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QUESTION 6 - 1 POINT
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 3 or more of the tickets have popcorn coupons? (Round your answer to 3 decimal places if necessary.)
Provide your answer below:
\[
P(X \geq 3)=
\]
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Step 1 ：The problem is asking for the probability of getting 3 or more successes (getting a popcorn coupon) in 10 trials (buying movie tickets), given the probability of success on each trial is 0.608. This is a binomial probability problem. We can solve it by summing up the probabilities of getting exactly 3, 4, 5, ..., 10 successes.

Step 2 ：Let's denote the probability of success (getting a popcorn coupon) as p, which is 0.608, and the number of trials (buying movie tickets) as n, which is 10.

Step 3 ：By calculating, we get the probability as 0.9893105888489067.

Step 4 ：Rounding to three decimal places, the final answer is 0.989.

Step 5 ：So, the probability that 3 or more of the tickets have popcorn coupons is \(\boxed{0.989}\).