Question

A survey finds customers are charged incorrectly for an item 20​% of the time. Suppose a customer purchases 14 items. Find the probability that the customer is charged incorrectly on at least 3 items.

Answer 1

Answer:

55.2% probability that the customer is charged incorrectly on at least 3 items.

Step-by-step explanation:

For each item, there are only two possible outcomes. Either it is charged incorrectly, or it is not. The probability of an item being charged incorrectly is independent from other items. 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.

A survey finds customers are charged incorrectly for an item 20​% of the time.

This means that $p = 0.2$

Find the probability that the customer is charged incorrectly on at least 3 items.

Either the customer is charged incorrectly on 2 or less items, or he is charged on at least 3. The sum of the probabilities of these events is 1.

So

$P(X < 3) + P(X \geq 3) = 1$

We want $P(X \geq 3)$

So

$P(X \geq 3) = 1 - P(X < 3)$

In which

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

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

$P(X = 0) = C_{14,0}.(0.2)^{0}.(0.8)^{14} = 0.0440$

$P(X = 1) = C_{14,1}.(0.2)^{1}.(0.8)^{13} = 0.1539$

$P(X = 2) = C_{14,2}.(0.2)^{2}.(0.8)^{12} = 0.2501$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0440 + 0.1539 + 0.2501 = 0.448$

$P(X \geq 3) = 1 - P(X < 3) = 1 - 0.448 = 0.552$

55.2% probability that the customer is charged incorrectly on at least 3 items.