SOLUTION: Suppose in a game of basketball, Bryan has a 70% probability of making a free throw. Over a game, he attempts 4 free throws. What is the probability that he will miss at least one
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-> SOLUTION: Suppose in a game of basketball, Bryan has a 70% probability of making a free throw. Over a game, he attempts 4 free throws. What is the probability that he will miss at least one
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Question 1178202: Suppose in a game of basketball, Bryan has a 70% probability of making a free throw. Over a game, he attempts 4 free throws. What is the probability that he will miss at least one of them? Found 3 solutions by ewatrrr, greenestamps, ikleyn:Answer by ewatrrr(24785) (Show Source):
Hi
Binomial distribution: p(success) = .70 and q = .3 (missing)
4 trials
Using TI or similarly an inexpensive calculator like a Casio fx-115 ES plus
probability that he will miss at least one of them?
Demonstrates must read problem carefully.
probability that he will miss at least one of them:
Using .3 as probability of missing:
P( x ≥ 1) = 1 - P(x = 0) = 1 - binompdf(4, .3, 0) = 1 - .2401 = .7599
Or
Using .7 as probability of making:
1 - P(x=4) = 1-.2401 = .7599
Wish You the Best in your Studies.
