Q.Two dice are thrown. If it is known that the sum of numbers on the dice was less than $6$, the probability of getting a sum $3$ is
(A) $\dfrac{1}{18}$
(B) $\dfrac{5}{18}$
(C) $\dfrac{1}{5}$
(D) $\dfrac{2}{5}$

Concept: Conditional Probability — we restrict the sample space to the given condition and count favourable outcomes within it.

Step 1: Identify the reduced sample space.

The condition is that the sum is less than 6. Possible sums are 2, 3, 4, 5. List all ordered pairs $(a,b)$ with $1 \le a,b \le 6$ that satisfy this:

• Sum 2: (1,1) → 1 outcome
• Sum 3: (1,2), (2,1) → 2 outcomes
• Sum 4: (1,3), (2,2), (3,1) → 3 outcomes
• Sum 5: (1,4), (2,3), (3,2), (4,1) → 4 outcomes

Total outcomes under condition = $1+2+3+4 = 10$.

Step 2: Count favourable outcomes.

Favourable event: sum exactly 3. From above, that gives 2 outcomes. …

The key here is conditional probability — we are not finding the probability of sum 3 in the full sample space of 36 outcomes. Instead, we are given extra information: the sum is less than 6. This reduces the set of possible outcomes, and we must find the probability within that restricted set.

Think of it this way: when you know the sum is less than 6, you are no longer considering all 36 pairs. You only care about the ones that satisfy that condition. The probability becomes:

Since sum 3 is automatically less than 6, the numerator is just the number of ways to get sum 3.

Let’s work it out step by step.

1. Full sample space: When two dice are thrown, each die shows 1 to 6. Total outcomes = $6 \times 6 = 36$. All pairs are equally likely.

2. Restricted condition — sum less than 6: List all pairs $(a,b)$ where $a+b \leq 5$ (since sum is an integer, less than 6 means 2, 3, 4, or 5).

   • Sum = 2: (1,1) → 1 outcome
   • Sum = 3: (1,2), (2,1) → 2 outcomes
   • Sum = 4: (1,3), (2,2), (3,1) → 3 outcomes
   • Sum = 5: (1,4), (2,3), (3,2), (4,1) → 4 outcomes Total outcomes with sum < 6 = $1 + 2 + 3 + 4 = 10$.

3. Favorable outcomes — sum exactly 3: From the list above, there are 2 outcomes: (1,2) and (2,1). …

Method: Conditional probability with a restricted sample space

Use this when a condition ("given the sum is less than 6", "given the number is even") is imposed and you must find the probability of a further event within that restricted set.

Steps

Step 1: Build the full equally-likely sample space.

For two dice this is all $6\times 6 = 36$ ordered pairs.

Step 2: Reduce to the outcomes satisfying the condition $B$.

Keep only the outcomes obeying the given condition and count them — this count becomes the denominator, replacing the full $36$.

Step 3: Count outcomes satisfying both the condition and the target event $A$. …

Common Mistakes

Mistake 1: Using the full $36$ outcomes as the denominator, $\tfrac{2}{36}=\tfrac1{18}$.

Why it's wrong: the condition "sum $< 6$" has already eliminated most pairs, so $36$ is not the conditioned total (this yields the distractor $\tfrac1{18}$). Correct approach: the denominator is the number of outcomes with sum $<6$, which is $10$, giving $\tfrac{2}{10}=\tfrac15$. …