Solution — 2017 Question 11 Probability

(a)

The total number of students is $80 + x .$

$\begin{matrix} P (M) & = \frac{48 + x}{80 + x} ∎ \end{matrix}$

$\begin{matrix} P (E) & = \frac{32 + x}{80 + x} ∎ \end{matrix}$

(b)

Since $M$ and $E$ are independent, $P (M \cap E) = P (M) \times P (E) .$

$\begin{matrix} \frac{16 + x}{80 + x} & = \frac{48 + x}{80 + x} (\frac{32 + x}{80 + x}) \\ (80 + x) (16 + x) & = (48 + x) (32 + x) \\ 1280 + 80 x + 16 x + x^{2} & = 1536 + 48 x + 32 x + x^{2} \\ 1280 + 96 x + x^{2} & = 1536 + 80 x + x^{2} \\ 16 x & = 256 \\ x & = 16 ∎ \end{matrix}$

(c)

$\begin{matrix} P (M \cup F^{'}) \\ = \frac{20 + 16 + 12 + 16 + 11}{80 + 16} \\ = \frac{25}{32} ∎ \end{matrix}$

Tip: shade the region M \cup F’ in the Venn diagram

(d)

$P (F | M)$ is the probability that the student studies Finance given that the student studies Marketing. $∎$

$\begin{matrix} P (F | M) & = \frac{P (F \cap M)}{P (M)} \\ = \frac{\frac{12 + 16}{96}}{\frac{20 + 16 + 12 + 16}{96}} \\ = \frac{7}{16} ∎ \end{matrix}$

(e)

With $x = 16,$ there are $33$ students studying exactly two subjects and $96$ students in total.

$\begin{matrix} P (required) \\ = \frac{33}{96} (\frac{32}{95}) (\frac{31}{94}) \\ = \frac{341}{8930} ∎ \\ = 0.0382 ∎ (to 3 s.f.) \end{matrix}$