Question

Identify the parameter p in the following binomial distribution scenario. The probability of buying a movie ticket with a popcorn coupon is 0.699, and the probability of buying a movie ticket without a popcorn coupon is 0.301. If you buy 24 movie tickets, we want to know the probability that more than 16 of the tickets have popcorn coupons. (Consider tickets with popcorn coupons as successes in the binomial distribution.)

Answer #1

Here, probability of success, p = 0.699

q = 0.301

n = 24

We need to find P(X > 16)

This is equals to

P(X > 16) = P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) + P(X = 21) + P(X = 22) + P(X = 23) + P(X = 24)

Now,

P(X = 17) = ^{24}C_{17} * 0.699^{17} *
(0.301)^{7} = 0.1759

Similarly for all others,

P(X = 18) = ^{24}C_{18} * 0.699^{18} *
(0.301)^{6} = 0.1589

P(X = 19) = ^{24}C_{19} * 0.699^{19} *
(0.301)^{5} = 0.1165

P(X = 20) = ^{24}C_{20} * 0.699^{20} *
(0.301)^{4} = 0.0676

P(X = 21) = ^{24}C_{21} * 0.699^{21} *
(0.301)^{3} = 0.0299

P(X = 22) = ^{24}C_{22} * 0.699^{22} *
(0.301)^{2} = 0.0094

P(X = 23) = ^{24}C_{23}* 0.699^{23} *
(0.301)^{1} = 0.0019

P(X = 24) = ^{24}C_{24} * 0.699^{24} *
(0.301)^{0} = 0.00018

So, **P(X > 16) = 0.56**

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