A die is rolled twice. what is the probability of getting a sum equal to 9
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Given :- A die is rolled twice what is the probability that the sum is 9 ?
Solution :-
when a dice is rolled twice total Possible outcomes are :-
- (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.
and, total Possible outcomes of sum is 9 is :-
- (3,6) , (6,3) => sum 9.
- (5,4) , (4,5) => sum 9.
- Total 4 .
then,
→ Required Probability = (total Possible outcomes of sum is 9 or 11) / (total Possible outcomes) = (4/36) = (1/9) (Ans.)
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