#### 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

#### Solution

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 = 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=4/9`

P (X = 2) = P (number greater than 4 on both the tosses)

`= 2/6xx2/6= 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) = 56 × 56 = 2536

P (Y = 1) = P (six appears on at least one of the dice) = 16 × 56 + 56 × 16 +16 × 16 =1136

Thus, the required probability distribution is as follows

Y | 0 | 1 |

P(Y) | 2536 | 1136 |