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

Answer:

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

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

Step-by-step 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.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

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.