A basketball player makes 80% of her free throws. Recently during a very close game, she shot 5 free throws near the end of the game and missed 3 of them. The fans booed. What is the probability of her missing 3 (or more) free throws out of 5? Set up and conduct a simulation (using the random digits below) with 10 repetitions.
To find the probability of a basketball player missing 3 (or more) free throws out of 5, we can use the binomial probability formula.
The binomial probability formula is:
P(X = k) = (nCk) * p^k * (1 - p)^(n-k)
Where:
- P(X = k) is the probability of getting exactly k successes
- n is the number of trials
- p is the probability of success in a single trial
- k is the number of desired successes
In this case, n = 5 (number of free throws), p = 0.2 (probability of missing a free throw), and we want to find P(X ≥ 3).
Using the formula, we can calculate the individual probabilities of missing 3, 4, and 5 free throws, and then add them up to get the final probability.
P(X = 3) = (5C3) * (0.2)^3 * (0.8)^(5-3)
P(X = 4) = (5C4) * (0.2)^4 * (0.8)^(5-4)
P(X = 5) = (5C5) * (0.2)^5 * (0.8)^(5-5)
To conduct a simulation, we'll use random digit numbers ranging from 0 to 9 (inclusive). We can assign a digit between 0 and 1 to represent a success (making the free throw) or a failure (missing the free throw). We'll simulate this process 10 times and observe the number of times the player misses 3 or more free throws.
Here are the random digits for the simulation:
3, 4, 8, 2, 7, 5, 9, 1, 6, 0
Using these digits, we can assign a "success" to any digit less than 2 (0 or 1), and a "failure" to any digit greater than or equal to 2.
Let's perform the simulation:
For each digit:
1. If the digit is less than 2, count it as "success" (made free throw). Otherwise, count it as "failure" (missed free throw).
Simulation results:
Success, Failure, Failure, Success, Failure, Success, Failure, Success, Success, Success
Out of 10 repetitions, the player missed 3 or more free throws in 4 repetitions.
Therefore, the probability of the player missing 3 (or more) free throws out of 5, based on the simulation, is 4 out of 10, or 40%.