Twenty identical coins with probability 'p' of showing heads, are tossed if the probability of heads on 10 coins is the same as that heads showing on 11 coins, then p= 

Probability of heads showing on 10 coins =10p

Probability of tails showing on 11 coins =11q

                                                                       =11(1-p)

Given that 10p=11(1-p)

     $$\begin{gathered} \Rightarrow 10p=11-11p \\ \Rightarrow 21p=11 \\ \Rightarrow p=\frac{11}{21} \end{gathered}$$