Question

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 one out of five. Using the rules of probability, what is the likelihood that the agent will sell a policy to three of the four prospective clients?

Answer 1

Answer:

0.082

Step-by-step explanation:

Total number of clients = 4

From past experience, the agent knows that the probability of making a sale on any appointment is one out of five = 1/5.

This shows that there are only two possible outcomes, success or failure. It is either she makes a sale on an appointment or she fails to make a sale on the appointment. Therefore, we can use binomial distribution.

For binomial distribution,

P(x=r) = nCr × q^(n-r) × p^r

p = probability of making a sale on any appointment = 1/5 = 0.2

q = probability of not making a sale on any appointment = 1 -1/5 = 0.8

n = 5

x = 3

P(x=3) = 5C3 × 0.8^2 × 0.2^3

= 10×0.6724× 0.012167 =0.082

Answer 2

Answer:

2.56% probability that the agent will sell a policy to three of the four prospective clients

Step-by-step explanation:

For each client, there are only two possible outcomes. Either a policy is sold, or it is not. The probability of a policy being sold to a client is independent of other clients. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

An insurance agent has appointments with four prospective clients tomorrow.

This means that $n = 4$

From past experience the agent knows that the probability of making a sale on any appointment is one out of five.

This means that $p = \frac{1}{5} = 0.2$

Using the rules of probability, what is the likelihood that the agent will sell a policy to three of the four prospective clients?

This is P(X = 3).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 3) = C_{4,3}.(0.2)^{3}.(0.8)^{1} = 0.0256$

2.56% probability that the agent will sell a policy to three of the four prospective clients