Question

Two numbers are selected at random (without replacement) from the first six positive integers. Let $X$ denote the smaller of the two numbers obtained. Calculate the mathematical expectation of $X$.

Hint:

To compute expected value from a probability distribution, multiply each outcome by its probability and sum the results.

Answer 1

First six positive integers are: 1, 2, 3, 4, 5, 6.
Total number of ways to select 2 distinct numbers from 6: $^6C_2 = 15$
List all unordered pairs and determine the smaller number $X$ in each:
(1,2):1, (1,3):1, (1,4):1, (1,5):1, (1,6):1 → contributes five 1s
(2,3):2, (2,4):2, (2,5):2, (2,6):2 → four 2s
(3,4):3, (3,5):3, (3,6):3 → three 3s
(4,5):4, (4,6):4 → two 4s
(5,6):5 → one 5
Now compute expected value of $X$ using: \[ E(X) = \frac{1}{15} \left(5.1 + 4.2 + 3.3 + 2.4 + 1.5\right) \]
\[ E(X) = \frac{1}{15}(5 + 8 + 9 + 8 + 5) = \frac{35}{15} = \frac{7}{3} \]
Wait — above counts are correct, but the final sum: $5 + 8 + 9 + 8 + 5 = 35$ is correct. So: \[ E(X) = \frac{35}{15} = \boxed{\dfrac{7}{3}} \]