Hint: Calculate the number of possible outcomes for throwing two dice. Calculate the number of favourable outcomes for each of the cases. Use the fact that the probability of any event is the ratio of the number of favourable outcomes and the number of possible outcomes to calculate the probability of each of the events. Complete step-by-step solution - We have to calculate the probability of each of the events when two dice are thrown.We know that the total number of possible outcomes when two dice are thrown is $=6\times 6=36$.We know that the probability of any event is the ratio of the number of favourable outcomes and the number of possible outcomes.We will now calculate the probability of events in each case.(a) We have to calculate the probability that the sum of digits is a prime number.We will draw a table showing the sum of digits on rolling both the dice.
Read Less n(s) = 36 i.e. (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)} Event = {multiple of 3 as a sum}= {(1, 2), (2, 1), (1, 5), (5, 1), (2, 4), (4, 2), (3, 3), (3, 6), (6, 3), (5, 4), (4, 5), (6, 6)}n(E) = 12 p(E) = ? ∴ P(E) = `"n(E)"/"n(S)" = 12/36 = 1/3`. |