**Question:**

Two dice are thrown together. The probability of getting a doublet is

(a) $\frac{1}{3}$

(b) $\frac{1}{6}$

(c) $\frac{1}{4}$

(d) $\frac{2}{3}$

**Solution:**

(2) $\frac{1}{6}$

Explanation:

When two dice are thrown simultaneously, all 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)

Number of all possible outcomes = 36

Let *E* be the event of getting a doublet.

Then the favourable outcomes are:

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

Number of favourable outcomes = 6

$\therefore P($ getting a doublet $)=P(E)=\frac{6}{36}=\frac{1}{6}$