Probability Question 378
Question: Let X be a set containing n elements. If two subsets A and B of X are picked at random, the probability that A and B have the same number of elements, is
Options:
A) $ \frac{^{2n}C_{n}}{2^{2n}} $
B) $ \frac{1}{^{2n}C_{n}} $
C) $ \frac{1\cdot 3\cdot 5…(2n+1)}{2^{n}n!} $
D) $ \frac{3^{n}}{4^{n}} $
Show Answer
Answer:
Correct Answer: A
Solution:
Required probability $ =\frac{\sum\limits_{r=0}^{n}{^{n}C_{r}{{\times }^{n}}C_{r}}}{2^{n}\times 2^{n}} $
$ =\frac{C_0^{2}+C_1^{2}+C_2^{2}+…+C_n^{2}}{4^{n}}=\frac{^{2n}C_{n}}{2^{2n}} $
BETA
