mary shots free throws 70% on average. She is fouled shooting a three point shot and therefore gets 3 free throws. What is the probability she:
*makes first misses seconds and third shots
*Makes at least 1 shot
*Makes at least 1 shot, given misses first
* misses first, given makes at least 1 shot
I did the first two
for the last two , you are dealing with conditional probability
Prob( B given A)
= prob(B|A) = prob(A and A)/prob(A)
To calculate the probabilities in this scenario, we'll need to use basic probability formulas and a little bit of conditional probability.
Before we proceed, let's define the probability of Mary making a free throw as M = 0.7 (70%) and missing a free throw as M' = 0.3 (30%).
1. To calculate the probability that Mary makes the first shot, misses the second shot, and makes the third shot, we simply multiply the probabilities of each event together:
P(make, miss, make) = M * M' * M = 0.7 * 0.3 * 0.7 = 0.147 or 14.7%.
2. To calculate the probability that Mary makes at least 1 shot, we need to subtract the probability of missing all three shots from 1:
P(At least 1 shot) = 1 - P(miss, miss, miss) = 1 - M' * M' * M' = 1 - 0.3 * 0.3 * 0.3 = 1 - 0.027 = 0.973 or 97.3%.
3. To calculate the probability that Mary makes at least 1 shot, given that she misses the first shot, we need to find the conditional probability using the following formula:
P(A | B) = P(A ∩ B) / P(B), where A is the event of making at least 1 shot and B is the event of missing the first shot.
P(make | miss) = P(make ∩ miss) / P(miss).
The probability of making at least 1 shot and missing the first shot is the same as the probability of making the second and third shots without considering the first shot:
P(make ∩ miss) = P(make, make) = M * M = 0.7 * 0.7 = 0.49 or 49%.
The probability of missing the first shot is M', so:
P(make | miss) = P(make ∩ miss) / P(miss) = 0.49 / 0.3 = 1.633 or 163.3%.
4. To calculate the probability that Mary misses the first shot, given that she makes at least 1 shot, we use the formula for conditional probability again:
P(B | A) = P(A ∩ B) / P(A), where A is the event of making at least 1 shot and B is the event of missing the first shot.
We already know that P(make, make) = P(make ∩ miss) = 0.49 or 49% and P(At least 1 shot) = 0.973 or 97.3%.
Thus, P(miss | At least 1 shot) = P(make ∩ miss) / P(At least 1 shot) = 0.49 / 0.973 = 0.503 or 50.3%.
Therefore, the probabilities are as follows:
1. P(make, miss, make) = 0.147 or 14.7%
2. P(At least 1 shot) = 0.973 or 97.3%
3. P(make | miss) = 1.633 or 163.3%
4. P(miss | At least 1 shot) = 0.503 or 50.3%.