Probability of getting sum 8 given that first die shows 3

Question

Two fair dice are thrown simultaneously. Find the probability of getting a sum of 8, given that the first die shows a 3.

Solution — Step by Step

Let:

Event A: The first die shows 3
Event B: The sum of both dice is 8

We need to find $P(B | A)$ — the probability of B given that A has already occurred.

P(B | A) = \frac{P(A \cap B)}{P(A)}

This formula asks: among all outcomes where A occurs, what fraction also have B?

Total outcomes when two dice are thrown: $6 \times 6 = 36$

Outcomes where first die = 3: $(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)$ — exactly 6 outcomes

P(A) = \frac{6}{36} = \frac{1}{6}

We need outcomes where the first die shows 3 AND the sum is 8. If first die = 3, then second die must = 5 to get sum = 8.

The outcome $(3, 5)$ : first die = 3 ✓, sum = 3 + 5 = 8 ✓

So $A \cap B = \{(3, 5)\}$ — just 1 outcome

P(A \cap B) = \frac{1}{36}

P(B | A) = \frac{P(A \cap B)}{P(A)} = \frac{\frac{1}{36}}{\frac{1}{6}} = \frac{1}{36} \times \frac{6}{1} = \frac{6}{36} = \frac{1}{6}

$P(\text{sum} = 8 \mid \text{first die} = 3) = \dfrac{1}{6}$

Why This Works

Conditional probability restricts our sample space. Once we know the first die shows 3, there are only 6 possible outcomes (not 36). Of these 6 outcomes — (3,1), (3,2), (3,3), (3,4), (3,5), (3,6) — exactly one gives a sum of 8: namely (3,5). So the probability is 1/6.

The formula $P(B|A) = P(A \cap B)/P(A)$ is just a formal way of doing this restriction.

Alternative Method

Reduced sample space method (often faster):

Given: first die = 3. Restricted sample space = $\{(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)\}$ — 6 equally likely outcomes.

Favourable outcome for sum = 8: only $(3, 5)$ — 1 outcome.

P = \frac{1}{6}

For CBSE Class 12 and JEE, both the formula method and the reduced sample space method are acceptable. The reduced sample space method is faster for simple problems. Use the formula method when the events are more complex or when you need to demonstrate your knowledge of conditional probability notation for full marks.

Common Mistake

Students sometimes look for ALL ways to get a sum of 8 (which is (2,6), (3,5), (4,4), (5,3), (6,2) — five outcomes) and divide by 36 to get 5/36. This is wrong because the question specifies the first die already shows 3. We must restrict to outcomes consistent with the given condition. The conditional probability 1/6 is different from the unconditional probability 5/36.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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