Question

Question 3 (4 marks)

Answer 1

Answer:

3.1 0.6912 = 69.12% probability that one or two out of the next four customers will make a purchase.

3.2 0.9744 = 97.44% probability that at least one out of the next four customers do not make a purchase

Step-by-step explanation:

For each customer, there are only two possible outcomes. Either they make a purchase, or they do not. The probability of a customer making a purchase is independent of any other customer. This means that the binomial probability distribution is used 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.

40% of all customers walking into their store will buy at least one item on that occasion.

This means that $p = 0.4$

4 customers:

This means that $n = 4$

3.1 What is the probability that one or two out of the next four customers will make a purchase?

This is:

$P(1 \leq X \leq 2) = P(X = 1) + P(X = 2)$

In which

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

$P(X = 1) = C_{4,1}.(0.4)^{1}.(0.6)^{3} = 0.3456$

$P(X = 2) = C_{4,2}.(0.4)^{2}.(0.6)^{2} = 0.3456$

So

$P(1 \leq X \leq 2) = P(X = 1) + P(X = 2) = 0.3456 + 0.3456 = 0.6912$

0.6912 = 69.12% probability that one or two out of the next four customers will make a purchase.

3.2. What is the probability that at least one out of the next four customers do not make a purchase?​

This is:

$P(X \leq 4) = 1 - P(X = 4)$

In which

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

$P(X = 4) = C_{4,4}.(0.4)^{4}.(0.6)^{0} = 0.0256$

$P(X \leq 4) = 1 - P(X = 4) = 1 - 0.0256 = 0.9744$

0.9744 = 97.44% probability that at least one out of the next four customers do not make a purchase