Question -
Answer -
Given two fair diceare rolled
When two fair dice arerolled then number of observations will be 6 × 6 = 36.
X denote the sum ofthe numbers obtained when two fair dice are rolled. Hence, X can take any valueof 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 or 12.
For X = 2, thepossible observations are (1, 1).
P (X) = 1/36
For X = 3, thepossible observations are (1, 2) and (2, 1)
P (X) = 2/36 = 1/18
For X = 4, thepossible observations are (1, 3), (2, 2) and (3, 1).
P (X) = 3/36 = 1/12
For X = 5, thepossible observations are (1, 4), (4, 1), (2, 3) and (3, 2)
P (X) = 4/39 = 1/9
For X = 6, thepossible observations are (1, 5), (5, 1), (2, 4), (4, 2) and (3, 3).
P (X) = 5/36
For X = 7, thepossible observations are (1, 6), (6, 1), (2,5), (5,2),(3,4) and (4,3).
P (X) = 6/36 = 1/6
For X = 8, thepossible observations are (2, 6), (6, 2), (3, 5), (5, 3) and (4, 4).
P (X) = 5/36
For X = 9, thepossible observations are (5, 4), (4, 5), (3, 6) and (6, 3)
P (X) = 4/36 = 1/9
For X = 10, thepossible observations are (5, 5), (4, 6) and (6, 4).
P (X) = 3/36 = 1/12
For X = 11, thepossible observations are (6, 5) and (5, 6)
P (X) = 2/36 = 1/18
For X = 12, thepossible observations are (6, 6).
P (X) = 1/36
Hence, the requiredprobability distribution is,
| X | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
| P (X) | 1/36 | 1/18 | 1/12 | 1/9 | 5/36 | 1/6 | 5/36 | 1/9 | 1/12 | 1/18 | 1/36 |
Therefore E(X) is:
= √5.833
= 2.415