Question

In a simultaneous throw of a pair of dice, find the probability of getting a doublet of prime numbers

Answer 1

In a throw of pair of dice, total no of possible outcomes = 36 (6 × 6) which 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)

E ⟶ event of getting a doublet of prime no’s

No. of favorable outcomes = 3 {(2, 2) (3, 3) (5, 5)}

Total no. of possible outcomes = 36

We know that, Probability P(E) =`"No.of favorable outcomes"/"Total no.of possible outcomes"`

P(E) =3/36 =1/12