Let S be a set of 5 elements and P(S) denote the power set of S .

Question

Let S be a set of 5 elements and $P(S)$ denote the power set of S . Let E be an event of choosing an ordered pair (A, B) from the set $P(S)	imes P(S)$ such that $A\cap B=\emptyset$. If the probability of the event $E$ is $\frac{{3}^{p}}{{2}^{q}}$, where $p,q\in N$, then $p+q$ is equal to

Answer: 15

Accepted Answer

$\begin{gathered} S={{a}_{1},{a}_{2},{a}_{3},{a}_{4},{a}_{5}} \ n(S)=5 \ P(S)=S \ nP(S)={2}^{n(S)}={2}^{5}=32 \ n(T)=n[P(S)	imes P(S)]=32	imes 32\, \end{gathered}$

$E=$ event that choosing ordered pair ( $A,B$ ) from $P(S)	imes P(S)$ such that $A\cap B=\phi $

for each element $x\in S,n(S)=5$

there are 4 possibilities such that $A\cap B=\phi $

1.$x\in A$ and $x
otin B$

2.$x
otin A$ and $x\in B$

3.$x
otin A$ and $x
otin B$

4.$x\in A$ and $x\in B$

for $A\cap B=\phi $, exclude the fourth.

so, for each of 5 elements of $S$ there are 3 choices.

The number of favourable outcome $n(E)=3	imes 3	imes 3	imes 3	imes 3$

$P(E)=\frac{}{}=\frac{{3}^{5}}{32	imes 32}=\frac{{3}^{5}}{{2}^{10}}$

It is given that $P(E)=\frac{{3}^{p}}{{2}^{q}}$

compare $p=5,q=10\implies p+q=15$ Ans

Solutions · 1