Question

Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as number greater than 4 appears on at least one die.

Answer 1

When a die is tossed twice, the sample space S has 6 × 6 = 36 sample points.

∴ n(S) = 36

Trial will be a success if the number on at least one die is 5 or 6.

Let X denote the number of dice on which 5 or 6 appears.

Then X can take values 0, 1, 2

When X = 0 i.e., 5 or 6 do not appear on any of the dice, then

X = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)}.

∴ n(X) = 16

∴ P(X = 0) =`(n(X))/(n(S)) = 16/36 = 4/9`

When X = 1, i.e. 5 or 6 appear on exactly one of the dice, then

X = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (6, 1), (6, 2), (6, 3), (6, 4)}

∴ n(X) = 16

∴ P(X = 1) = `(n(X))/(n(S)) = 16/36 = 4/9`

When X = 2, i.e. 5 or 6 appear on both of the dice, then

X = {(5, 5), (5, 6), (6, 5), (6, 6)}

∴ n(X) = 4

∴ P(X = 2) =`(n(X))/(n(S)) = 4/36 = 1/9`

∴ The required probability distribution is

X	0	1	2
P(X = x)	`4/9`	`4/9`	`1/9`

Answer 2

Success is defined as a number greater than 4 appears on at least one die

Let X denote the number of successes.

∴ Possible values of X and 0, 1, 2.

Let P(getting a number greater than 4) = p

= `2/6`

= `1/3`

∴ q = 1 – p

= `1 - 1/3`

= `2/3`

∴ P(X = 0) = P(no success)

= qq

= q²

= `4/9`

P(X = 1) = P(one success)

= qp + pq = 2pq

=  `2 xx 1/3 xx 2/3`

= `4/9`

P(X = 2) = P(two successes)

= pp

= p²

= `1/9`

∴ Probability distribution of X is as follows:

X	0	1	2
P(X = x)	`4/9`	`4/9`	`1/9`