SOLUTION: a basketball player makes 70% of the free throws he shoots. if he tries 13 throws what is the probability that he will make more than five throws?

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Question 609927: a basketball player makes 70% of the free throws he shoots. if he tries 13 throws what is the probability that he will make more than five throws?
Found 2 solutions by ewatrrr, htmentor:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi

Note: The probability of x succesP(0)ses in n trials is: 

P = nCx* p%5Ex%2Aq%5E%28n-x%29 where p and q are the probabilities of success and failure respectively. In this case p= .70 & q = .30 and n = 13

nCx = n%21%2F%28x%21%28n-x%29%21%29

P( making >5) = 1 - [P(0)+P(1)+  P(2) + P(3) + P(4)+ P(5)]

P = 1-(+%28.7%29%5E0%2A%28.3%29%5E%2813%29+ 13%28.7%29%5E1%2A%28.3%29%5E%2812%29++78%28.7%29%5E2%2A%28.3%29%5E%2811%29+286%28.7%29%5E3%2A%28.3%29%5E%2810%29+715%28.7%29%5E4%2A%28.3%29%5E%289%29+1287%28.7%29%5E5%2A%28.3%29%5E%288%29)= .9960

Recommend using TI or stattrek.com for statistical calculations.

Above is an explanatory presenation.  Hope this helps.

Answer by htmentor(1343) About Me  (Show Source):
You can put this solution on YOUR website!
This problem involves binomial probabilities
The binomial probability distribution is written as
P(x) = C(n,x)*p^x*(1-p)^(n-x) where n = the number of trials, x = the number of successes, and p is the probability of success
More than 5 throws means we need to compute the probability of 6 or more
P(6 or more) = 1 - P(5 or less)
P(5 or less) = P(5) + P(4) + ... P(0)
P(5) = C(13,5)*0.7^5*0.3^8 = 0.0142
.
.
.
Adding up all the probabilities and subtracting from 1, we get:
P(6 or more) = 1 - 0.018 = 0.982