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


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,


X



0



1



2



P(X)



4/9



4/9



1/9



(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,


X



0



1



P(X)



25/36



5/18



5
1