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 to A. 10 B. 11 C. 12 D. 13
ANSWER
Verified
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.
Complete step-by-step answer: 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.