SOLUTION: A pro basketball player is a poor​ free-throw shooter. Consider situations in which he shoots a pair of free throws. The probability that he makes the first free throw is 0.4

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Question 1049372: A pro basketball player is a poor​ free-throw shooter. Consider situations in which he shoots a pair of free throws. The probability that he makes the first free throw is 0.49. Given that he makes the​ first, suppose the probability that he makes the second is 0.65. Given that he misses the​ first, suppose the probability that he makes the second one is 0.41. Complete parts a through c below.
possiblity of him making both free throws is .49*.65 .3185
B. Find the probability that he makes one of the two free throws using the multiplicative rule with the two possible ways he can do this.

Answer by Aevans52(1) About Me  (Show Source):
You can put this solution on YOUR website!
P(3 Free Throws Made) = P(One Free Throw Made AND One Free Throw Made AND One Free Throw Made)

P(3 Free Throws Made) = P(One Free Throw Made) * P(One Free Throw Made) * P(One Free Throw Made)

P(3 Free Throws Made) = (7/10) * (7/10) * (7/10)

P(3 Free Throws Made) = (7*7*7)/(10*10*10)

P(3 Free Throws Made) = 343/1000

P(3 Free Throws Made) = 0.343

So the probability is 343/1000 or 0.343 (which is a 34.3% chance) to make all 3 free throws.