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

Answer 1

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`