Question

Let X = {1,2,3,…,50}. A subset A of X is chosen at random. The set X reconstructed by replacing the elements of A, and another set B of X is chosen at random. The probability that A B contains exactly 5 elements is

• A. $50C_5\frac{3^{45}}{4^{50}}$
• B. $50C_5{\left(\frac{4}{5}\right)}^{45}$
• C. $50C_5{\left(\frac{1}{2}\right)}^{50}$
• D. $50C_5{\left(\frac{3}{4}\right)}^{45}$

Answer 1

The correct option is A $50C_5\frac{3^{45}}{4^{50}}$

5 elements can be selected from the set $A\bigcap B$ in $50C_5$ ways. Let a be an element from remaining 45 elements. Then we have following cases.
$aϵA;aϵB\left(i\right)$
$aϵA;a\notin B\left(ii\right)$
$aϵA;aϵB\left(iii\right)$
$a\notin A;a\notin B\left(iv\right)$
(ii), (iii), (iv) are favorable since $a\notin A\bigcap B.$ thus 3 cases are favorable for each of the reamaining 45 elements. Hence the number of favorable ways $=50C_5{3}^{45}$
Also, the total number of ways. i.e. ways to form subset A and B $=2^{50}{2}^{50}$
Hence required probability $=\frac{50C_5{3}^{45}}{4^{50}}$