Solution — 2018 Question 9 Probability

(a)

\begin{matrix} P (B) \\ = \frac{112}{300} \\ = \frac{28}{75} ∎ \end{matrix}

(b)

\begin{matrix} P (F \cup B) \\ = P (F) + P (B) - P (F \cap B) \\ = \frac{140}{300} + \frac{112}{300} - \frac{40}{300} \\ = \frac{53}{75} ∎ \end{matrix}

(c)

\begin{matrix} P (M \cap A) \\ = \frac{54}{300} \\ = \frac{9}{50} ∎ \end{matrix}

(d)

\begin{matrix} P (M) P (A) \\ = \frac{160}{300} (\frac{90}{300}) \\ = \frac{4}{25} \end{matrix}

Since $P (M \cap A) \neq P (M) P (A),$ the events $M$ and $A$ are not independent $∎$

(e)

\begin{matrix} P (exactly 2 out of 3 study Art) \\ = P (A A A^{'}) + P (A A^{'} A) + P (A A A^{'}) \\ = \frac{90}{300} (\frac{89}{299}) (\frac{210}{298}) + \frac{90}{300} (\frac{89}{299}) (\frac{210}{298}) + \frac{90}{300} (\frac{89}{299}) (\frac{210}{298}) \\ = 0.189 ∎ \end{matrix}