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.
8323460278436012763012608726876805665109324646108127541745017491243217468017649817480716408712807408783402746237416207
48648148631085738
DO I USED THE NUMBERS 0 - 7 OR 1 - 8
0 1 2 3 4 5 6 7 OR 1 2 3 4 5 6 7 8
BELOW I USED 1-7 AND DID NOT USED ZERO HAS A FREE THROW
83234=MHHHH ALL HIT
60278=HMHHM MISS TWO
43601=HHHMH MISS ONE
27630=HHHHM MISS ONE
12608=HHHMM MISS TWO
72687=HHHMH HIT ALL
68056=HMMHH MISS ONE
6 5109=HHHMM MISS TWO
32464=HHHHH HIT ALL
61081=HHMMH MISS ONE
5 of 5 3
4 of 5 4
3 of 5 3
2 of 5 0
1 of 5
0 fo 5
To calculate the probability of the basketball player missing 3 (or more) free throws out of 5, we can use a simulation approach with the provided random digits.
First, let's use the random digits to simulate the outcomes of the free throws. When using the random digits for simulations, it is common to interpret them as numbers between 0 and 9, inclusive. So we will use the numbers 0-7 or 1-8 according to your preference.
Using the random digits you provided, we can simulate 10 repetitions of shooting 5 free throws. Below are the outcomes for each repetition:
1. 83234 - All hit (MHHHH)
2. 60278 - Miss two (HMHHM)
3. 43601 - Miss one (HHHMH)
4. 27630 - Miss one (HHHHM)
5. 12608 - Miss two (HHHMM)
6. 72687 - All hit (HHHMH)
7. 68056 - Miss one (HMMHH)
8. 6 5109 - Miss two (HHHMM)
9. 32464 - All hit (HHHHH)
10. 61081 - Miss one (HHMMH)
Next, we count the number of simulations where 3 or more free throws were missed out of 5. Looking at the outcomes above, we can see that in the following repetitions, the player missed 3 or more free throws:
- In the second repetition: Missed two (HMHHM)
- In the fifth repetition: Missed two (HHHMM)
- In the eighth repetition: Missed two (HHHMM)
So, out of the 10 repetitions, there were 3 simulations where the player missed 3 or more free throws.
To find the probability, we divide the number of simulations where the player missed 3 or more free throws by the total number of repetitions:
Probability = 3/10 = 0.3
Therefore, the probability of the basketball player missing 3 (or more) free throws out of 5, based on the provided simulation, is 0.3 or 30%.