Probability: Question 9

Worked solution

Step 1: List the sample space

Rolling two dice gives $36$ equally likely outcomes.

Step 2: Find P(A), the sum is 9

The pairs (first, second) giving a sum of $9$ are $(3,6), (4,5), (5,4), (6,3)$. That is $4$ outcomes: $P(A) = \frac{4}{36} = \frac{1}{9}$

Step 3: Find P(B), a double

The doubles are $(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)$. That is $6$ outcomes: $P(B) = \frac{6}{36} = \frac{1}{6}$

Step 4: Check whether A and B are mutually exclusive

Every double $(n,n)$ has an even sum ($2,4,6,8,10,12$), but $9$ is odd, so no double can also give a sum of $9$. Hence $A$ and $B$ share no outcomes: $P(A \cap B) = 0$, and the events are mutually exclusive.

Step 5: Apply the addition rule

Since $A$ and $B$ are mutually exclusive, the probabilities simply add: $P(A \text{ or } B) = P(A) + P(B) = \frac{1}{9} + \frac{1}{6} = \frac{2}{18} + \frac{3}{18} = \frac{5}{18} \approx 0.278$

Check by listing directly

The outcomes satisfying $A$ or $B$ are the $4$ sum-of-$9$ outcomes plus the $6$ double outcomes, with none shared, giving $4+6=10$ outcomes out of $36$: $P(A \text{ or } B) = \frac{10}{36} = \frac{5}{18}$ which agrees exactly with Step 5.

Why the other options are wrong

• B ($\frac{2}{15}$): this comes from incorrectly adding numerators and denominators separately, $\frac{1+1}{9+6}$, rather than using a common denominator.
• C ($\frac{1}{9}$): this is only $P(A)$, ignoring event $B$ entirely.
• D ($\frac{1}{6}$): this is only $P(B)$, ignoring event $A$ entirely.

Final answer

• $P(A \text{ or } B) = \boxed{\dfrac{5}{18}}$, option A.