Question -
Answer -
Given a die is tossedtwo times.
When a die is tossedtwo times then the number of observations will be (6 × 6) = 36.
Now, let X is a randomvariable which represents the success.
(i) Here success isgiven as the number greater than 4.
Now
P (X = 0) = P (number≤ 4 in both tosses) = 4/6 × 4/6 = 4/9
P (X = 1) = P (number≤ 4 in first toss and number ≥ 4 in second case) + P (number ≥ 4 in first tossand number ≤ 4 in second case) is
= (4/6 × 2/6) + (2/6 ×4/6) = 4/9
P (X = 2) = P (number≥ 4 in both tosses) = 2/6 × 2/6 = 1/9
Hence, the requiredprobability distribution is,
| X | 0 | 1 | 2 |
| P (X) | 4/9 | 4/9 | 1/9 |
(ii) Here success isgiven as six appears on at least one die.
Now P (X = 0) = P (sixdoes not appear on any of die) = 5/6 × 5/6 = 25/36
P (X = 1) = P (sixappears at least once of the die) = (1/6 × 5/6) + (5/6 × 1/6) = 10/36 = 5/18
Hence, the requiredprobability distribution is,