Help ASAP Please check to see if I have the last part correct. Thanks
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
0-7 hit and 0,8,9 miss
83234 mhhhh
60278 hmhhm
43601 hhhmh
27630 hhhhm
12608 hhhmm
72687 hhhmh
68056 hmmhh
65109 hhhmh
32464 hhhhh
61081 hhmmh
In 10 repetitions, there was know instance where 3 or more out of 5 free throw missed. So the probability her missing 3 or more out of 5 free throw is 0/10=0% or 0.0%
5 of 5
4 of 5
3 of 5
2 of 5
1 of 5
0 of 5
To calculate the probability of the basketball player missing 3 or more free throws out of 5, we can rely on a simulation using the provided random digits.
Let's go through the process of conducting the simulation step-by-step:
1. Start by grouping the random digits into pairs. Since we need 10 repetitions, we will have 10 pairs of random digits.
Pair 1: 83
Pair 2: 23
Pair 3: 46
Pair 4: 02
Pair 5: 78
Pair 6: 43
Pair 7: 60
Pair 8: 27
Pair 9: 63
Pair 10: 01
2. We know that a hit corresponds to random digits 0-7, and a miss corresponds to random digits 8 and 9.
3. For each pair of random digits, use them to determine whether the basketball player made or missed a free throw.
Pair 1 (83): Miss
Pair 2 (23): Miss
Pair 3 (46): Miss
Pair 4 (02): Miss
Pair 5 (78): Miss
Pair 6 (43): Miss
Pair 7 (60): Miss
Pair 8 (27): Miss
Pair 9 (63): Miss
Pair 10 (01): Hit
Based on this simulation, the basketball player missed all 5 free throws. Therefore, the probability of her missing 3 or more free throws out of 5 is 1 (since all 5 were missed) out of 10 (the total number of repetitions), which can be simplified to 1/10 or 0.1 or 10%.
Note: Your previous statement that "In 10 repetitions, there was no instance where 3 or more out of 5 free throws were missed" is incorrect, as per the simulation provided.
To find the probability of the basketball player missing 3 or more free throws out of 5, we can simulate the process using the given random digits. The random digits provided are used to determine whether each free throw is a hit (0, 1, 2, 3, 4, 5, 6, or 7) or a miss (8 or 9).
By analyzing the simulation results, we can determine the number of instances where the player misses 3 or more out of 5 free throws.
Let's go through the simulation results you provided:
1st repetition: 83234 - This means the player missed the first three free throws and hit the last two.
2nd repetition: 60278 - In this case, the player missed the 2nd, 4th, and 5th free throws.
3rd repetition: 43601 - The player missed the 1st, 4th, and 5th free throws.
4th repetition: 27630 - Here, the player missed the last three free throws.
5th repetition: 12608 - The player missed the 3rd, 4th, and 5th free throws.
6th repetition: 72687 - In this repetition, the player missed the 3rd, 4th, and 5th free throws.
7th repetition: 68056 - Here, the player missed the 2nd and 3rd free throws.
8th repetition: 65109 - The player missed the 3rd, 4th, and 5th free throws.
9th repetition: 32464 - In this case, the player missed all five free throws.
10th repetition: 61081 - Finally, the player missed the 3rd and 4th free throws.
From the 10 repetitions, we can see that there were no instances where the player missed 3 or more out of 5 free throws. Therefore, the probability of her missing 3 or more out of 5 free throws based on this simulation is 0/10, which is equal to 0% or 0.0%.
To summarize, the basketball player has a 0% chance of missing 3 or more free throws out of 5 based on the given simulation results.