Question

Find the probability that
$P_i\cap P_j=\varphi fori\ne jandi,j=1,2,....m$

• A. $P\left(E_1\right)=\frac{{\left(m\right)}^n}{2^{mn+1}}$
• B. $P\left(E_1\right)=\frac{{(m+1)}^n}{2^{mn+1}}$
• C. $P\left(E_1\right)=\frac{{\left(m\right)}^n}{2^{mn}}$
• D. $P\left(E_1\right)=\frac{{(m+1)}^n}{2^{mn}}$

Answer 1

The correct option is D $P\left(E_1\right)=\frac{{(m+1)}^n}{2^{mn}}$

Let $A=\{a_1,a_2,...a_n\}$
Let $S$ be the sample space and $E_1$ be the event that $P_i\cap P_j=\varphi$ for $i\ne j$ and $E_2$ be the event that $P_1\cap P_2\cap ...\cap P_m=\varphi$

Therefore numbers of subsets of $A=2^n$.

Therefore each $P_1,P_2,...P_m$ can be selected in $2^n$ ways.

$\therefore n\left(S\right)=$ total number of selection of $P_1,P_2,...P_m={\left(2^n\right)}^m=2^{nm}$

When $P_i\cap P_j=\varphi$ for $i\ne j$, each element of A either does not belong to any of the subsets, or it belong to at most one of them.

Therefore there are $m+1$ choices for each of the $n$ elements
$\therefore n\left(E_1\right)={(m+1)}^n$

Therefore required probability
$P\left(E_1\right)=\frac{n\left(E_1\right)}{n\left(S\right)}=\frac{{(m+1)}^n}{2^{nm}}$