Question

The numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers obtained. Find the probability distribution of the random variable X and hence find the means of the distribution.

Answer 1

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

First six numbers are $1,2,3,4,5,6.$

X is bigger number among two numbers.

If X $=2$ for P(X) $=$ probability of event that bigger of the two chosen number is $2.$

So, Cases $=(1,2)$

So, $P\left(X\right)=\frac{1}{{}^6{C}_2}=\frac{1}{15}$---(1)

If X $=3$

So, favourable cases are $=(1,3),(2,3)$

$P\left(X\right)=\frac{2}{{}^6{C}_2}=\frac{2}{15}$---(2)

If X $=4$

So, favourable cases are $=(1,4),(2,4),(3,4)$

$P\left(X\right)=\frac{3}{{}^6{C}_2}=\frac{3}{15}$---(3)

If X $=5$

So, favourable cases are $=(1,5),(2,5),(3,5),(4,5)$

$P\left(X\right)=\frac{4}{{}^6{C}_2}=\frac{4}{15}$---(4)

If X $=6$

So, favourable cases are $=(1,6),(2,6),(3,6),(4,6),(5,6)$

$P\left(X\right)=\frac{5}{{}^6{C}_2}=\frac{5}{15}$---(5)

We can put all value of $P\left(X\right),$

Required mean = $2.\left(\frac{1}{15}\right)+3.\left(\frac{2}{15}\right)+4.\left(\frac{3}{15}\right)+5.\left(\frac{4}{15}\right)+6.\left(\frac{5}{15}\right)$

$=\frac{70}{15}\Rightarrow \frac{14}{3}$