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): You can put this solution on YOUR website!

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.


Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


The "opposite" of missing at least one is making all four. The sum of the probabilities of those two cases must be 1.

P(miss at least one) = 1 minus P(make all four)



Use a calculator....


Answer by ikleyn(52803)   (Show Source): You can put this solution on YOUR website!
.

The post  (the "solution")  by  @ewatrrr is  INCORRECT.

Follow to correct solution by  @greenestamps.


Ignore the post by @ewatrrr,  for your safety.



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