Q.Consider the experiment of throwing a die, if a multiple of $3$ comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event 'the coin shows a tail', given that 'at least one die shows a $3$'.

Let $A$ = 'at least one die shows a $3$' and $B$ = 'the coin shows a tail'. We want $P(B\mid A)=\dfrac{P(A\cap B)}{P(A)}$.

The experiment tosses a coin only when the first die is $1,2,4,$ or $5$ (a non-multiple of $3$). A die can show a $3$ only in the branch where the die is thrown again, and those outcomes $\{(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(6,3)\}$ involve no coin at all. …

Understand the two-stage experiment

Throw a die once.

• If it is a multiple of $3$ (a $3$ or a $6$), throw the die again.
• Otherwise (a $1,2,4,$ or $5$), toss a coin.

So every outcome is either die--die or die--coin, never both.

Outcomes that contain a 3

A die can show a $3$ only in the die--die branch. Those outcomes are

Every one of these came from throwing the die twice — no coin was tossed in any of them.

Outcomes with a tail

A coin (and hence a tail) appears only in the die--coin branch: …

Method: Conditional probability in a multi-stage (tree) experiment

Use this when the experiment branches — the outcome of a first action decides what the second action is (throw again, or toss a coin) — and you must condition one branch-dependent event on another.

Steps

Step 1: Build the sample space as a tree.

Draw the first stage, then from each first-stage result draw the second action it triggers. Here a multiple of $3$ ($3$ or $6$) leads to a second die throw, while $1,2,4,5$ lead to a coin toss. Each complete path (e.g. $(3,4)$ or $(2,T)$) is a full outcome.

Step 2: Write each event as the set of paths that realise it.

List exactly which leaf paths satisfy the condition $A$ and which satisfy the event $B$. Crucially, note which branch each requires: a die can only show a value in the die–die branch; a coin face can only appear in the die–coin branch.

Step 3: Check whether the two events can share a path. …

Common Mistakes

Mistake 1: Assuming a coin tail can coexist with a die showing $3$.

Why it's wrong: the coin is tossed only when the first die is $1,2,4,$ or $5$; a $3$ can appear only in the throw-again branch, which never tosses a coin. So "tail" and "a die shows $3$" cannot happen together, giving $P(A\cap B)=0$ and $P(B|A)=0$. Correct approach: map out the two branches of the experiment before counting.

Mistake 2: Building one flat die-plus-coin sample space. …