Question

A survey finds customers are charged incorrectly for 4 out of every 10 items. suppose a customer purchases 11 items. find the probability that the customer is charged incorrectly on at least 3 items. the probability that the customer is charged incorrectly on at least 3 items is nothing

Answer 1

The probability that the customer is charged incorrectly on at least 3 items is 88.11%

Explanation:

For each item, there are only two possible outcomes. Either they are charged correctly, or they are not. The probability of an item being charged incorrectly is independent of 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) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

A survey finds customers are charged incorrectly for 4 out of every 10 items.

This means that
`p = (4)/(10) = 0.4`

Suppose a customer purchases 11 items.

This means that
`n = 11`

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

Either he is charged incorrectly on less than 3 items, or at least 3. The sum of the probabilities of these events is decimal 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_(11,0).(0.4)^(0).(0.6)^(11) = 0.0036`

`P(X = 1) = C_(11,1).(0.4)^(1).(0.6)^(10) = 0.0266`

`P(X = 2) = C_(11,2).(0.4)^(2).(0.6)^(9) = 0.0887`

`P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0036 + 0.0266 + 0.0887 = 0.1189`

`P(X \geq 3) = 1 - P(X < 3) = 1 - 0.1189 = 0.8811`

88.11% probability that the customer is charged incorrectly on at least 3 items