A bag contains $n$ red and $n$ white balls. Two balls at a time are drawn at random from the bag till all the ball are drawn. Find the probability that in each draw there is one white and one red ball.
-
$\displaystyle \dfrac{(2^{n}){n!}}{^{2n}C_{n}}$
-
$\displaystyle \dfrac{4^{n}}{^{2n}C_{n}}$
-
$\displaystyle \dfrac{2^{n}}{^{2n}C_{n}}$
-
$\displaystyle \dfrac{(2^{n})/n!}{^{2n}C_{n}}$
C
Correct answer
Explanation
The total number of ways to draw 2n balls in pairs is (2n)! / (2^n). The number of ways to draw one red and one white in each of n draws is n! * n!. The probability is (n! * n!) / ( (2n)! / 2^n ) = (2^n * n! * n!) / (2n)!. This simplifies to 2^n / (2nCn).