Question

Identify the parameter p in the following binomial distribution scenario. The probability of buying a movie ticket with a popcorn coupon is 0.405 and without a popcorn coupon is 0.595. If you buy 12 movie tickets, we want to know the probability that exactly 2 of the tickets have popcorn coupons. (Consider tickets with popcorn coupons as successes in the binomial distribution.) Do not include 'p

Answer 1

`p = 0.405` is the probability of a ticket having popcorn coupon

6.02% probability that exactly 2 of the tickets have popcorn coupons.

Explanation:

For each movie ticket, there are only two possible outcomes. Either it has a popcorn coupon, or it does not. The probability of a movie ticket having a popcorn coupon is independent of other movie tickets. 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.

If you buy 12 movie tickets, we want to know the probability that exactly 2 of the tickets have popcorn coupons.

So p is the probability of a ticket having popcorn coupon.

The probability of buying a movie ticket with a popcorn coupon is 0.405

So
`p = 0.405`

This probability is P(X = 2) when n = 12. So

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

`P(X = 2) = C_(12,2).(0.405)^(2).(0.595)^(10) = 0.0602`

6.02% probability that exactly 2 of the tickets have popcorn coupons.