Q.A die is thrown three times. Events $A$ and $B$ are defined as below: $A$ : 4 on the third throw $B$ : 6 on the first and 5 on the second throw Find the probability of $A$ given that $B$ has already occurred.

Concept: Conditional Probability — we want $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$.

Step 1: Since the die is fair and throws are independent, each outcome has probability $\frac{1}{6}$.

Step 2: Event $B$ requires a specific result on the first two throws (6 then 5), so

$P(B) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$.

Step 3: Event $A \cap B$ means: first throw = 6, second throw = 5, third throw = 4. …

Why conditional probability works this way here

The key idea is independence. When you roll a fair die, each throw is completely unaffected by the others. The outcome of the third throw has no connection to what happened on the first or second throw.

Events $A$ and $B$ involve different throws — $A$ only cares about the third throw, $B$ only about the first two. Because the throws are independent, knowing that $B$ happened tells you nothing new about whether $A$ will happen.

This is a special case where $P(A \mid B) = P(A)$. Most conditional probability problems aren't this simple — but when the events involve separate independent trials, the condition drops out.

Step-by-step reasoning

1. Identify the sample space.

   A die thrown three times has $6 \times 6 \times 6 = 216$ equally likely outcomes. Each outcome is an ordered triple $(x_1, x_2, x_3)$ where each $x_i \in \{1,2,3,4,5,6\}$.

2. Define the events clearly.

   • $A$: "4 on the third throw" means $x_3 = 4$.
   • $B$: "6 on the first and 5 on the second throw" means $x_1 = 6$ and $x_2 = 5$.

3. Recognize independence across throws.

   The outcome of any one throw does not influence any other throw. So the event $A$ (which depends only on the third throw) is independent of the event $B$ (which depends only on the first two throws). …

Method: Using independence to shortcut a conditional probability

Use this when the conditioning event and the target event depend on separate independent trials (different throws, different coins). Recognising independence saves the full calculation.

Steps

Step 1: Check whether the two events involve disjoint, independent trials.

If event $A$ depends only on one trial (the third throw) and event $B$ only on other trials (the first two), and the trials are independent, then $A$ and $B$ are independent.

Step 2: Apply the independence consequence for conditioning.

For independent events, conditioning changes nothing:

$P(A\mid B) = P(A).$ …

Common Mistakes

Mistake 1: Grinding through the full formula and mis-handling $P(B)$.

Why it's wrong: while $P(A\mid B)=P(A\cap B)/P(B)$ works, students often compute $P(B)$ or $P(A\cap B)$ with the wrong power of $\tfrac16$ and slip. Correct approach: notice the third throw is independent of the first two, so $P(A\mid B)=P(A)=\tfrac16$ directly.

Mistake 2: Thinking the condition $B$ must change $P(A)$. …