Q.Determine $P(E|F)$. A die is thrown three times, $E$ : 4 appears on the third toss, $F$ : 6 and 5 appears respectively on first two tosses.

Concept: Conditional Probability — we want the probability of $E$ given that $F$ has already occurred.

Step 1: The sample space for three tosses of a die has $6^3 = 216$ equally likely outcomes.

Step 2: Event $F$ fixes the first two tosses as 6 and 5. So $F$ contains exactly $6$ outcomes (the third toss can be any of 1 through 6).

Thus $P(F) = \frac{6}{216}$. …

Why Conditional Probability Works Here

When we ask for $P(E|F)$, we are restricting our attention only to those outcomes where $F$ has happened. In this problem, $F$ tells us exactly what happened on the first two tosses: a 6 then a 5. The event $E$ only cares about the third toss. Because each toss of a fair die is independent, knowing the first two results gives us no information about the third. So the conditional probability should equal the simple probability of rolling a 4 on any single toss.

Let’s verify this formally.

1. Define the sample space.

   When a die is thrown three times, each toss has 6 possible outcomes. The total number of equally likely outcomes is $6 \times 6 \times 6 = 216$. Each outcome is an ordered triple like $(a, b, c)$ where $a, b, c \in \{1,2,3,4,5,6\}$.

2. Identify the events.

   • $E$: "4 appears on the third toss." This means the third coordinate is fixed as 4. The first two can be anything. So $|E| = 6 \times 6 \times 1 = 36$.
   • $F$: "6 and 5 appear respectively on first two tosses." This means the first toss is 6 and the second is 5. The third can be anything. So $|F| = 1 \times 1 \times 6 = 6$.

3. Find the intersection $E \cap F$.

   For both $E$ and $F$ to happen, the first toss must be 6, the second must be 5, and the third must be 4. That is exactly one outcome: $(6,5,4)$. So $|E \cap F| = 1$.

4. Apply the conditional probability formula. …

Method: Conditioning on independent trials of a die

Use this when a die (or coin) is thrown several times and the conditioning event $F$ fixes the results of some throws while the event $E$ depends only on other throws.

Steps

Step 1: Count the full sample space with the multiplication principle.

For $k$ throws of a die there are $6^k$ equally likely ordered outcomes (e.g. $6^3=216$ for three throws), so you may again use

Step 2: Count $n(F)$ by fixing the constrained throws.

Each throw that $F$ pins down contributes a factor $1$; each free throw contributes a factor $6$. Fixing the first two throws leaves $1\cdot1\cdot6=6$ outcomes in $F$.

Step 3: Count $n(E\cap F)$ the same way.

Now additionally fix the throw that $E$ constrains. If $E$ pins the remaining throw to one value, $n(E\cap F)=1$. …

Common Mistakes

Mistake 1: Dividing by $216$ instead of by $6$.

Why it's wrong: given $F$ fixes the first two tosses as $6$ then $5$, only $6$ outcomes remain (one for each value of the third die), not all $216$. Correct approach: $P(E|F)=\frac{n(E\cap F)}{n(F)}=\frac{1}{6}$, giving $\frac16$, not $\frac{1}{216}$ or $\frac{1}{36}$.

Mistake 2: Thinking the earlier tosses change the third toss. …