SOLUTION: The probability that a basketball player makes his free throws is 0.7. What is the probability that in his next 6 free throw attempts, he makes at least two? Assume independence
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Question 1170716: The probability that a basketball player makes his free throws is 0.7. What is the probability that in his next 6 free throw attempts, he makes at least two? Assume independence of the free throw attempts.
Answer by Solver92311(821) (Show Source): You can put this solution on YOUR website!
The probability of 
successes in 
independent trials where the probability of success on any given trial is 
is given by:
\ =\ {{n}\choose{k}}\,(p)^k(1-p)^{n-k})
"At least two" out of six would be:
\ +\ P(3,6,0.7)\ +\ P(4,6,0.7)\ + P(5,6,0.7)\ +\ P(6,6,0.7))
Or more succinctly put:
\ =\ \sum_{r=2}^6\,{{6}\choose{r}}\,(0.7)^r(0.3)^{6-r})
Or you could save yourself some arithmetic by realizing that:
\ =\ \sum_{r=0}^n\,{{n}\choose{r}}\,(p)^r(1-p)^{n-r}\,\equiv\ 1)
So:
\ =\ 1\ -\ P(< 2,6,0.7)\ =\ 1\ -\ \sum_{r=0}^1\,{{6}\choose{r}}\,(0.7)^r(0.3)^{6-r})
In either case, you can do your own arithmetic.
John
My calculator said it, I believe it, that settles it
From
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