Question

A bag contains $2n$ coins out of which $n-1$ are unfair with heads on both sides and the remaining are fair. One coin is picked from the bag at random and tossed. If the probability that head falls in the toss is $\frac{41}{56}$, then the number of unfair coins in the bag is

• A. 18
• B. 15
• C. 13
• D. 14

Answer 1

Answer: (C)

13

Answer 2

According to given informations,
$\frac{(n-1) \times 1+(n+1) \frac{1}{2}}{2 n}=\frac{41}{56}$
$\Rightarrow \frac{2 n-2+n+1}{4 n} =\frac{41}{56}$
$\Rightarrow (3 n-1) 14=4 1 n$
$\Rightarrow 42 n-14=41 n$
$\Rightarrow n=14$
$\therefore$ Number of unfair coins in the bag is $(n-1)=13$.