Q.Consider the experiment of tossing a coin. If the coin shows head, toss it again but if it shows tail, then throw a die. Find the conditional probability of the event that 'the die shows a number greater than 4' given that 'there is at least one tail'.

Idea: The outcomes here are NOT equally likely, so weight each by its true probability before using $P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}$.

Outcomes and their probabilities:

• Head first, then coin: HH, HT, each $\tfrac12\cdot\tfrac12=\tfrac14$.
• Tail first, then die: T1,…,T6, each $\tfrac12\cdot\tfrac16=\tfrac{1}{12}$.

Events:

• $A$ = die shows $>4$ = $\{\text{T5},\text{T6}\}$, so $P(A)=\tfrac{1}{12}+\tfrac{1}{12}=\tfrac16$.
• $B$ = at least one tail = everything except HH, so $P(B)=1-\tfrac14=\tfrac34$. …

Why the outcomes are not equally likely

The experiment branches on the first toss:

• If the coin shows head, we toss the coin again, giving HH or HT.
• If it shows tail, we throw a die, giving T1, T2, T3, T4, T5, T6.

The first toss is fair, so $P(H)=P(T)=\tfrac12$. On the head branch the second toss is fair, so

On the tail branch the die is fair, so each of the six outcomes has

These probabilities sum to $2\cdot\tfrac14+6\cdot\tfrac{1}{12}=\tfrac12+\tfrac12=1$, as they should. The eight outcomes are therefore NOT equally likely, so we must use probabilities, not raw counts.

Defining the events

• $A$: the die shows a number greater than $4$, i.e. $5$ or $6$. A die is thrown only after a tail, so $A=\{\text{T5},\text{T6}\}$.
• $B$: there is at least one tail. Every outcome has a tail except HH, so $B=\{\text{HT},\text{T1},\text{T2},\text{T3},\text{T4},\text{T5},\text{T6}\}$.

Computing the probabilities

$P(B)$: easiest via the complement. The only "no tail" outcome is HH with probability $\tfrac14$, so …

Method: Conditional probability when outcomes are NOT equally likely

Use this for branching experiments (a coin decides whether you toss again or throw a die) where the final outcomes carry different probabilities. Counting outcomes fails here — you must weight by probability.

Steps

Step 1: Build the tree and attach a probability to each leaf.

Multiply along each branch. A "head then head/tail" leaf has probability $\tfrac12\cdot\tfrac12=\tfrac14$, while a "tail then die face" leaf has $\tfrac12\cdot\tfrac16=\tfrac{1}{12}$. Confirm all leaf probabilities sum to $1$.

Step 2: Add probabilities (not counts) for each event. …

Common Mistakes

Mistake 1: Treating all eight leaves as equally likely.

Why it's wrong: a "head then head/tail" leaf has probability $\tfrac14$, but each "tail then die" leaf has only $\tfrac{1}{12}$; equal weighting wrongly gives $\tfrac27$. Correct approach: weight each leaf by its branch product before adding, giving $\tfrac29$.

Mistake 2: Counting outcomes instead of adding probabilities. …