Probability: Question 3

Worked solution

Step 1: Identify the two independent events

Let

• $R$ = “the spinner lands on red”, with $P(R)=\dfrac{2}{5}$,
• $G$ = “the dice shows a number greater than $4$”.

The two pieces of apparatus are stated to operate independently, so the outcome of the spinner does not affect the dice and vice versa.

Step 2: Find the probability for the dice

The dice is fair with faces $1,2,3,4,5,6$. “Greater than $4$” means the outcomes $\{5,6\}$, that is, 2 favourable outcomes out of $6$ equally likely ones:

$P(G)=\frac{2}{6}=\frac{1}{3}.$

Note that “greater than $4$” does not include $4$ itself.

Step 3: Combine with “AND” by multiplying

For independent events, the probability that both occur is the product of their individual probabilities (the AND-rule / multiplication rule):

$P(R \text{ and } G)=P(R)\times P(G)=\frac{2}{5}\times\frac{1}{3}.$

Multiply numerators and denominators:

$P(R \text{ and } G)=\frac{2\times 1}{5\times 3}=\frac{2}{15}.$

Step 4: Check against a tree diagram

A tree diagram confirms the method. The first set of branches is the spinner ($P(R)=\frac{2}{5}$, $P(\text{not red})=\frac{3}{5}$); each then splits into the dice branches ($P(G)=\frac{1}{3}$, $P(\text{not }G)=\frac{2}{3}$). The required outcome follows the single path red → greater than 4, and along any path you multiply:

$\frac{2}{5}\times\frac{1}{3}=\frac{2}{15}.$

As a sanity check, the four path probabilities are $\frac{2}{15},\ \frac{4}{15},\ \frac{3}{15},\ \frac{6}{15}$, which sum to $\frac{15}{15}=1$, as they must.

Why not add?

Adding, $\frac{2}{5}+\frac{1}{3}=\frac{6}{15}+\frac{5}{15}=\frac{11}{15}$, would answer a different question: roughly the chance of red or a high dice score (and even then it would need a correction for double counting). For “AND” of independent events you multiply.

Final answer

$\boxed{P(R\text{ and }G)=\dfrac{2}{15}}$

This is option A.