X = k) = (n choose k) * p^k * (1-p)^(n-k), we compute the probability P(X = 3) as follows:
- Calculate the combination for 3 successes out of 4 trials: (4 choose 3) = 4
- Calculate the probability of 3 successes: p^3 = (0.20)^3
- Calculate the probability of 1 failure: (1-p)^1 = (0.80)^1
- Multiply them together: 4 * (0.20)^3 * (0.80)^1
The result gives us the answer 4 * (0.008) * (0.80) = 0.0256.
Therefore, the likelihood (or probability) that the agent will sell policies to exactly 3 out of the 4 clients is 0.0256, which is option B.