Question

For each element in a set of size 2n, an unbiased coin is tossed. The 2n coin tossed are independent. An element is chosen if the corresponding coin toss were head. The probability that exactly n elements are chosen is

• A. $\frac{\left(2n\right)!}{2^n}$
• B. 3${}^n$
• C. $\frac{\left(\begin{array}{c}2n\\ n\end{array}\right)}{4^n}$
• D. $\left(\begin{array}{c}2n\\ n\end{array}\right)$

Answer 1

The correct option is C $\frac{\left(\begin{array}{c}2n\\ n\end{array}\right)}{4^n}$

The probability that exactly n elements are chosen = The probability of getting n heads out of 2n tosses
${=}^{2n}{C}_n\left(1/2{)}^n\right(1/2{)}^{2n-n}$
(Binominal formula)
${=}^{2n}{C}_n\left(1/2{)}^n\right(1/2{)}^n$
$=\frac{{}^{2n}{C}_n}{2^{2n}}=$ $\frac{{}^{2n}{C}_n}{(2^2{)}^n}=$ $\frac{{}^{2n}{C}_n}{4^n}$