Expected order of an element in $A_5$

Question

Suppose we pick every element from the Alternating group $A_5$ equally likely. What would be the expected order of the element?

My answer is $\frac{241}{60}$. This is because, we have $1$ element of order $1$(the identity), $15$ elements of order $2$(the product of two disjoint transpositions),$10$ elements each of order $3$ and $6$ (the three element cycles and products of three element cycles with a disjoint transposition), and $24$ elements of order $5$(the five cycles). So, using the definition of expectation, I should get the expected value as $\frac{1}{60}+\frac{15\times2}{60}+\frac{10\times3}{60}+\frac{10\times6}{60}+\frac{24\times5}{60}=\frac{241}{60}$.

Is the calculation right? Because, I saw the answer key to the problem saying that the answer is $\frac{211}{60}$. So, where am I wrong (or is the key wrong?) Thanks beforehand.

Answer 1

With the corrections mentioned in the comments we indeed obtain the "desired result": $$\frac{1\times 1}{60}+\frac{15\times2}{60}+\frac{20\times3}{60}+\frac{24\times5}{60}=\frac{211}{60}$$ I would like to add that one can determine the order of elements in the alternating groups as follows:

Orders of the elements in the alternating group $A_n$

Answer 2

As shown by the comments of @DietrichBurde and @AnneBauval, the number of elements of order $3$ is $20$(taking permutation among the three cycles into account) and the product of three cycles with disjoint transposition is not an even permutation. Therefore, the revised calculations would be the expected order$=\frac{1\times1}{60}+\frac{15\times2}{60}+\frac{20\times3}{60}+\frac{24\times5}{60}=\frac{211}{60}$ as desired by the answer key.