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Abraxas
New Member
USA
1 Posts |
 Posted - 10/26/2009 : 23:53:35
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Test message.
Also, my statistics class just started probability, and when the teacher explained an example about half the class decided to question his work. My common sense also told me he was wrong, but I'd like some clarification.
The question says that there is a 70% probablity a basketball player will make a free throw.
(a) Describe how to simulate a single shot if the probability of making a shot is .7. Then describe how to simulate 5 consecutive shots.
This part is pretty straightforward (I think). 0-6 are the chance of making a shot, 7-9 are misses. Using a random number table or random number generator to get results, one digit at a time for single shots, 5 digits at a time for the 5 consecutive shots.
(b) Simulate 50 repetitions of the 5 shots and record the number of missed on each repetition. What is the approximate likelihood that the player will miss 3 or more of the five shots?
This is the part that half the class questioned. For simplicity's sake I will only post 10 repetitions using the random number table from my book.
# of trial Simulation
1 8,6,7,4,6
2 1,2,1,4,9
3 3,7,8,2,3
4 7,1,8,6,8
5 1,8,4,4,2
6 3,5,1,1,9
7 6,2,1,0,3
8 3,9,2,4,4
9 9,6,9,2,7
10 1,9,9,3,1
My teacher says that you find the trials in which 3 or more shots were missed (4,9) and add up the total number of missed shots in these trials. In this case it is 6, so the probability of missing 3 or more shots is 6/50, or .12.
I and part of my class felt that it should be the number of trials that missed 3 or more shots divided by the total number of trials. Thus the probability using this logic is 2/10 (in this example), or .2.
Any help would be greatly appreciated since probability is one are I've always had some trouble with, and this debate we had in class didn't really help.
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Probability
