Question

A bag contain $(n+1)$ coins. It is known that one of these has a head on both sides whereas the other coins are normal. One of these coins is selected at random & tossed. If the probability that the toss results in head is $\frac{7}{12}$, then the value of $n$ is

• A. $5$
• B. $6$
• C. $4$
• D. $3$

Answer 1

Answer: (A)

$5$

The total number of ways one coin can be picked up from the bag of $(n+1)$ coins = ${}^{(}n+1)C_1$ = $(n+1)$

In the bag there are 1 coins with both sides have heads and therefore the number of fair coins = $n+1-1$

Fair coins = n

Let A be the event that the unfair coins are tossed and that gives a head.

Number of ways one can pick an unfair coin is ${}^1{C}_1=1$ so, the probability of getting an unfair coin is $\frac{1}{(n+1)}$.

Since the unfair coin tossed will always give a head

Therefore,

$P\left(A\right)=1\times \frac{1}{(n+1)}$

Now, let us consider in the same way the event B of getting a head from a fair coin.

In this case the probability of getting a fair coin is $\frac{n}{(n+1)}$

However, in the fair coin the probability of getting a head is $\frac{1}{2}$.

So, $P\left(B\right)=\frac{1}{2}\times \frac{\left(n\right)}{(n+1)}$

Now, events A and B are mutually exclusive and the probability of getting a head from either of the event is

$P(A\cup B)=P\left(A\right)+P\left(B\right)=\frac{7}{12}$

Therefore,

$\frac{1}{(n+1)}\times 1+\frac{n}{2(n+1)}=\frac{7}{12}$

$\frac{2+n}{2(n+1)}=\frac{7}{12}$

$24+12n=14n+14⟹2n=10⟹n=5$