Q79 / 160 · AP EAMCET 2013 · Mathematics

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

  • Correct answer
    A
    $10$
  • B
    $11$
  • C
    $12$
  • D
    $13$

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. $$ \begin{array}{lc} \therefore & 1-\frac{n+1}{2(2 n+1)}=\frac{31}{42} \\ \Rightarrow & \frac{4 n+2-n-1}{4 n+2}=\frac{31}{42} \\ \Rightarrow & \frac{3 n+1}{4 n+2}=\frac{31}{42} \\ \Rightarrow & 126 n+42=124 n+62 \\ \Rightarrow & 2 n=20 \\ \Rightarrow & n=10 \end{array} $$

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

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