Solution — 2025 Question 11 Probability

(a)

(i)

If $C$ and $D$ are mutually exclusive events, then $P (C \cup D) = 0.6 + 0.7 = 1.3$ which is impossible. Hence $C$ and $D$ cannot be mutually exclusive $∎$

(ii)

If $C$ and $D$ are independent,

\begin{matrix} P (C \cap D) & = P (C) P (D) \\ = 0.6 (0.7) \\ = 0.42 \end{matrix}

(b)

(i)

Let $P (F) = x$

\begin{matrix} P (E | F^{'}) & = \frac{7}{25} \\ \frac{P (E \cap F^{'})}{P (F^{'})} & = \frac{7}{25} \\ \frac{P (E \cup F) - P (F)}{1 - P (F)} & = \frac{7}{25} & (*) \\ \frac{\frac{89}{125} - x}{1 - x} & = \frac{7}{25} \\ 25 (\frac{89}{125} - x) & = 7 (1 - x) \\ \frac{89}{5} - 25 x & = 7 - 7 x \\ 18 x & = \frac{54}{5} \\ x & = \frac{3}{5} \\ P (F) & = \frac{3}{5} ∎ \end{matrix}

Use a Venn diagram to see that $P (E \cap F^{'})$ is $P (E \cup F) - P (F)$

(ii)

\begin{matrix} P (E^{'} \cup F) \\ = P (F) + P (E^{'} \cap F^{'}) & (*) \\ = \frac{3}{5} + (1 - P (E \cup F)) & (*) \\ = \frac{3}{5} + 1 - \frac{89}{125} \\ = \frac{111}{125} ∎ \end{matrix}

A Venn diagram is also very useful here. We have tagged the steps where the Venn diagram is crucial

(iii)

Let $P (E \cap F) = y$

\begin{matrix} P (E^{'} \cap F) & = P (F) - P (E \cap F) & (*) \\ = \frac{3}{5} - y \end{matrix}

\begin{matrix} P (E^{'}) & = P (E^{'} \cap F) + P (E^{'} \cap F^{'}) & (*) \\ = \frac{3}{5} - y + (1 - P (E \cup F)) & (*) \\ = \frac{3}{5} - y + (1 - \frac{89}{125}) \\ = \frac{111}{125} - y \end{matrix}

\begin{matrix} P (F | E^{'}) & = \frac{19}{55} \\ \frac{P (F \cap E^{'})}{P (E^{'})} & = \frac{19}{55} \\ \frac{P (F) - P (E \cap F)}{P (E^{'})} & = \frac{19}{55} & (*) \\ \frac{\frac{3}{5} - y}{\frac{111}{125} - y} & = \frac{19}{55} \\ 55 (\frac{3}{5} - y) & = 19 (\frac{111}{125} - y) \\ 33 - 55 y & = \frac{2109}{125} - 19 y \\ 36 y & = \frac{2016}{125} \\ y & = \frac{56}{125} \\ P (E \cap F) & = \frac{56}{125} ∎ \end{matrix}

Another part where a Venn diagram is crucial to understand some of the manipulations above. We have tagged the steps where the Venn diagram is crucial