SOLUTION: I need help reducing/increasing likelihood/probabilities. Here is the question I'm attempting to answer: A cereal manufacturer places one movie ticket in every tenth box of cereal.

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Question 866758: I need help reducing/increasing likelihood/probabilities. Here is the question I'm attempting to answer: A cereal manufacturer places one movie ticket in every tenth box of cereal. They are then randomly distributed for sale. One family buys eight boxes. Now, use the binomial table to find the probability that the family will find a) exactly three movie tickets, b) at least one movie ticket.
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!


Hi

p(ticket) = 1/10 = .1,   n = 8

P(x=3) = binompdf(8, .1, 3)= .0331

P(at least one) = 1 - P(none) = 1 - (.9)^8 = .5695 

Note: typos