# Question-1

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

(i) An even number as the sum.

(ii) The sum as a prime number.

(iii) A total of at least 10.

(iv) A doublet of even number.

(i) An even number as the sum.

(ii) The sum as a prime number.

(iii) A total of at least 10.

(iv) A doublet of even number.

**Solution:**

Elementary events associated to the random experiment of throwing two dice 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 elementary events = 6 x 6 = 36.

(i) Let A be the event of getting an even number as the sum i.e., 2, 4, 6, 8, 10, 12.

Elementary vents favorable to event A are : {(1, 1), (1, 3), (3, 1), (2, 2), (1, 5), (5, 1), (2, 4), (4, 2), (3, 3), (2, 6), (6, 2), (4, 4), (5, 3), (3, 5), (5, 5), (6, 4), (4, 6) and (6, 6)}.

Clearly favorable number of elementary events n(A) = 18.

Hence, required probability P(A) = = .

(ii) Let A be the event of getting the sum as a prime number i.e., 2, 3, 5, 7, 11.Elementary events favorable to event A are:(1, 1), (1, 2), (2, 1), (1, 4),

(4, 1), (2, 3), (3, 2), (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3), (6, 5) and (5, 6).

Favorable number of elementary events n(B)= 15.

Hence, required probability P(B) = = .

(iii) Let A be the event of getting a total of at least 10 i.e., 10, 11, 12. Then, the elementary events favorable to A are:

(6, 4), (4, 6), (5, 5), (6, 5), (5, 6) and (6, 6).

Favorable number of elementary events n(C) = 6.

Hence, required probability P(C) = = .

(iv) Let A be the event of getting a doublet of even number. Then, the elementary events favorable to A are

(2, 2), (4, 4) and (6, 6).

**âˆ´**Favorable number of elementary events n(D) = 3.

Hence, required probability P(D)= = .