Q.A black and a red dice are rolled.

Concept: Conditional Probability — $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$, where we restrict the sample space to the given condition.

(a) Let $B$ be the event: black die shows 5.

Given black = 5, the possible outcomes for the red die are $\{1,2,3,4,5,6\}$.

We need sum $> 9$, i.e. red die must be $5$ or $6$ (since $5+5=10$, $5+6=11$). …

Why Conditional Probability Works This Way

When we say "given that" something has happened, we are no longer looking at all possible outcomes — we only care about the outcomes that satisfy the condition. The formula is:

But in problems with equally likely outcomes (like dice rolls), it's often easier to count: just count the outcomes in the condition, then count how many of those also satisfy the event.

Part (a): Sum > 9, given black die shows 5

1. Identify the condition.

The black die shows a 5. So the possible outcomes are:

That's 6 equally likely outcomes.

2. Identify the event.

We want the sum to be greater than 9. With black = 5, the red die must be such that $5 + \text{red} > 9$, i.e., red > 4. So red can be 5 or 6.

That gives exactly 2 outcomes: $(5,5)$ and $(5,6)$.

3. Compute the probability.

Part (b): Sum = 8, given red die shows a number less than 4

1. Identify the condition.

Red die shows a number less than 4, so red can be 1, 2, or 3. For each red value, black can be anything from 1 to 6. So the total outcomes satisfying the condition are:

$3 \times 6 = 18$ …

Method: Conditioning on a row or block of the two-dice grid

Use this whenever two distinguishable dice are rolled and the condition fixes one die's value (or restricts it to a range), and you want the probability of a sum event given that condition.

Steps

Step 1: Picture the $6\times6$ grid.

Two dice give $36$ equally likely ordered pairs. The conditioning event selects a sub-block of this grid, and within that block outcomes stay equally likely, so

Step 2: Count the conditioning block $n(B)$.

If one die is fixed to a single value, $B$ has $6$ outcomes; if a die is restricted to $m$ allowed values, $B$ has $6m$ outcomes (e.g. "red $<4$" allows $3$ values, so $6\times3=18$).

Step 3: Inside that block, count the outcomes meeting the sum condition. …

Common Mistakes

Mistake 1: Using $36$ in the denominator.

Why it's wrong: "given the black die shows $5$" restricts the world to the $6$ outcomes $(5,1),\dots,(5,6)$, so in (a) the answer is $\frac{2}{6}=\frac13$, not $\frac{2}{36}$. Correct approach: count favourable outcomes only within the conditioned set.

Mistake 2: In part (b), forgetting that "red $<4$" allows three values. …