Bob is a high school basketball player. He is a 90 % free throw shooter. That
means his probability of making a free throw is 0.90. During the season, what is
the probability that Bob makes his third free throw on his fifth shot?
We first calculate the number of ways he makes exactly 2 free throws on his
first 4, which is what he must do first in order to make his third free throw
on his fifth shot.
Of his 1st, 2nd, 3rd and 4th throws, that's 4 throws CHOOSE 2 or 4C2 = 6 ways
he can make exactly 2 free throws.
[They are these 6 ways: YYNN, YNYN, YNNY, NYYN, NYNY, NNYY]
For each of those, to be successful, he must make his fifth shot. So there are
the above 5 ways, with a Y on the end of each of them.
[So they are these 6 ways YYNNY, YNYNY, YNNYY, NYYNY, NYNYY, NNYYY]
So the numerator for the probability is 6.
Now we calculate the denominator for the probability, which is the number of
possibilities for the 5 shots.
There are 2 ways to choose the outcome of the first throw, Y or N.
There are 2 ways to choose the outcome of the second throw, Y or N.
There are 2 ways to choose the outcome of the third throw, Y or N.
There are 2 ways to choose the outcome of the fourth throw, Y or N.
There are 2 ways to choose the outcome of the fifth throw, Y or N.
That's 2∙2∙2∙2∙2 = 25 = 32 ways.
So the probability is 6 ways out of 32 or 6/32 which reduces to 3/16 = 0.1875.
Edwin
