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

The Correct Option is A

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)=∑X_iP(X_i)
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) = \(\frac {14}{3}\)