Question

A set $A$ has $22$ elements a subset $P$ of $A$ is selected at random. After replacing the elements, again a subset $Q$ of $A$ is selected. The probability that $P$ and $Q$ have exactly $5$ elements in common is

• A. ${\left(\frac{3}{4}\right)}^{22}$
• B. ${{}^{22}C}_5\cdot \left(\frac{3^{17}}{4^{22}}\right)$
• C. ${{}^{22}C}_{17}\cdot \frac{3^{17}}{4^{22}}$
• D. ${{}^{22}C}_5\cdot \frac{3^5}{2^{44}}$

Answer 1

The correct option is D ${{}^{22}C}_5\cdot \frac{3^5}{2^{44}}$

Since, set A contains 22 elements.

So, it has $2^{22}$ subsets.

Set P can be chosen in $2^{22}$ ways,

similarly set Q can be chosen $2^{22}$ ways.

$P$ and $Q$can be chosen in $\left(2^{22}\right)\left(2^{22}\right)=4^{22}$ ways.

Suppose, P contains 5 elements.

Then, P can be chosen in ${}^{22}{C}_5$ ways,for 0 to be disjoint from A,

it should be chosen from the set of ali subsets of set consisting of remaining $(22-r)$ elements.

This can be done in $2^{22-r}$ ways.

$P$ and $Q$ can be chosen in ${}^{22}{C}_5.2^{22-r}$ ways.

But, P can vary from 0 to 5.

Total number of disjoint sets $P$ and $Q$

=${\sum }_{r=0}^2{2}^{2}2C_5{.2}^{22-17}{=}^{22}{C}_5(1+2{)}^5{=}^{22}{C}_5{.3}^5$

Hence required probability=${}^{22}{C}_5.\frac{3^5}{2^{44}}$