Question

if two ludu-dice are thrown once. then what is the probability of getting 9 as the sum of the two numbers shown by both the dice ?

Answer 1

total no of possible events = 6²=36
and number of favorable events = (3,6),(4,5),(5,4) and (6,3)=4
then the probability=4/36
=1/9
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Answer 2

Question :- If two balanced dice are tossed once, what is the probability of the event that the sum of the integers coming on the upper sides of the two dice is 9 ?

Solution :-

we know that, when two dice are tossed once, then possible events will be

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

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

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

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

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

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

Total 36 .

so,

→ Total possible events where sum of upper side of the dice is 9 is = (3,6), (4,5), (5,4) and (6,3) = Total 4 .

therefore,

→ Required Probability = Favourable outcomes / Total number of outcomes = 4/36 = (1/9) (Ans.)

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