Q.A bag contains $(2n + 1)$ coins. It is known that $n$ of these coins have a head on both sides whereas the rest of the coins are fair. A coin is picked up at random from the bag and is tossed. If the probability that the toss results in a head is $\dfrac{31}{42}$, determine the value of $n$.

Concept: Conditional Probability (Law of Total Probability)

The bag has $2n+1$ coins:

• $n$ two-headed coins (probability of head = 1)
• $(n+1)$ fair coins (probability of head = $\frac12$)

Step 1 – Probability of picking each type

$P(\text{two-headed}) = \frac{n}{2n+1}$,

$P(\text{fair}) = \frac{n+1}{2n+1}$.

Step 2 – Apply Law of Total Probability

$P(\text{Head}) = P(\text{two-headed})\cdot 1 + P(\text{fair})\cdot \frac12$

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

Step 3 – Simplify and equate to given probability …

We have a bag with $2n+1$ coins. Of these, $n$ are two-headed (always show heads) and the remaining $(2n+1) - n = n+1$ are fair coins (one head, one tail). A coin is chosen at random and tossed once. The overall probability of getting a head is given as $\frac{31}{42}$. We need to find $n$.

The key idea is conditional probability — the total probability of heads is the weighted average of the head probabilities from each type of coin, where the weights are the probabilities of picking that type.

1. Define the events

Let:

• $A$ = event that the chosen coin is two-headed.
• $B$ = event that the chosen coin is fair.
• $H$ = event that the toss shows a head.

We know:

• $P(A) = \frac{n}{2n+1}$ (since $n$ out of $2n+1$ coins are two-headed).
• $P(B) = \frac{n+1}{2n+1}$ (the rest are fair).
• $P(H \mid A) = 1$ (a two-headed coin always gives heads).
• $P(H \mid B) = \frac{1}{2}$ (a fair coin gives heads half the time).

2. Apply the law of total probability

The total probability of heads is:

Substitute the values:

3. Simplify the expression

Combine the terms over the common denominator $2n+1$:

Write the first term with denominator $2(2n+1)$:

4. Set equal to the given probability …

Method: Law of total probability to build an equation in an unknown

Use this when a mixture (some biased coins, some fair) has an overall event probability given, and you must solve for an unknown count.

Steps

Step 1: Express each type's count and probability in terms of the unknown

If there are $2n+1$ coins and $n$ are two-headed, the rest are $n+1$ fair coins. Selection probabilities are $\tfrac{n}{2n+1}$ and $\tfrac{n+1}{2n+1}$.

Step 2: Total probability of the event

$P(H)=P(\text{two-headed})\cdot 1+P(\text{fair})\cdot\tfrac12.$ …

Common Mistakes

Mistake 1: Taking the number of fair coins as $n$

Why it's wrong: with $2n+1$ coins total and $n$ two-headed, the fair coins number $(2n+1)-n=n+1$. Correct approach: subtract to get $n+1$ fair coins.

Mistake 2: Wrong head probability for a coin type

Why it's wrong: a two-headed coin gives a head with probability $1$ (not $\tfrac12$), and only the fair coin gives $\tfrac12$. Correct approach: assign $1$ and $\tfrac12$ to the right types. …