Question

A bag contains $(2n+1)$ coins. It is known that $n$ of these have a head on both the sides, where as 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 $31/42$, then value of n is

• A. $10$
• B. $8$
• C. $6$
• D. $25$

Answer 1

Answer: (A)

$10$

$P$(head when fair coin is chosen) $=1/2$
$P$(head when fair coin is not chosen)$=1$
$P$(choosing fair coin)$=n+1/(2n+1)$
$P$(not choosing fair coin)$=n/(2n+1)$
$\frac{n}{2n+1}\ast 1+\frac{n+1}{2n+1}\ast 1/2=31/42$
$\Rightarrow n=10$