Q.There are three coins. One is a two headed coin (having head on both faces), another is a biased coin that comes up heads $75\%$ of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin ?

The key idea is Conditional Probability — we update the probability of each coin being chosen given that the toss resulted in heads.

Step 1: Define events and prior probabilities.

Let $C_1$ = two-headed coin, $C_2$ = biased coin (heads 75%), $C_3$ = unbiased coin.

Each coin is equally likely to be chosen:

Step 2: Likelihood of heads given each coin.

Step 3: Apply Bayes' Theorem.

Total probability of heads: …

Why Bayes' theorem works here

We are given three coins with different head-probabilities:

• Coin A (two-headed): $P(H) = 1$
• Coin B (biased): $P(H) = 0.75 = \frac{3}{4}$
• Coin C (unbiased): $P(H) = \frac{1}{2}$

One coin is chosen at random, so each has prior probability $\frac{1}{3}$. Then we toss it and see heads. The question is: given that outcome, what is the chance it was Coin A?

This is a textbook case of inverse probability — we know the effect (heads) and want the cause (which coin). Bayes' theorem reverses the conditional probability:

The denominator $P(H)$ is the total probability of heads, found by summing over all three coins (law of total probability).

Step-by-step solution

1. Write down the prior probabilities.

Each coin is equally likely to be chosen:

2. Write down the likelihoods — probability of heads given each coin.

• For the two-headed coin: $P(H \mid A) = 1$
• For the biased coin (75% heads): $P(H \mid B) = \frac{3}{4}$
• For the unbiased coin: $P(H \mid C) = \frac{1}{2}$

3. Compute the total probability of heads, $P(H)$.

Using the law of total probability:

Substitute:

Factor $\frac{1}{3}$:

Add the fractions inside: $1 = \frac{4}{4}$, $\frac{3}{4}$, $\frac{1}{2} = \frac{2}{4}$. Sum = $\frac{4+3+2}{4} = \frac{9}{4}$.

Thus: …

Method: Bayes' Theorem with several equally likely causes

Use this when one of three (or more) objects is chosen at random and you observe an outcome, then must judge which object it was.

Steps

Step 1: List every candidate and its prior.

With $n$ objects picked at random, each hypothesis $H_i$ has prior $P(H_i) = \tfrac{1}{n}$.

Step 2: Write each object's likelihood of the observed outcome.

$P(E \mid H_i)$ is the chance of the outcome (heads) for that particular object — $1$ for a two-headed coin, its bias for a biased coin, $\tfrac12$ for a fair coin.

Step 3: Total probability of the outcome. …

Common Mistakes

Mistake 1: Using $P(\text{heads}\mid \text{two-headed}) = \tfrac12$.

Why it's wrong: a coin with heads on both faces always shows heads, so this likelihood is $1$. Correct approach: set $P(H\mid C_1)=1$, $P(H\mid C_2)=\tfrac34$, $P(H\mid C_3)=\tfrac12$.

Mistake 2: Forgetting to divide by $P(H)$.

Why it's wrong: the likelihoods alone are not posteriors; they must be normalised by the total chance of heads. Correct approach: $P(H)=\tfrac13(1+\tfrac34+\tfrac12)=\tfrac34$, then divide. …