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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.
Solution
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 =
Probability of not getting a total of 9 =
\[P\left( Y = 1 \right) = \frac{1}{9} \times \frac{8}{9} + \frac{8}{9} \times \frac{1}{9} = \frac{16}{81}\]
\[P\left( Y = 2 \right) = \frac{1}{9} \times \frac{1}{9} = \frac{1}{81}\]
| Y | P(Y) |
| 0 |
\[\frac{64}{81}\]
|
| 1 |
\[\frac{16}{81}\]
|
| 2 |
\[\frac{1}{81}\]
|
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