Question

A bag contains $2n+1$ 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:

• A. $10$
• B. $11$
• C. $12$
• D. $13$

Answer 1

The correct option is A

$10$

Explanation for the correct option:
Finding the value of $n$.

Let $S$ be the event of tossing all the coins.

$\Rightarrow n\left(S\right)=2n+1$

Let $A$ be the event of obtaining tail tossing the fair coins.

$\Rightarrow n\left(A\right)=\frac{1}{2}\left(n+1\right)$

$\begin{array}{rcl}& \Rightarrow & P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}\\ & =& \frac{{\displaystyle \frac{n+1}{2}}}{2n+1}\\ & =& \frac{n+1}{2\left(2n+1\right)}\end{array}$

Let $B$ be the event of obtaining head tossing the fair coins.

$\begin{array}{rcl}P\left(B\right)& =& 1-P\left(A\right)\\ & =& 1-\frac{n+1}{2(2n+1)}\\ & =& \frac{4n+2-n-1}{4n+2}\\ & =& \frac{3n+1}{4n+2}\end{array}$

From the given information,

$$\begin{gathered} \frac{3n+1}{4n+2}=\frac{31}{42} \\ \Rightarrow 126n+42=124n+62 \\ \Rightarrow 2n=20 \\ \Rightarrow n=10 \end{gathered}$$

Hence, the correct answer is option A.