Q.Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die?

Concept: Conditional Probability — we need $P(\text{die} \in \{1,2,3,4\} \mid \text{exactly one head})$.

Step 1: Define events.

Let $E_1$ = die shows 5 or 6, $E_2$ = die shows 1,2,3,4.

$P(E_1) = \frac{2}{6} = \frac{1}{3}$, $P(E_2) = \frac{4}{6} = \frac{2}{3}$.

Step 2: Probability of exactly one head under each case.

• Under $E_1$: 3 coin tosses. Exactly one head: $\binom{3}{1} \left(\frac{1}{2}\right)^3 = \frac{3}{8}$.
• Under $E_2$: 1 coin toss. Exactly one head: $\frac{1}{2}$.

Step 3: Apply Bayes' theorem.

$P(E_2 \mid \text{1 head}) = \frac{P(E_2) \cdot P(\text{1 head} \mid E_2)}{P(E_1) \cdot P(\text{1 head} \mid E_1) + P(E_2) \cdot P(\text{1 head} \mid E_2)}$ …

Why Bayes’ theorem?

We are asked: Given that exactly one head was obtained, what is the probability that the die showed 1, 2, 3, or 4?

This is a “reverse” conditional probability — we know the result (one head) and want the probability of a particular cause (die in {1,2,3,4}). Bayes’ theorem is the natural tool for such problems.

Step 1: Define the events clearly

Let $E$ be the event that the die shows 1, 2, 3, or 4.

Let $F$ be the event that the die shows 5 or 6.

Let $A$ be the event that exactly one head is obtained.

We want $P(E \mid A)$.

Step 2: Find the prior probabilities

A fair die:

$P(E) = \frac{4}{6} = \frac{2}{3}$

$P(F) = \frac{2}{6} = \frac{1}{3}$

Step 3: Find $P(A \mid E)$ — probability of exactly one head given die showed 1–4

If the die shows 1–4, she tosses a coin once.

Possible outcomes: H or T. Exactly one head means the single toss is H.

So $P(A \mid E) = \frac{1}{2}$.

Step 4: Find $P(A \mid F)$ — probability of exactly one head given die showed 5 or 6

If the die shows 5 or 6, she tosses a coin three times.

Number of heads in 3 tosses follows a binomial distribution with $n=3$, $p=\frac{1}{2}$.

Exactly one head:

$P(A \mid F) = \binom{3}{1} \left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^2 = 3 \cdot \frac{1}{8} = \frac{3}{8}$.

Step 5: Apply Bayes’ theorem

Bayes’ theorem states:

Substitute the values:

Simplify numerator: …

Method: Bayes' theorem for a two-stage ("cause then result") experiment

Use this whenever an experiment happens in two stages — a first random choice decides which second experiment is run — and you observe the final result, then are asked for the probability of a particular first-stage cause.

Steps

Step 1: List the mutually exclusive first-stage "causes" and their prior probabilities.

Partition the first stage into events that cannot overlap and together cover everything (e.g. "die shows 1–4" vs "die shows 5 or 6"), and write down each prior probability.

Step 2: For each cause, find the likelihood of the observed result.

Compute $P(\text{result}\mid \text{cause})$ separately under each branch — the second-stage experiment is different in each, so these likelihoods differ. …

Common Mistakes

Mistake 1: Forgetting the die outcomes split into unequal groups.

Why it's wrong: "$1,2,3,4$" has prior $\tfrac46=\tfrac23$ while "$5$ or $6$" has $\tfrac26=\tfrac13$; treating them as equal wrecks the weighting. Correct approach: use the true priors $\tfrac23$ and $\tfrac13$ in Bayes' theorem.

Mistake 2: Using only one branch in the denominator. …