Calculating mixed strategy equilibrium in battle of sexes 0
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In the following game, the mixed strategy equilibrium is found as player 2 playing B 3 out of 5 times and A 2 out of 5 times but I am not able to get it. Please find the mistake in the following calculation: 
                         Player 2 
                    A                      B

         A     3,2                    1,1 


Player 1

         B     0,0                    2,3  


Payoff for player 1 if he plays A: 3q+1(1-q)  assuming player 2 plays A with probability q. And if he plays B payoff is 0q + 2(1-q)
For an equilibrium both payoffs should equal so 3q+1(1-q)=2(1-q) which gives q=1/4 i.e. probability of player 2 playing A which is not correct.

Thanks in advance.
Your calculations are correct. If we take this game: 

\begin{array}{| r | r |}
  \hline 
\hline                       
  &amp; A&amp; B\\
  \hline                      
A&amp;3,2&amp;1,1\\
B&amp;0,0&amp;2,3\\
  \hline 
\end{array}

then the mixed strategy probability used by the column player must make the row player indifferent: $3q+1(1-q)=0q+2(q-1)$, implying $q=1/4$.



From your question, I infer that you have a solution claiming that the answer should be $q=2/5$. In that case, it looks to me like there has been a typo somewhere. Indeed, if we take the following game:

\begin{array}{| r | r |}
  \hline 
\hline                       
  &amp; A&amp; B\\
  \hline                      
A&amp;3,2&amp;0,0\\
B&amp;0,0&amp;2,3\\
  \hline 
\end{array}

We find the mixed strategy equilibrium implies the column player using probability $q$, with $3q=2(1-q)$ so $q=2/5$.

Note that this second, modified game is symmetric (as textbook battle of the sexes games usually are), further strengthening my suspicion that the confusion has been caused by a typo in the description of the game.