Question -
Answer -
Given first sixpositive integers.
Two numbers can beselected at random (without replacement) from the first six positive integer in6 × 5 = 30 ways.
X denote the larger ofthe two numbers obtained. Hence, X can take any value of 2, 3, 4, 5 or 6.
For X = 2, thepossible observations are (1, 2) and (2, 1)
P (X) = 2/30 = 1/15
For X = 3, thepossible observations are (1, 3), (3, 1), (2, 3) and (3, 2).
P (X) = 4/30 = 2/15
For X = 4, thepossible observations are (1, 4), (4, 1), (2,4), (4,2), (3,4) and (4,3).
P (X) = 6/30 = 1/5
For X = 5, thepossible observations are (1, 5), (5, 1), (2,5), (5,2), (3,5), (5,3) (5, 4) and(4,5).
P (X) = 8/30 = 4/15
For X = 6, thepossible observations are (1, 6), (6, 1), (2,6), (6,2), (3,6), (6,3) (6, 4),(4,6), (5,6) and (6,5).
P (X) = 10/30 = 1/3
Hence, the requiredprobability distribution is,
| X | 2 | 3 | 4 | 5 | 6 |
| P (X) | 1/15 | 2/15 | 1/5 | 4/15 | 1/3 |
Therefore E(X) is:
