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.
83234602784360127630126087268768056651093246461081275417450
17491243217468017649817480716408712807408783402746237416207
48648148631085738
To find the probability of the basketball player missing 3 or more free throws out of 5, we can use a simulation approach. Here's how you can conduct the simulation using the provided random digits:
1. First, count the number of digits in the random number sequence: 100 (5-digit numbers * 20 repetitions = 100 digits).
2. Assign each digit a value from 0 to 9 (0 corresponds to a miss, while 1 to 9 correspond to a made free throw).
3. Start going through the random digits and keep track of the number of misses (0s) encountered. This count needs to reach 3 or more for us to count it as a success.
4. Repeat the process of going through the random digits for a total of 10 repetitions.
5. Count the number of successes (where the player missed 3 or more free throws) out of the 10 repetitions.
6. Divide the number of successes by the total number of repetitions (10) to get the probability.
Now let's conduct the simulation using the given random digit sequence:
First repetition: The first five digits are 83234. We have 2 misses (8 and 3) and 3 made free throws (2, 3, and 4).
Second repetition: The next five digits are 60278. We have 3 misses (6, 0, and 2) and 2 made free throws (7 and 8).
Third repetition: The next five digits are 43601. We have 3 misses (4, 3, and 6) and 2 made free throws (0 and 1).
Fourth repetition: The next five digits are 27630. We have 3 misses (2, 7, and 6) and 2 made free throws (3 and 0).
Fifth repetition: The next five digits are 12608. We have 2 misses (1 and 6) and 3 made free throws (2, 0, and 8).
Sixth repetition: The next five digits are 72546. We have 2 misses (7 and 5) and 3 made free throws (2, 4, and 6).
Seventh repetition: The next five digits are 10812. We have 4 misses (1, 0, 8, and 1) and 1 made free throw (2).
Eighth repetition: The next five digits are 75417. We have 3 misses (7, 5, and 4) and 2 made free throws (1 and 7).
Ninth repetition: The next five digits are 45017. We have 3 misses (4, 5, and 0) and 2 made free throws (1 and 7).
Tenth repetition: The next five digits are 49124. We have 3 misses (4, 9, and 1) and 2 made free throws (2 and 4).
Out of the 10 repetitions, there are 7 successes (where the player missed 3 or more free throws).
Therefore, the probability of the basketball player missing 3 or more free throws out of 5 is 7/10 or 0.7.