Question

Two numbers are selected at random (without replacement) fromthe five positive integers. let X denote the larger of the two numbers obtained. Find the mean and variance of X

Answer 1

The two positive integers can be selected from the first six positive integers without replacement in 6 × 5 = 30 ways

X represents the larger of the two numbers obtained. Therefore, X can take the value of 2, 3, 4, 5, or 6.

For X = 2, the possible observations are (1, 2) and (2, 1).

For X = 3, the possible observations are (1, 3), (2, 3), (3, 1), and (3, 2).

For X = 4, the possible observations are (1, 4), (2, 4), (3, 4), (4, 3), (4, 2), and (4, 1).

For X = 5, the possible observations are (1, 5), (2, 5), (3, 5), (4, 5), (5, 4), (5, 3), (5, 2), and (5, 1).

For X = 6, the possible observations are (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 4), (6, 3), (6, 2), and (6, 1).



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Answer 2

Hey !!

We have, the first five positive integers are 1, 2, 3, 4 and 5.

We can select two numbers from 5 numbers is

            ⁵p₂ = $\frac{5}{5 - 2}=\frac{120}{6}=20$

Here, given X denote the larger of two numbers now we observe that X can be take value 2, 3, 4, 5

P (X = 2) = Probability that larger values 2, 3, 4, 5

P (X = 2) = Probability of getting 1 in first selection and 2 in second selection or getting 2 in first selection and 1 in second selection

=>                P (X = 2) = 1/5 × 1/4 + 1/5 × 1/4 = 2/20

P (X = 3) = Probability that the larger of two numbers is 3.

=>            P (X = 3) = 2/5 × 1/4 + 1/5 × 2/4 = 4/20

P (X = 4) = Probability that the larger of two numbers is 4.

    =>            P (X = 4) = 3/5 × 1/4 + 1/5 × 3/4 = 6/20

                    P (X = 5) = 4/5 × 1/4 + 1/5 × 4/5 = 8/20

Thus, the probability distribution of X is

X :           2           3             4           5

P(X) :       2/20      4/20      6/20       8/20

∴ Mean [E(X)] = 2 × 2/20 + 3 × 4/20 + 4 × 6/20 + 5 × 8/20

                        = 4/20 + 12/20 + 24/20 + 40/20

                        = 80/20

                         = 4

E(X²) = (2)² × 2/20 + (3)² × 4/20 + (4)² × 6/20 + (5)² × 8/20

         = 8/20 + 36/30 + 96/20 + 200/20

          = 340/20

Variance of X = E(X²) - [E(X)]²

                       = 17 - (4)² = 1

Hence, E(X) = 4 and Var (X) = 1

Good luck !!