Question

Two numbers are selected at random from the first six positive integers. If X denotes the larger of two numbers, then Var (X) =?

• A. $\frac {14}{3}$
• B. $\frac {14}{9}$
• C. $\frac {7}{3}$
• D. $\frac {19}{3}$

Answer 1

Answer: (A)

$\frac {14}{3}$

Answer 2

For X = 2, the possible observations are (1, 2) and (2,1),
∴P(X=2)= $\frac {2}{30}$=$\frac {1}{15}$
For X = 3, the possible observations are (1, 3), (3,1), (2,3) and (3, 2).
∴P(X=3)=$\frac {4}{30}$=$\frac {2}{15}$
For X = 4, the possible observations are (1, 4), (4, 1), (2,4), (4,2), (3,4) and (4,3).
∴P(X=4)=$\frac {6}{30}$=$\frac {1}{5}$
For X = 5, the possible observations are (1, 5), (5, 1), (2,5), (5,2), (3,5), (5,3) (5, 4) and (4,5).
∴P(X=5)=$\frac {8}{30}$=$\frac {4}{15}$
For X = 6, the possible 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=6)=10/30=1/3.
Therefore, the required probability distribution is as follows.
Then, E(X)=∑XiP(Xi)
E(X) = $\frac {2\times 1}{15}$+$\frac {3\times 2}{15}$+$\frac {4\times 1}{5}$+$\frac {5\times 4}{15}$+$\frac {6\times 1}{3}$
E(X) = $\frac {2}{15}$+$\frac {6}{15}$+$\frac {4}{5}$+$\frac {20}{15}$+2
E(X) = $\frac {2+6+12+20+30}{15}$
E(X) = $\frac {70}{15}$
E(X) = $\frac {14}{3}$