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

Given the first six positive integers: 1, 2, 3, 4, 5, 6.
There are $^6C_2 = 15$ distinct ways to choose 2 numbers from these 6.
We list all unordered pairs and identify the minimum value $X$ for each pair:
Pairs with 1 as the minimum: (1,2), (1,3), (1,4), (1,5), (1,6) - 5 instances of $X=1$
Pairs with 2 as the minimum: (2,3), (2,4), (2,5), (2,6) - 4 instances of $X=2$
Pairs with 3 as the minimum: (3,4), (3,5), (3,6) - 3 instances of $X=3$
Pairs with 4 as the minimum: (4,5), (4,6) - 2 instances of $X=4$
Pairs with 5 as the minimum: (5,6) - 1 instance of $X=5$
The expected value of $X$ is calculated as:
\[E(X) = \frac{1}{15} \left(5.1 + 4.2 + 3.3 + 2.4 + 1.5\right)\]
\[E(X) = \frac{1}{15}(5 + 8 + 9 + 8 + 5) = \frac{35}{15} = \frac{7}{3}\]
The sum $5 + 8 + 9 + 8 + 5 = 35$ is correct, leading to the expected value:
\[E(X) = \frac{35}{15} = \boxed{\dfrac{7}{3}}\]