in a throw of two dice,find the probability of getting a sum of 10
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36
Total outcomes are n(s) 36
Sum equal to 10 are n(a) 3={4,6 5,5 6,4}
Therefore probability of an event is n(a) /n(s) 3/36=1/12
Sum equal to 10 are n(a) 3={4,6 5,5 6,4}
Therefore probability of an event is n(a) /n(s) 3/36=1/12
Answered by
31
total possible outcomes = 6x6 = 36
favourable outcomes = (4,6) ( 6,4) ( 5,5) = 3
p roblablity = f outcomes / total possible outcomes
p = 3 / 36
p = 1/12
favourable outcomes = (4,6) ( 6,4) ( 5,5) = 3
p roblablity = f outcomes / total possible outcomes
p = 3 / 36
p = 1/12
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