in a single throw with two dice find the probabiloty that a sum is a multiple either of 3 or

answer is 5/9
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total favourable events=20
total events=36
probability=20/36=5/9
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probability = 20/36 = 0.55
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for sum of multiple of 3 and 4
getting 3,4,6,8,9,12
3-> (1,2),(2,1)
4-> (1,3),(2,2),(3,1)
6-> (1,5),(2,4),(3,3),(4,2),(5,1)
8-> (2,6),(3,5),(4,4),(5,3),(6,2)
9-> (3,6),(4,5),(5,4),(6,3)
12->(6,6)
total possibility = 2+3+5+5+4+1 = 20
probability = 20/36 =5/9
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20/36= 5/9
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getting mul of 3 is = 12/36
getting mul of 4 is = 8/36
here they give clue as "OR" so we can add both the things
so,we get (12/36)+(8/36)=(20/36)=(5/9)
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sum is multiple of 3----3,6,9,12
sum is multiple of 4----4,8,12
3-----[21]
6-----[15][24][33]
9-----[63][54]
12----[66]
4-----[13][22]
8-----[26][53][44]
so combinations to get on 2 dice = 8*(2!)+4=20
probability=20/(6*6)=5/9
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getting the sum as multiple 3 & 4

(1,2),(1,3),(1,5)
(2,1),(2,2),(2,4),(2,6)
(3,1),(3,3),(3,5),(3,6)
(4,2),(4,4),(4,5)
(5,1),(5,3),(5,4)
(6,2),(6,3),(6,6)

total no of possibility=3+4+4+3+3+3=20
probability=20/36=5/9
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this is a single throw so fav events will be 12 and total is 36 bhence probability is 12/36................can any1 clarify how twenty fav events when there is single throw
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4/36 sice 1,2 2,1 1,3 3,1
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in a single throw with two dice,find the probabiloty that a sum is a multiple either of 3 or 4