Question

A person calls people to ask if they would like to extend their automobile insurance beyond the normal3years. The probability that the respondent says yes is about37%. If she calls11people, find the probability that the first person to say yes will occur with the seventh customer.

Answer 1

Using the binomial distribution, it is found that there is a 0.0231 = 2.31% probability that the first person to say yes will occur with the seventh customer.

For each person, there are only two possible outcomes, either they say yes, or they say no. The probability of a person saying yes is independent of any other person, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

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

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

The parameters are:

• x is the number of successes.
• n is the number of trials.
• p is the probability of a success on a single trial.

In this problem:

• The probability that the seventh person is the first to say yes is P(X = 0) when n = 6(first six say no) multiplied by 0.37(probability the seventh say yes).
• 37% say yes, hence $p = 0.37$

Then:

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

$P(X = 0) = C_{6,0}.(0.37)^{0}.(0.63)^{6} = 0.0625$

$p = 0.37(0.0625) = 0.0231$

0.0231 = 2.31% probability that the first person to say yes will occur with the seventh customer.

A similar problem is given at brainly.com/question/24863377