Question

A subset $A$ of the set $X=1,2,.....100$ 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. ${{}^{100}C}_{10}{\left(\frac{3}{4}\right)}^{90}$
• B. ${{}^{100}C}_{10}{\left(\frac{1}{1}\right)}^{100}$
• C. ${{}^{100}C}_{10}\frac{3^{90}}{4^{100}}$
• D. none of these

Answer 1

Answer: (C)

${{}^{100}C}_{10}\frac{3^{90}}{4^{100}}$

The number of ways in which 10 elements will be common:
$C_{10}^{100}\ast 3^{90}$
{$C_{10}^{100}$ for selecting the 10 common elements and $3^{90}$ for the other 90 elements which can be either in A or in B or in neither}.
Hence, probability = $\frac{C_{10}^{100}\ast 3^{90}}{4^{100}}$
{$4^{100}$ is used since each elements has 4 options: be either in A or in B or in both or in neither}
Hence, (C) is correct.