Every time Jim shoots a free throw in basketball, he has an 80% chance of making the shot. He is given 3 free throws. Find the probability of each event.
a) misses all 3 shots
b) misses the first two shots and makes the third.
Can someone please explain to me how to get the answer? Thank you
They are independent events. Multiply the probabilities
prob making=.8 probab missin=.2
a. mass all three: .2*.2*.2
misstwo, make last: .2*.2*.8
To find the probability of each event, we will use the concept of independent events.
a) The probability of missing a shot is 1 - the probability of making a shot since Jim has an 80% chance of making a shot. So, the probability of missing a shot is 1 - 0.80 = 0.20. Since each shot is independent, we can multiply the probabilities together.
Therefore, the probability of missing all 3 shots is (0.20) * (0.20) * (0.20) = 0.008 or 0.8%.
b) In this case, Jim needs to miss the first two shots and make the third shot.
The probability of missing the first shot is 0.20, and the probability of missing the second shot is also 0.20. However, since Jim needs to make the third shot, the probability of making it is 0.80.
Since the shots are independent events, we multiply the probabilities together.
The probability of missing the first two shots and making the third shot is (0.20) * (0.20) * (0.80) = 0.032 or 3.2%.