Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Probability

Question:

A is a set containing n elements. A subset P of A is chosen at random. The set A is reconstructed by replacing the elements of P. A subset Q is again chosen at random. The probability that P and Q have equal number of elements, is

Options:

$\left(\frac{3}{4}\right)^n$

$\frac{^{2n}C_n}{4^n}$

$\frac{^{2n}C_{n-1}}{4^n}$

$\frac{n^2}{4^n}$

Correct Answer:

$\frac{^{2n}C_n}{4^n}$

Explanation:

The set A has n elements. So, it has $2^n$ subsets.

Therefore, set P can be chosen in ${^{2n}C}_1$ ways. Similarly, set Q can also be chosen in ${^{2n}C}_1$ ways.

∴ Sets P and Q can be chosen in ${^{2n}C}_1 × {^{2n}C}_1= 2^n  ×2^n = 4^n $ ways.

Let the subset P of A contains r elements, with ) ≤ r < n. Then, the number of ways of choosing P is ${^nC}_r$.

Similarly, Q can be chosen in ${^nC}_r$ ways.

So, P and Q can be chosen in ${^nC}_r  × {^nC}_r $ ways. But, r can vary from 0 to n.

∴ P and Q can be chosen in $\sum\limits^{n}_{r=0} {^nC}_r  × {^nC}_r $ ways, = ${^{2n}C}_n$ ways.

∴ Required probability $=\frac{^{2n}C_n}{4^n}$