Question 339175: A basketball player has a 70% free throw shooting average, which can be interpreted to mean that the probability of his hitting any single free throw is 0.7 or 7/10. Assume that free throws are independent events, so that what happens on one free throw doesn't affect the outcome of the next. What is the probability the player makes his first three free throws?
What is the probability that in shooting five free throws he hist at least one?
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! What is the probability the player makes his first three free throws?
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)
Note: this equation is possible since the events are independent.
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.
Note: In real life, these events are likely to be dependent. Eg, say the player misses the first two. If this happens, his/her confidence may be lowered which may contribute to missing the third one (ie also contributing to lowering the probability).
I'll let you answer the second question.
You either have to break up the problem into cases (ie chances of the player hitting all 5 + chances of the player hitting 4 shots + chances of the player hitting 3 shots + chances of the player hitting 2 shots + chances of the player hitting 1 shot), or you can think of it like 1-P(missing all 5)
If you need more help, email me at jim_thompson5910@hotmail.com
Also, feel free to check out my tutoring website.
Jim
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