A die is rolled twice Write the indicated event in set notat

A die is rolled twice. Write the indicated event in set notation. The sum of the rolls is 9. {(1, 4), (2, 3), (3, 2), (4, 1)} {(3, 2), (4, 1)} {(2, 3), (4, 1)} {(2, 3), (3, 2)} Find the indicated probability. Find the probability that the sum is at least 7 when two fair dice are rolled. 1/18 1/3 5/36 1/12

Solution

3) When a die is rolled out two times the out comes 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)}

The set of outcomes for which the sum of the rolls is 9 ={(3.6), (4,5), (5,4), (6,3)}

There is no correct option given in A, B, C, D of your problem (3).

If you are sure that one of the option must be correct then the question (event) is to be modified as \"sum of the rolls is 5\" and the corresponding out comes are {(1,4), (2,3), (3,2), (4,1)}. In this case option \"A\" is correct.

4) The sum is atleast seven means it may be 7 or 8 or 9 or 10 or 11 or 12

The favourable outcomes are {(1.6), (2,5), (2,6), (3,4), (3,5), (3,6), (4,3), (4,4), (4,5), (4,6), (5,2), (5,3), (5,4), (5,5)(5,6), (6.1), (6,2), (6,3), (6,4), (6,5), (6,6)}

Number of Favourable cases for our event = 21

Total number of out comes in rolling a die twice =36

Therefore the probability for sum to be atleast 7 =21/36 =7/12

Again there is no correct option given. please check your questions.