Q.In a class, 60% of the students are boys and 40% are girls. It is known that 5% of the boys and 10% of the girls are left-handed. A student is selected at random and found to be left-handed. What additional insight does the multiplication theorem give about the probability that this student is a boy AND left-handed?

Apply the multiplication theorem $P(A\cap B)=P(A)\times P(B\mid A)$ with $A=$ 'boy' and $B=$ 'left-handed', using the given conditional percentage.

Step 1 — Identify the given probabilities. $P(\text{boy}) = 0.60$; $P(\text{left-handed}\mid\text{boy}) = 0.05$ (5% of boys are left-handed — this is already a conditional probability, given as boys specifically).

Step 2 — Apply the multiplication theorem.

Similarly, for girls: $P(\text{girl AND left-handed}) = P(\text{girl})\times P(\text{left-handed}\mid\text{girl}) = 0.40\times0.10=0.04$.

Step 3 — Interpret. Out of every 100 students in the class, we expect $3$ to be left-handed boys and $4$ to be left-handed girls — a total of $7$ left-handed students overall (i.e. $P(\text{left-handed})=0.03+0.04=0.07$, using the addition theorem for these two mutually exclusive groups, since a student is either a boy or a girl, never both). …

Multiplying $P(\text{boy})$ by the overall left-handedness rate (mixing boys' and girls' rates together) instead of by the boy-specific conditional rate of 5% is a common error — the multiplication theorem requires the …