For each element in a set of size 2n, an unbiased coin is tossed. All 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. \[\frac{{\left( {\begin{array}{*{20}{c}} {2{\text{n}}} \\ {\text{n}} \end{array}} \right)}}{{{4^{\text{n}}}}}\] B. \[\frac{{\left( {\begin{array}{*{20}{c}} {2{\text{n}}} \\ {\text{n}} \end{array}} \right)}}{{{2^{\text{n}}}}}\] C. \[\frac{1}{{\left( {\begin{array}{*{20}{c}} {2{\text{n}}} \\ {\text{n}} \end{array}} \right)}}\] D. $$\frac{1}{2}$$

” option2=”\[\frac{{\left( {\begin{array}{*{20}{c}} {2{\text{n}}} \\ {\text{n}} \end{array}} \right)}}{{{2^{\text{n}}}}}\]” option3=”\[\frac{1}{{\left( {\begin{array}{*{20}{c}} {2{\text{n}}} \\ {\text{n}} \end{array}} \right)}}\]” option4=”$$\frac{1}{2}$$” correct=”option1″]

The correct answer is $\boxed{\frac{{\left( {\begin{array}{*{20}{c}} {2{\text{n}}} \ {\text{n}} \end{array}} \right)}}{{{2^{\text{n}}}}}}$.

To see why, let’s consider the following example. Suppose we have a set of 4 elements, $A$, $B$, $C$, and $D$, and we toss a coin for each element. There are $2^4 = 16$ possible outcomes, corresponding to all the possible combinations of heads and tails. Of these outcomes, only 4 of them (HHHT, HTHH, HTH, and THHH) will result in exactly 2 heads being chosen.

In general, for a set of $2n$ elements, there are $2^{2n}$ possible outcomes. Of these, there are $\binom{2n}{n}$ outcomes that will result in exactly $n$ heads being chosen. Therefore, the probability of exactly $n$ heads being chosen is $\frac{\binom{2n}{n}}{2^{2n}}$.

We can also calculate this probability using a more formal argument. Let $X$ be the number

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