Given: A die is tossed two times.

When a die is tossed two times then the number of observations will be (6 × 6) = 36.

Now, let X is a random variable which represents the success.

(i) Here success is given as the number greater than 4.

Now

P(X = 0) = P(number ≤ 4 in both tosses)

P(X = 1) = P(number ≤ 4 in first toss and number ≥ 4 in second case) + P(number ≥ 4 in first toss and number ≤ 4 in second case) is:

P(X = 2) = P(number ≥ 4 in both tosses)

Hence, the required probability distribution is,

(ii) Here success is given as six appears on at least one die.

Now

P(X = 0) = P(six does not appear on any of die)

P(X = 1) = P(six appears atleast once of the die)

Hence, the required probability distribution is,