Each time Parker shoots a free throw in basketball, he has an 80% chance of making the shot. Suppose he is given 3 free throws. Find the probability of each event.
a) Makes the basket, misses the basket, makes the baset
b) makes all 3 shots
Please help having hard time with this
A. Assuming the shots can only be in that order, then P(Make, miss, make)=.8*.2*.8
b. prob=.8*.8*.8
To find the probability of each event, we can multiply the probabilities of the individual outcomes. In this case, we need to use the probability of making a shot, which is 80%, and the probability of missing a shot, which is the complement of making a shot, 100% - 80% = 20%.
a) The probability of making the first shot is 80%, the probability of missing the second shot is 20%, and the probability of making the third shot is 80%.
Therefore, the probability of each event is as follows:
a) P(makes, misses, makes) = P(makes) * P(misses) * P(makes) = 0.8 * 0.2 * 0.8 = 0.128 (or 12.8%)
b) The probability of making all three shots is 0.8 * 0.8 * 0.8 = 0.512 (or 51.2%)
Again, to calculate the probability of each event, you need to multiply the probabilities of individual outcomes.