Q.Given that the two numbers appearing on throwing two dice are different. Find the probability of the event ‘the sum of numbers on the dice is 4’.

Concept: Conditional Probability — we restrict the sample space to only those outcomes where the two dice show different numbers.

Step 1: Total outcomes when two dice are thrown

$6 \times 6 = 36$.

Step 2: Restrict to outcomes with different numbers

Number of outcomes where dice are different = $36 - 6 = 30$ (subtract the 6 doublets).

Step 3: Favorable outcomes for sum = 4 with different numbers …

Let $A$ be the event "sum is $4$" and $B$ the event "the two numbers are different". We want $P(A\mid B)=\dfrac{n(A\cap B)}{n(B)}$.

1. Condition $B$ (different numbers). Of the $36$ ordered outcomes, the $6$ doubles $(1,1),(2,2),\dots,(6,6)$ are excluded, leaving

$n(B)=36-6=30.$ …

Method: Conditioning by restricting the two-dice space with a complement count

Use this when two dice are rolled and the condition is a property of the pair as a whole — "the two numbers are different", "the numbers are equal", "one is larger" — rather than a fixed value.

Steps

Step 1: Start from the full grid and count the condition with a complement.

Two dice give $36$ equally likely ordered pairs. It is often easiest to size the conditioning event $B$ by subtracting the easy-to-count opposite case: "different numbers" $= 36 - 6 = 30$, removing the six doublets $(1,1),\dots,(6,6)$. Then

Step 2: List the outcomes for the target event $A$.

Enumerate all ordered pairs giving the required sum (for a sum $s$, the pairs $(i,s-i)$ with both in $1$–$6$). …

Common Mistakes

Mistake 1: Dividing by $36$ instead of $30$.

Why it's wrong: the condition "the two numbers are different" removes the $6$ doublets $(1,1),\dots,(6,6)$, so the conditioned sample space has only $30$ outcomes. Correct approach: use $P=\frac{n(A\cap B)}{30}$, not $\frac{n}{36}$.

Mistake 2: Counting $(2,2)$ among the sum-$4$ outcomes. …