A bag contains (N+1) coins -N fair coins, and one coin with 'Head' on both sides.

Question

A bag contains $(N+1)$ coins -N fair coins, and one coin with 'Head' on both sides. A coin is selected at random and tossed. If the probability of getting 'Head' is $\frac{9}{16}$, then N is equal to:

Accepted Answer

Let ${E}_{1}$ be the event of selecting a fair coin.

Let ${E}_{2}$ be the event of selecting the two-headed coin.

Let $A$ be the event of getting a 'Head' when the selected coin is tossed.

We are given that there are $N$ fair coins and $1$ two-headed coin.

Hence, the total number of coins in the bag is $N+1$.

Since a coin is selected at random, the probability of choosing a fair coin is:

$P({E}_{1})=\frac{N}{N+1}$

The probability of choosing the two-headed coin is:

$P({E}_{2})=\frac{1}{N+1}$

Now, let us find the conditional probabilities of getting a 'Head' for each type of coin:

If a fair coin is selected, the probability of getting a 'Head' is:

$P(A|{E}_{1})=\frac{1}{2}$

If the two-headed coin is selected, the probability of getting a 'Head' is a certainty (since both sides are 'Head'):

$P(A|{E}_{2})=1$

According to the Law of Total Probability, the total probability of getting a 'Head', denoted as $P(A)$, is given by:

$P(A)=P({E}_{1})\cdot P(A|{E}_{1})+P({E}_{2})\cdot P(A|{E}_{2})$

We are given that the probability of getting a 'Head' is $\frac{9}{16}$. Let us substitute the known values into the equation:

$\frac{9}{16}=(\frac{N}{N+1}\cdot \frac{1}{2})+(\frac{1}{N+1}\cdot 1)$

Let us simplify the equation:

$\frac{9}{16}=\frac{N}{2(N+1)}+\frac{1}{N+1}$

To add the terms on the right side, let us take the common denominator as $2(N+1)$:

$\frac{9}{16}=\frac{N+2(1)}{2(N+1)}$

$\frac{9}{16}=\frac{N+2}{2N+2}$

Now, we can solve for $N$ by cross-multiplying:

$9\cdot (2N+2)=16\cdot (N+2)$

$18N+18=16N+32$

Transpose the terms containing $N$ to one side and the constant terms to the other side:

$18N-16N=32-18$

$2N=14$

Divide both sides by $2$:

$N=7$

Therefore, the value of $N$ is equal to $7$.

Option B is the correct answer.

Solutions · 1