The probability of buying a movie ticket with a popcorn coupon is 0.608. If you buy 10 movie tickets, what is the probability that more than 3 of the tickets have popcorn coupons? (Round your answer to 3 decimal places if necessary.)

Answer: the probability that more than 3 of the tickets have popcorn coupons is 0.951

Step-by-step explanation:

We would assume a binomial distribution for the event of buying a movie ticket with a popcorn coupon. The formula is expressed as

P (x = r) = nCr * p^r * q^ (n - r)

Where

x represent the number of successes.

p represents the probability of success.

q = (1 - r) represents the probability of failure.

n represents the number of trials or sample.

From the information given,

p = 0.608

q = 1 - p = 1 - 0.608

q = 0.392

n = 10

P (x > 3) = 1 - P (x ≤ 3)

P (x ≤ 3) = P (x = 0) + P (x = 1) + P (x = 2) + P (x = 3)

Therefore,

P (x = 0) = 10C0 * 0.608^0 * 0.392^ (10 - 0) = 0.000086

P (x = 1) = 10C1 * 0.608^1 * 0.392^ (10 - 1) = 0.0013

P (x = 2) = 10C2 * 0.608^2 * 0.392^ (10 - 2) = 0.0093

P (x = 3) = 10C3 * 0.608^3 * 0.392^ (10 - 3) = 0.038

P (x ≤ 3) = 0.000086 + 0.0013 + 0.0093 + 0.038 = 0.049

Therefore,

P (x > 3) = 1 - 0.049 = 0.951