Michael Jordan’s career free throw percentage is 74%. That means that he makes 74% of his free throws. Let’s consider his next four free throws. Calculate the following probabilities, Round your answers to 3 decimal places.
A. Pimakes the next 4)
B. Pimisses the next 4)-
C. Pimakes at least one of the next 4)
Which of the following contains the correct answers in the correct order.
0.005 0.008 and 0.678
0.300.0005 and 0.995
0.300,0.005, and 0.912
0.300,0.023, and 0.995
The Correct Answer and Explanation is :
To solve this problem, we use the binomial probability formula:
[
P(X = k) = \binom{n}{k} p^k (1 – p)^{n-k}
]
where:
- ( p = 0.74 ) (probability of making a free throw),
- ( 1 – p = 0.26 ) (probability of missing a free throw),
- ( n = 4 ) (total free throws attempted),
- ( k ) is the number of successful free throws.
A. Probability of making all 4 free throws
This means making every attempt, so:
[
P(4) = (0.74)^4
]
[
P(4) = 0.29986576 \approx 0.300
]
B. Probability of missing all 4 free throws
This means missing every attempt, so:
[
P(0) = (0.26)^4
]
[
P(0) = 0.00456976 \approx 0.005
]
C. Probability of making at least one free throw
This is the complement of missing all free throws, so:
[
P(\text{at least 1}) = 1 – P(0)
]
[
P(\text{at least 1}) = 1 – 0.005 = 0.995
]
Correct Answer
The correct choice is:
[
0.300, 0.005, 0.995
]
which corresponds to option B.
Explanation
Michael Jordan’s free throw performance follows a binomial probability model, where each attempt is independent. Since he makes free throws at a 74% rate, the likelihood of making all four is calculated by multiplying the probabilities of four successful attempts.
Similarly, missing all four is determined by multiplying the probabilities of four consecutive failures. Finally, since probabilities must sum to 1, we calculate the chance of making at least one free throw by subtracting the probability of missing all four from 1.
This is a common probability problem in sports analytics, often used to predict player performance over multiple attempts.