Answer:
(1 - 0.82)^5 or 0.00019
Step-by-step explanation:
We know that 18% of the time, he'll miss his first shot (100%−82%=18%).
Then 18% of the time he misses his first shot, he will also miss his second shot, and 82% of the time he misses his first shot, he will make his second shot.
Notice how we can completely ignore the rightmost section of the line now, because those were the times that he made the first free throw, and we only care about if he misses the first and the second. So the chance of missing two free throws in a row is 18% of the times that he missed the first shot, which happens 18% of the time in general.
This is 18%*18%, or 0.18*0.18= 0.032
18% of 3% is 0.18*0.032=0.006, or about 1%
There is a pattern here: the chance of missing two free throws in a row was 0.18⋅0.18, and the probability of missing three in a row was 0.18*0.032=0.18* (0.18 * 0.18)= 0.18^3.
In general, you can continue in this way to find the probability of missing any number of shots.
The probability of missing 5 free throws in a row is 0.18^5= (1 - 0.82)^5
