a) Given, two dice are thrown simultaneously.
We have to find the probability that the sum of the numbers appearing on the dice is a prime number.
When 2 dice are thrown at the same time, the overall 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 number of possible outcomes = 36
Favourable outcomes = {(1,1) (1,2) (1,4) (1,6) (2,1) (2,3) (2,5) (3,2) (3,4) (4,1) (4,3) (5,2) (5,6) (6,1) (6,5)}
Number of favourable outcomes = 15
Number of possible outcomes = 36
Probability = number of favourable outcomes / number of possible outcomes
Probability = 15/36
= 5/12
Therefore, the probability of getting a sum of a prime number is 5/12.
b) When two dice are rolled together then total outcomes are 36 and Sample space is
[ (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) ]
So, pairs with sum 7 are (1, 6) (2, 5) (3, 4) (4, 3) (5, 2) (6, 1) i.e. total 6 pairs
Total outcomes = 36
Favorable outcomes = 6
Probability of getting the sum of 7 = Favorable outcomes / Total outcomes
= 6 / 36 = 1/6
So, P(sum of 7) = 1/6.
c) Getting a total of at most 5 means getting a total of 5 or total equal to less than 5.
⇒ Number of outcomes with total of 5 or less than 5 on two dice = {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)} = 10
⇒ Probability of getting a total of at most 5 = 10/36 = 5/18
∴ Probability of getting a total of at most 5 is 5/18.
Hope it helps you :)