Question

Three numbers are selected at random(without replacement ) from first six positive integers . if X denotes the smallest of the three numbers obtained ,find the probability distribution of X. also find the mean and variance of the distribution

Answer 1

No. of ways of selecting 3 integers drom a set of 6 integers $=6C_3=\frac{6!}{3!3!}=20$
If 'X' denotes smallest of the 3 members.
The probability distribution of 'X' is
For (x =1) (1,2,3), (1,2,4), (1,2,5), (1,2,6)
(1,3,4), (1,3,5), (1,3,6), (1,4,5)
(1,4,6), (1,5,6)
P(x = 1) $=\frac{10}{20}=\frac{1}{2}=0.5$
For (x = 2)
(2,3,4), (2,3,5), (2,3,6), (2,4,5)
(2,4,6), (2,5,6)
P(x = 2) $=\frac{6}{20}=\frac{3}{10}=0.3$
For (x = 3) (3,4,5), (3,4,6), (3,5,6)
$P(x=3)=\frac{3}{20}=0.15$
For (r = 4) (4, 5, 6)
P(x = 4) = $\frac{1}{20}$ = 0.05
5 & 6 cannot be smallest
P(x = 5) = P(x = 6) = 0
$\begin{array}{ccccccc}X& 1& 2& 3& 4& 5& 6\\ P\left(x\right)& 0.5& 0.3& 0.15& 0.05& 0& 0\end{array}$
$Mean=EXP\left(x\right)=(1\times 0.5)+(2\times 0.3)$
(E(x)) + (3 $\times$ 0.15) + 4(0.05)
+ 5(0) + 6(0)
= 0.5 + 0.6 + 0.45 + 0.2
= 1.1 + 0.65
= 1.75
$E\left(X^2\right)=1^2\left(0.5\right)+2^2\left(0.3\right)+3^2\left(0.15\right)+4^2\left(0.05\right)+5^2\left(0\right)+6^2\left(0\right)$
= 0.5 + 1.2 + 1.35 + 0.8 + 0
= 3.85
VArience = Var(x) = $E\left(x^2\right)-\left(E\right(x){)}^2$
= 3.85 - ${1.75}^2$
= 3.85 - 3.0625
= 0.7875