Question

Two numbers are selected at random (without placement),from the first six positive integers.Let \(X\) denotes the larger of two numbers obtained.Find \(E(X).\)

Answer 1

The correct answer is: \(\frac{14}{3}\)
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).
\(\therefore P(X=2)=\frac{2}{30}=\frac{1}{15}\)
For X = 3, the possible observations are (1, 3), (2, 3), (3, 1), and (3, 2).
\(\therefore P(X=3)=\frac{4}{30}=\frac{2}{15}\)
For X = 4, the possible observations are (1, 4), (2, 4), (3, 4), (4, 3), (4, 2), and (4, 1).
\(\therefore P(X=4)=\frac{6}{30}=\frac{1}{5}\)
For X = 5, the possible observations are (1, 5), (2, 5), (3, 5), (4, 5), (5, 4), (5, 3), (5, 2), and (5, 1).
\(\therefore P(X=5)=\frac{8}{30}=\frac{4}{15}\)
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).
\(\therefore P(X=6)=\frac{10}{30}=\frac{1}{3}\)
Therefore, the required probability distribution is as follows.

X	2	3	4	5	6
P(X)	\(\frac{1}{15}\)	\(\frac{2}{15}\)	\(\frac{1}{5}\)	\(\frac{4}{15}\)	\(\frac{1}{3}\)

Then,\(E(X)=\sum X_iP(X_i)\)
\(=2.\frac{1}{15}+3.\frac{2}{15}+4.\frac{1}{5}+5.\frac{4}{15}+6.\frac{1}{3}\)
\(=\frac{2}{15}+\frac{2}{5}+\frac{4}{5}+\frac{4}{3}+2\)
\(=\frac{70}{15}\)
\(=\frac{14}{3}\)