A bag contains $2 n+1$ coins. It is known that $n$ of these coins have a head on both sides, whereas the…
A bag contains $2 n+1$ coins. It is known that $n$ of these coins have a head on both sides, whereas the remaining $n+1$ coins are fair. A coin is picked up at random from the bag and tossed. If the probability that the toss results in a head is $\frac{31}{42}$, then $n$ is equal to
$10$
$11$
$12$
$13$
Solution
The probability that the toss results is a tail $=\frac{(n+1)}{2(2 n+1)}$
$\therefore 1-\frac{(n+1)}{2(2 n+1)}$ is the probability that the toss result is a head.
$
\begin{array}{lc}
\therefore & 1-\frac{n+1}{2(2 n+1)}=\frac{31}{42} \\
\Rightarrow & \frac{4 n+2-n-1}{4 n+2}=\frac{31}{42} \\
\Rightarrow & \frac{3 n+1}{4 n+2}=\frac{31}{42} \\
\Rightarrow & 126 n+42=124 n+62 \\
\Rightarrow & 2 n=20 \\
\Rightarrow & n=10
\end{array}
$