Question -
Answer -
Given a die is throwntwo 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 and is given as six appears on at leastone 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
P (X = 2) = P (sixdoes appear on both of die) = 1/6 × 1/6 = 1/36
Hence, the requiredprobability distribution is,
| X | 0 | 1 | 2 |
| P (X) | 25/36 | 5/18 | 1/36 |
Therefore Expectationof X E (X):
