Two dice are thrown simultaneously. Find the probability of getting:

Explanation:

If we throw two dice then total number of sample points = 6² = 36

Sample Space = {(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)}

(i) The sum of two face numbers is a prime, then the sum can be = 2, 3, 5, 11

The sum is 2 when the outcome is {(1,1)} → 1outcome

The sum is 3 when the outcomes are {(1,2), (2,1)} → 2 outcomes

The sum is 5 when the outcomes are {(1,4), (2,3), (3,2), (4,1)} → 4 outcomes

The sum is 7 when the outcomes are {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)} → 6 outcomes

The sum is 11 when the outcomes are {(5,6), (6,5)} → 2outcomes

The number of favorable cases = 1 + 2 + 4 + 6 + 2 = 15

The required probability = 15/36 = 5/12 = 15

(ii) A total of at least 10 that is the sum can be = 10, 11 or 12

The sum is 10 when the outcomes are {(4,6), (5,5), (6.4)} → 3outcomes

The sum is 11 when the outcomes are {(5,6), (6,5)} → 2outcomes

The sum is 12 when the outcomes are {(6,6)} → 1 outcomes

The number of favorable outcomes = 3 + 2 + 1 = 6

Then required probability = 6/36 = 1/6 = 6

(iii) A doublet of even numbers i.e. outcomes are = {(2,2), (4,4), (6,6)}

The number of favorable case = 3

The required probability = 3/36 = 1/12

(iv) One dice shows the multiple of 2, that are = 2, 4 or 6

Other shows the multiple of 3, that are = 3 or 6

Then the favorable outcomes = {(2,3), (2,6), (3,2), (3,4), (3,6), (4,3), (4,6), (6,2), (6,3), (6,4), (6,6)}

The number of favorable case = 11

Then the required probability = 11/36

(v) Multiple of 3 as the sum then the sum can be = 3, 6, 9, or 12

The sum is 3 when the outcomes are {(1,2), (2,1)} → 2 outcomes

The sum is 6 when the outcomes are {(1,5), (2,4), (3,3), (4,2), (5,1)} → 5 outcomes

The sum is 9 when the outcomes are {(3,6), (4,5), (5,4), (6,3)} → 4 outcomes

The sum is 12 when the outcomes are {(6, 6)} → 1 outcomes

The number of favorable outcomes = 2 + 5 + 4 + 1 = 12

Then the required probability = 12/36 =1/3

Final Answer:

Hence the required probability that

(i) The sum as a prime number is = 5/12

(ii) Total of at least 10 is = 1/6

(iii) A doublet of even numbers = 1/12

(iv) A multiple of 2 on one dice and multiple of 3 on another is = 11/36

(v) A multiple of 3 as the sum is = 1/3