Question 1172584: When a golfer plays any hole, he will take 3, 4, 5, 6, or 7 strokes with probabilities of 1/10, 1/5, 2/5, 1/5, and 1/10 respectively. He never takes more than 7 strokes.
Find the probability of the following events:
a) scoring 4 on each of the first three holes.
b) scoring 3, 4 and 5 (in that order) on the first three holes.
c) scoring a total of 28 for the first four holes.
d) scoring a total of 10 for the first three holes.
e) scoring a total of 20 for the first three holes.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! this is the probability of scoring a 4 (0.2) cubed=0.008.
In that order is the product of 0.1*0.2*0.4=0.008
Score of 28 in 4 holes is 0.1^4, the probability of a 7^4=0.0001
10 for the first three holes
Can't have a 5,6,7
Can have a 4 and two 3s, and there are 3 ways that can happen; so probability is 3*0.2*0.01=0.06
Score of 20 can be 6,7,7 and there are 3 ways that can happen, so the probability is 3*0.2*0.01=0.06
The last two should be the same and are, because the distribution is symmetrical.
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