Question

suppose that a factory manufactures greeting cards that are sold in packs of 10 cards. Assume that each card is torn during printing with probability .01, and that damage occurs or does not occur independently on different cards. suppose that you are about to purchase a pack of cards. find the probability that you will have at least two torn cards in your pack.

Answer 1

0.42% probability that you will have at least two torn cards in your pack.

Explanation:

For each card, there are only two possible outcomes. Either it is torn, or it is not. The probability of a card being torn is independent of other cards. So 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) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

10 cards

This means that
`n = 10`

Each card is torn during printing with probability .01

This means that
`p = 0.01`

Find the probability that you will have at least two torn cards in your pack.

Either you have less than two cards, or you have at least two. The sum of the probabilities of these events is decimal 1. So

`P(X < 2) + P(X \geq 2) = 1`

We want
`P(X \geq 2)`. So

`P(X \geq 2) = 1 - P(X < 2)`

In which

`P(X < 2) = P(X = 0) + P(X = 1)`

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 0) = C_(10,0).(0.01)^(0).(0.99)^(10) = 0.9044`

`P(X = 1) = C_(10,1).(0.01)^(1).(0.99)^(9) = 0.0914`

`P(X < 2) = P(X = 0) + P(X = 1) = 0.9044 + 0.0914 = 0.9958`

`P(X \geq 2) = 1 - P(X < 2) = 1 - 0.9958 = 0.0042`

0.42% probability that you will have at least two torn cards in your pack.