Three dice are thrown at the same time. Find the probability of getting three two’s, if it is known that the sum of the numbers on the dice was six.

$\frac{1}{10}$

The correct answer is Option (2) → $\frac{1}{10}$ ##

On a throw of three dice, we have the of sample space i.e., $[n(S)] = 6^3 = 216$

Let $E_1$ is the event when the sum of numbers on the dice was six and $E_2$ is the event when three two’s occurs.

$\Rightarrow E_1 = \{(1, 1, 4), (1, 2, 3), (1, 3, 2), (1, 4, 1), (2, 1, 3), (2, 2, 2), (2, 3, 1), (3, 1, 2), (3, 2, 1), (4, 1, 1)\}$

$\Rightarrow n(E_1) = 10$ and $E_2 = \{2, 2, 2\}$

$\Rightarrow n(E_2) = 1$

Also, $(E_1 \cap E_2) = 1$

$∴P(E_2 | E_1) = \frac{P(E_1 \cap E_2)}{P(E_1)} = \frac{1/216}{10/216} = \frac{1}{10}$