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Question
Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as
(i) number greater than 4
(ii) six appears on at least one die
Solution
When a die is tossed two times, we obtain (6 × 6) = 36 number of observations.
Let X be the random variable, which represents the number of successes.
Here, success refers to the number greater than 4.
P (X = 0) = P (number less than or equal to 4 on both the tosses) = `4/6xx4/6 = 16/36 = 4/9`
P (X = 1) = P (number less than or equal to 4 on first toss and greater than 4 on second toss) + P (number greater than 4 on first toss and less than or equal to 4 on second toss)
`= 4/6xx2/6+4/6xx2/6=8/36 + 8/36 = 16/36 =4/9`
P (X = 2) = P (number greater than 4 on both the tosses)
`= 2/6xx2/6= 4/36 =1/9`
Thus, the probability distribution is as follows.
| X | 0 | 1 | 2 |
| P(X) | `4/9` | `4/9` | `1/9` |
(ii) Here, success means six appears on at least one die.
P (Y = 0 ) = P (six appears on none of the dice) = `5/6 xx 5/6 = 25/36`
P(Y = 1) = P (six appears on none of the dice x six appears on at least one of the dice ) + P (six appears on none of the dice x six appears on at least one of the dice)
`= 1/6 xx 5/6 + 1/6 xx 5/6 = 5/36 + 5/36 = 10/36`
P (Y = 2) = P (six appears on at least one of the dice) = `1/6 xx 1/6 =1/36`
Thus, the required probability distribution is as follows
| Y | 0 | 1 | 2 |
| P(Y) | `25/36` | `10/36` | `1/36` |
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