QUESTION

# A bag contains $(2n+1)$ coins, it is known that n of these coins have heads on both sides whereas the remaining $(n+1)$ coins are fair. A coin is selected at random from the bag and tossed once. If the probability that the toss results in a head is $\dfrac{31}{42}$, then $n$ is equal toA. 10B. 11C. 12D. 13

Hint: Here, we are considering the conditional probability and for that choose two events ${{E}_{1}}$ and ${{E}_{2}}$ and apply the total probability theorem $P(A)=P\left( {{E}_{1}} \right)P\left( A|{{E}_{1}} \right)+P\left( {{E}_{2}} \right)P\left( A|{{E}_{2}} \right)$. Since, $P(A)$ is given as $\dfrac{31}{42}$ and from the given data you can calculate all other values and then find the value of n.

Here, we are given that the bag contains $(2n+1)$ coins where n of these coins have head on both sides and the remaining $(n+1)$ coins are fair. A coin is selected at random from the bag and tossed once.
We are also given that the probability that a toss results in a head is $\dfrac{31}{42}$.
Now, we have to find the value of n.
Let ${{E}_{1}}$ be the event that the coin is chosen from the fair $(n+1)$ coins.
Let ${{E}_{2}}$ be the event that the coin is chosen from biased n coins.
Let A be the event that the tossing results in head.
Here, the total number of coins is $(2n+1)$.
By the theorem of total probability we have the formula:
$P(A)=P\left( {{E}_{1}} \right)P\left( A|{{E}_{1}} \right)+P\left( {{E}_{2}} \right)P\left( A|{{E}_{2}} \right)+......+P\left( {{E}_{n}} \right)P\left( A|{{E}_{n}} \right)$
Here, there are two events, so we can write:
$P(A)=P\left( {{E}_{1}} \right)P\left( A|{{E}_{1}} \right)+P\left( {{E}_{2}} \right)P\left( A|{{E}_{2}} \right)$ ….. (1)
Now, consider
$P(A)$ = Probability (getting a head)
$P(A)=\dfrac{31}{42}$
$P\left( A|{{E}_{1}} \right)$ = Probability (getting a head from the fair $(n+1)$ coins)
$P\left( A|{{E}_{1}} \right)=\dfrac{{}^{n+1}{{C}_{1}}}{{}^{2n+1}{{C}_{1}}}$
$P\left( {{E}_{1}} \right)$ = Probability (choosing the coin from fair $(n+1)$ coins)
$P\left( {{E}_{1}} \right)=\dfrac{1}{2}$
$P\left( A|{{E}_{2}} \right)$ = Probability (getting a head from the biased n coins)
$P\left( A|{{E}_{2}} \right)=\dfrac{{}^{n}{{C}_{1}}}{{}^{2n+1}{{C}_{1}}}$
$P\left( {{E}_{2}} \right)$ = Probability (choosing the coin from biased n coins)
$P\left( {{E}_{2}} \right)=1$
Hence, by substituting all the above values in equation (1) we get:
$P(A)=\dfrac{{}^{n+1}{{C}_{1}}}{{}^{2n+1}{{C}_{1}}}\times \dfrac{1}{2}+\dfrac{{}^{n}{{C}_{1}}}{{}^{2n+1}{{C}_{1}}}\times 1$ …… (2)
We know that,${}^{n}{{C}_{1}}=n$
Hence, we can say that:
${}^{n+1}{{C}_{1}}=n+1$
${}^{2n+1}{{C}_{1}}=2n+1$
Therefore, by applying this values in equation (2) we obtain:
$P(A)=\dfrac{n+1}{2n+1}\times \dfrac{1}{2}+\dfrac{n}{2n+1}\times 1$
Next, by taking LCM we obtain:
\begin{align} & P(A)=\dfrac{n+1+2n}{2(2n+1)} \\ & P(A)=\dfrac{3n+1}{2(2n+1)} \\ \end{align}
Since, $P(A)=\dfrac{31}{42}$, we have:
$\dfrac{31}{42}=\dfrac{3n+1}{2(2n+1)}$
Next, by cross multiplication we get:
$\dfrac{31\times 2}{42}=\dfrac{(3n+1)}{(2n+1)}$
By cancellation we get:
$\dfrac{31}{21}=\dfrac{(3n+1)}{(2n+1)}$
Now, again by cross multiplication we obtain:
\begin{align} & 31\times (2n+1)=21\times (3n+1) \\ & 31\times 2n+31\times 1=21\times 3n+21\times 1 \\ & 62n+31=63n+21 \\ \end{align}
In the next step, take constants to one side and the variables to the other side. i.e. 21 from left side goes to right side and becomes -21 and 62n from left side goes to right side and becomes -62n.
\begin{align} & 31-21=63n-62n \\ & 10=n \\ \end{align}
Hence, we can say that the value of $n=10$.
Therefore, the correct answer for this question is option (a).

Note: Here, $P\left( {{E}_{1}} \right)=\dfrac{1}{2}$, since, it is the probability of a getting a head from fair $(n+1)$coins. It is because the probability of getting a head is $\dfrac{1}{2}$. Similarly, $P\left( {{E}_{2}} \right)=1$, since we are choosing it from biased n coins with total probability 1. It is only biased towards the head so there is no point in getting a tail.