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 |

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