Solution — 2020 Question 6 Probability

(a)

\begin{matrix} P (Year 2 and Bicycle) \\ = \frac{80}{1600} \\ = \frac{1}{20} ∎ \end{matrix}

(b)

\begin{matrix} P (not Bus) \\ = 1 - P (Bus) \\ = 1 - \frac{320}{1600} \\ = \frac{4}{5} ∎ \end{matrix}

(c)

\begin{matrix} P (Year 1 | Foot) \\ = \frac{P (Year 1 \cap Foot)}{P (Foot)} \\ = \frac{400}{512} \\ = \frac{25}{32} ∎ \end{matrix}

(d)

Let $C$ be the event that ‘the student travels by car’

Let $Y_{1}$ be the event that ‘the student is in Year 1’

\begin{matrix} P (C \cap Y_{1}) & = \frac{432}{1600} \\ = 0.27 \end{matrix}

\begin{matrix} P (C) P (Y_{1}) & = \frac{576}{1600} (\frac{1200}{1600}) \\ = 0.27 \end{matrix}

Since $P (C \cap Y_{1}) = P (C) P (Y_{1}),$ the events are independent $∎$