Question 1078225: A die is rolled twice. What is the probability of getting either a multiple of 3 on the first roll or a total of 8 for both rolls?
I got 2/6+5/36, which is 17/36. But that doesn't match an available answer.
Found 2 solutions by stanbon, natolino_2017:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A die is rolled twice. What is the probability of getting either a multiple of 3 on the first roll or a total of 8 for both rolls?
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P(mult of 3) = [2+4+4+1]/36
P(sum of 8) = 5/36
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P(mult of 3 or sum of 8) = (11+5)/36 = 16/36
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Cheers,
Stan H.
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Answer by natolino_2017(77) (Show Source):
You can put this solution on YOUR website! Total Universe=6^2=36 cases-
First Condition only (First is multiple of 3 and the sum isn't 8) :
(3,1),(3,2),(3,3),(3,4),(3,6),(6,1),(6,3),(6,4),(6,5),(6,6).
Favorable cases= 10.
Second Condition only (First is not a multiple of 3 and the sum is 8) :
(2,6),(4,4),(5,3)
Favorable cases=3.
First and Second condition at the same time:
(3,5),(6,2)
favorable cases=2.
Total favorable cases=10+3+2=15.
So P()= 15/36 =5/12
#natolino
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