2023 H2 Mathematics Paper 2 Question 5

Probability

Answers

{{A \text{ and } B. }}

{{A \text{ and } D.}}

(aii)

{{ \mathrm{P}\left( A \cap C \right) = \mathrm{P}\left( A \right) \mathrm{P}\left( C \right).}}

{{B \text{ and } C.}}

(bi)

{{A \text{ and } B. }}

{{A \text{ and } D.}}

(bii)

Not independent.

Full solutions

(ai)

\begin{gather*} A \text{ and } B \; \blacksquare \\ A \text{ and } D \; \blacksquare \end{gather*}

(aii)

\begin{align*} & \mathrm{P}\left( A \cap C \right) \\ & = \mathrm{P}\left( 3,9,15,21,27,33 \right) \\ & = \frac{6}{36} \\ & = \frac{1}{6} \end{align*}

\begin{align*} & \mathrm{P}\left( A \right) \times \mathrm{P}\left( C \right) \\ & = \frac{18}{36} \times \frac{12}{36} \\ &= \frac{1}{6} \end{align*}

Since ${{ \mathrm{P}\left( A \cap C \right) = \mathrm{P}\left( A \right) \mathrm{P}\left( C \right),}}$

{ A \text{ and } C \text{ are independent } \; \blacksquare }

(bi)

\begin{gather*} A \text{ and } B \; \blacksquare \\ A \text{ and } D \; \blacksquare \end{gather*}

(bii)

\begin{align*} & \mathrm{P}\left( A \cap C \right) \\ & = \mathrm{P}\left( 3,9,15,21,27,33 \right) \\ & = \frac{6}{35} \\ & = \frac{6}{35} \end{align*}

\begin{align*} & \mathrm{P}\left( A \right) \times \mathrm{P}\left( C \right) \\ & = \frac{18}{35} \times \frac{11}{35} \\ &= \frac{198}{1225} \end{align*}

Since ${{ \mathrm{P}\left( A \cap C \right) \neq \mathrm{P}\left( A \right) \mathrm{P}\left( C \right),}}$

{ A \text{ and } C \text{ are not independent} \; \blacksquare }

Question Commentary

For part (aii), to find the other pair of independent events, the fastest way is to observe that ${{A}}$ and ${{B}}$ are complementary events, so if ${{A}}$ and ${{C}}$ are independent, then ${{B}}$ and ${{C}}$ are also independent.

The longer method is to manually calculate all the individual probabilities, their products and the probabilities of the intersections.

The use of the discriminant to justify the number of points of intersection is also tested in part (b).