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# Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers obtained.

Q. 12    Two numbers are selected at random (without replacement) from the first six positive integers. Let $\inline X$ denote the larger of the two numbers obtained. Find $\inline E(X).$

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Two numbers are selected at random (without replacement) from the first six positive integers in $6\times 5=30$ ways.

$\inline X$ denote the larger of the two numbers obtained.

X can be 2,3,4,5,6.

X=2, obsevations : $(1,2),(2,1)$

$P(X=2)=\frac{2}{30}=\frac{1}{15}$

X=3, obsevations : $(1,3),(3,1),(2,3),(3,2)$

$P(X=3)=\frac{4}{30}=\frac{2}{15}$

X=4, obsevations : $(1,4),(4,1),(2,4),(4,2),(3,4),(4,3)$

$P(X=4)=\frac{6}{30}=\frac{3}{15}$

X=5, obsevations : $(1,5),(5,1),(2,5),(5,2),(3,5),(5,3),(4,5),(5,4)$

$P(X=5)=\frac{8}{30}=\frac{4}{15}$

X=6, obsevations : $(1,6),(6,1),(2,6),(6,2),(3,6),(6,3),(4,6),(6,4),(5,6),(6,5)$

$P(X=6)=\frac{10}{30}=\frac{1}{3}$

Probability distribution is as follows:

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

$E(X)=2\times \frac{1}{15}+3\times \frac{2}{15}+4\times \frac{3}{15}+5\times \frac{4}{15}+6\times \frac{1}{3}$

$E(X)= \frac{2}{15}+ \frac{2}{5}+ \frac{4}{5}+ \frac{4}{3}+ \frac{2}{1}$

$E(X)= \frac{70}{15}$

$E(X)= \frac{14}{3}$

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