Question

A subset $A$ of $X=$ {1, 2, 3, ...., 100} is chosen at random. The set $X$ is reconstructed by replacing the elements of $A$ and another subset $B$ of $X$ is chosen at random. The probability that $A\cap B$ contains exactly 10 elements is

• A. $\frac{1}{10}$
• B. ${}^{100}{C}_{10}(\frac{1}{2}{)}^{100}$
• C. ${}^{100}{C}_{10}(\frac{1}{4}{)}^{100}$
• D. ${}^{100}{C}_{10}\frac{3^{90}}{4^{100}}$

Answer 1

The correct option is D ${}^{100}{C}_{10}\frac{3^{90}}{4^{100}}$

The $10$ common elements may be selected from the $100$ elements in ${}_{10}^{100}C$ ways.
For the remaining $90$ elements, there are $3$ possible cases:
i) Belongs to neither $A$ nor $B$
ii) Belongs to only $A$
iii) Belongs to only $B$
So, total cases = $3^{90}$
Hence, favorable cases = ${}_{10}^{100}C\ast 3^{90}$
Total number of ways of forming subsets $A$ and $B$ = $2^{100}\ast 2^{100}=4^{100}$
Therefore, required probability = $\frac{{}_{10}^{100}C\ast 3^{90}}{4^{100}}$
Hence, (d) is correct.