Twenty identical coins each with probability $p$ of showing heads are tossed. If the probability of heads showing on $10$ coins is same as that of tails showing on $11$ coins, then $p$ is

The probability of heads showing on $10$ coins $=10p$
The probability of tails showing on $11$ coins  $=11(1-p)$
 according to question both are same => $10p=11(1-p)$
                                                            =>      $p=\frac{11}{21}$