Question

Find the probability distribution of Y in two throws of two dice, where Y represents the number of times a total of 9 appears.

Answer 1

It is given that Y denotes the number of times a total of 9 appears on throwing the pair of dice.

When the dice is thrown 2 times, the possibility of getting a total of 9 is possible only for the given combinations:

(3, 6) (4, 5) (5, 4) (6, 3)

So, the total number of outcomes is 36 and the total number of favourable outcomes is 4.

Probability of getting a total of 9 =

\[\frac{4}{36} = \frac{1}{9}\]

Probability of not getting a total of 9 =

\[1 - \frac{1}{9} = \frac{8}{9}\]

If Y takes the values 0, 1 and 2, then

\[P\left( Y = 0 \right) = \frac{8}{9} \times \frac{8}{9} = \frac{64}{81}\]
\[P\left( Y = 1 \right) = \frac{1}{9} \times \frac{8}{9} + \frac{8}{9} \times \frac{1}{9} = \frac{16}{81}\]
\[P\left( Y = 2 \right) = \frac{1}{9} \times \frac{1}{9} = \frac{1}{81}\]

Thus, the probability distribution of X is given by

Y	P(Y)
0	\[\frac{64}{81}\]
1	\[\frac{16}{81}\]
2	\[\frac{1}{81}\]