Let S = {1, 2, . . . , n} and suppose that A and B are, independently, equally likely to be any of the 2n subsets (including the null set and S itself) of S.

(a) Show that

P{A $\subset$B} =${\left(\begin{array}{c}\frac{3}{4}\end{array}\right)}^n$

Hint: Let N(B) denote the number of elements in B. Use

P{A $\subset$B} =$\sum _{i=0}^n$P{A ($\subset$B|N(B) = i}P{N(B) = i}

Show that P{AB = Ø} =${\left(\begin{array}{c}\frac{3}{4}\end{array}\right)}^n$

Given Information:

$S=\{1,2,......,n\}$

Suppose that $\mathrm{A}$ and $\mathrm{B}$are singly inversely likely to be any of the $2\mathrm{n}$subsets of $\mathrm{S}$.

Computation :

Before procedding with the problem, we will prove a many individualities .

Using Binomial theorem,

$(1+x{)}^n=n_{C_8}+n_{C_1}x+n_{C_2}{x}^2+\dots \dots \dots +n_{C_n}{x}^n$

cover $x=1$,

$n_{C_0}+n_{C_1}+n_{C_2}+{\dots }_{{\dots }_1}+n_{C_n}=2^n-\left(1\right)$

Substitute $x=2,n_{C_0}+2n_{C_1}+2^2{n}_{C_2}+\dots \dots +2^n{n}_{C_n}=3^n-\left(2\right)$

Now,

Let $N\left(B\right)$denote the number of rudiments in B

$P[A\subseteq B]=\sum _{i=0}^{n}P\{A\subseteq B\mid N(B)=i\}P\left\{N\right(B)=i\}$

$=\frac{m_2}{x}$

role="math" localid="1646726529192" $$\begin{gathered} P\left[A\underline{\underline{\subset }}B\right|N(B=I)]=P \\ =\frac{i{c}_0+i{c}_1+...+i_Q}{x} \\ =\frac{z^2}{2} \end{gathered}$$

Thus,

role="math" localid="1646726550497" $$\begin{gathered} P\left\{A\underline{\subset }B\right\}=\sum _{i=0}^n\frac{2^i}{2^n}\times \frac{n_{Ci}}{2^n} \\ =\frac{1}{4^{*}}\sum _{i=0}^n{2}^i\times n_{Ci} \\ =\frac{1}{4^4}\times 3^n \\ ={\left(\frac{3}{4}\right)}^n \end{gathered}$$

Let $N\left(B\right)$denote the number of rudiments in $B$

$P\{AB=\varphi \}=\sum _{i=0}^{n}P\{AB=\varphi N(B)=i\}-P\left\{N\right(B)=i\}$

$P\left\{N\right(B)=i\}=\frac{n_C}{2}$

$P\{AB=\Phi N(B)=i\}=P$

$P\{AB=\varphi \}=\sum _{i=0}^n\frac{2^{n-i}}{2^n}\times \frac{n_{C_C}}{2^n}$

$=\frac{1}{4^{*}}\sum _{i=0}^n{2}^{n-i}{n}_{Ci}$

$=\frac{1}{4^n}\left[n_{C_1}{2}^n+n_{C_1}{2}^{n-1}+\dots \dots +n_{C_0}{2}^0\right]$

$=\frac{1}{4^n}\left[n_{C_{-}}{2}^n+n_{C_{-1}}{2}^{n-1}+\dots \dots +n_{C_0}{2}^0\right]$

$\sin ce,{\mathrm{n}}_{C,}=n_{C_s}$

$=\frac{1}{4^2}\times 3^n$

role="math" localid="1646727165171" $P\{AB=\varphi \}={\left(\frac{3}{4}\right)}^{\pi }$