Q68 / 180 · KCET 2023 · Mathematics

A bag contains $2 n+1$ coins. It is known that $n$ of these coins have head on both sides whereas, the other $n+1$ coins are fair. One coin is selected at random and tossed. If the probability that toss results in heads is $\frac{31}{42}$, then the value of $n$ is

  • A
    $6$
  • B
    $8$
  • Correct answer
    C
    $10$
  • D
    $5$

Solution

Given, $n$ coins have head on both the sides and $(n+1)$ coins are fair coins. Total coins $=2 n+1$ Let events $E_1, E_2$ be the following $E_1=$ Event that an unfair coin is selected $E_2=$ Event that a fair coin is selected $\therefore \quad P\left(E_1\right)=\frac{n}{2 n+1}$ and $P\left(E_2\right)=\frac{n+1}{2 n+1}$ From the law of total probability, $\therefore \quad P(E)=P\left(E_1\right) \times P\left(\frac{E}{E_1}\right)+P\left(E_2\right) \times P\left(\frac{E}{E_2}\right)$ $\begin{array}{ll}\Rightarrow \quad & \frac{31}{42}=\frac{n}{2 n+1} \times 1+\frac{n+1}{2 n+1} \times \frac{1}{2} \\ \Rightarrow \quad & \frac{31}{42}=\frac{2 n+n+1}{2(2 n+1)} \\ \Rightarrow & 31 \times 2(2 n+1)=42(3 n+1) \\ \Rightarrow & 124 n+62=126 n+42 \\ \Rightarrow & 2 n=20 \\ & n=10\end{array}$

Hence, the correct answer is option: 'C'.

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