A general problem-solving technique video — not a specific exam question.
Problem
Two dice are rolled once. Find the probability of getting an even number on the first die or a total of 8.
Key insight. “Or” between two events means their probabilities can’t simply be added — some outcomes belong to both events and would be counted twice. Identifying exactly which outcomes give both an even first die and a total of 8 is the key step the addition rule needs.
Watch this explained step by step → · More on the Shiwam’s Classes channel
Approach
Let $A$ = “even number on the first die” and $B$ = “total of 8.” Find $P(A)$ and $P(B)$ directly by counting favourable outcomes out of the 36 equally likely outcomes of rolling two dice. Find $P(A\cap B)$ by identifying outcomes satisfying both conditions, then apply the addition rule $P(A\cup B)=P(A)+P(B)-P(A\cap B)$.
Solution
Step 1 — Find P(A): even number on the first die
The first die can be $2$, $4$, or $6$ (3 choices), with the second die free (6 choices):
$$P(A) = \frac{3\times6}{36} = \frac{18}{36}$$
Step 2 — Find P(B): total of 8
The outcomes summing to $8$ are $(2,6), (3,5), (4,4), (5,3), (6,2)$ — 5 outcomes:
$$P(B) = \frac{5}{36}$$
Step 3 — Find P(A ∩ B): even first die AND total of 8
Checking each of the 5 outcomes from Step 2 for an even first die: $(2,6)$ ✓, $(3,5)$ ✗, $(4,4)$ ✓, $(5,3)$ ✗, $(6,2)$ ✓ — 3 outcomes satisfy both:
$$P(A\cap B) = \frac{3}{36}$$
Step 4 — Apply the addition rule
$$P(A\cup B) = P(A)+P(B)-P(A\cap B) = \frac{18}{36}+\frac{5}{36}-\frac{3}{36} = \frac{20}{36}$$
Step 5 — Simplify
$$\frac{20}{36} = \frac{5}{9}$$
Answer
$$\frac{5}{9}$$
Common mistakes
- Simply adding $P(A)+P(B)$ without subtracting the overlap. This double-counts the 3 outcomes that satisfy both conditions, giving an inflated (wrong) probability of $\frac{23}{36}$.
- Miscounting which outcomes satisfy both conditions. It’s easy to assume none of the “total of 8” outcomes have an even first die, but three of the five actually do — checking each one explicitly avoids this error.
Practise next
- Two dice are rolled once. Find the probability of getting an odd number on the second die or a total of 7, using the same addition-rule approach.
Show answer
$\dfrac{7}{12}$. Let $A$ be ‘second die odd’ and $B$ be ‘total $7$’.
$P(A)=\tfrac12$ and $P(B)=\tfrac{6}{36}=\tfrac16$. For the overlap, the pairs totalling $7$ with an odd second die are $(6,1),(4,3),(2,5)$ — three of them, so $P(A\cap B)=\tfrac{3}{36}=\tfrac1{12}$.
$$P(A\cup B)=\tfrac12+\tfrac16-\tfrac1{12}=\tfrac{7}{12}.$$
Exactly half the total-$7$ outcomes have an odd second die, which is why the overlap is not simply $P(A)P(B)$ by accident here — the two events are in fact independent, but the intersection still has to be counted.
Doubts are answered by Shiwam, usually within a day. Ask about this question specifically — a general question about the chapter is better asked in class.