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.)

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

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