a die is rolled twice what is the probability that the sum is 9 or11?
Answers
Given :- A die is rolled twice what is the probability that the sum is 9 or 11 ?
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 or 11 is :-
- (3,6) , (6,3) => sum 9.
- (5,4) , (4,5) => sum 9.
- (5,6) , (6,5) => sum 11.
- Total 6 .
then,
→ Required Probability = (total Possible outcomes of sum is 9 or 11) / (total Possible outcomes) = (6/36) = (1/6) (Ans.)
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SOLUTION
GIVEN
A die is rolled twice
TO DETERMINE
The probability that the sum is 9 or 11
EVALUATION
A die is rolled twice
So the total number of possible outcomes
Let A be the event that the sum is 9
Then event points for the event A is (3,6),(4,5),(5,4),(6,3)
So the total number of possible outcomes for the event A is 4
So the probability of the event A
= P(A)
Let B be the event that the sum is 11
Then event points for the event B is (5,6),(6,5)
So the total number of possible outcomes for the event B is 2
So the probability of the event B
= P(B)
Now
A ∩ B is the event that the sum is 9 and 11
Now A and B are mutually exclusive
∴ A ∩ B = Φ
∴ P(A ∩ B) = 0
Hence the required probability that the sum is 9 or 11
= P(A ∪ B)
= P(A) + P(B) - P(A ∩ B)
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