Question

Two unbiased dice are rolled once. Find the probability of getting

(i) a doublet (equal numbers on both dice)

(ii) the product as a prime number

(iii) the sum as a prime number

(iv) the sum as 1

Answer 1

(i) Doublet = {(1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6,6)}

Total number of outcomes = 6 x 6

n(S) = 36

Number of favourable outcomes = 6

P(doublet) = \(\frac{6}{36}=\frac{1}{6}\)

(ii) Number of favourable outcomes = 6 as favourable outcomes = (1, 2), (2, 1), (1, 3), (3, 1), (1, 5),and (5, 1)

P(prime number as product) 

(iii) Sum as prime numbers = {(1, 1), (1, 2), (2, 3), (1, 4), (1, 6), (4, 3), (5, 6)}

Number of favourable outcomes = 7

⇒ Probability = \(\frac{7}{36}\)

(iv) With two dice, minimum sum possible = 2

∴ Prob (sum as 1) = 0 [Impossible event]