Solution — 2018 Question 9 Probability

$$\begin{array}{cc}& \mathrm{P}\left(B\right)\\ & =\frac{112}{300}\\ & =\frac{28}{75}∎\end{array}$$

$$\begin{array}{cc}& \mathrm{P}(F\cup B)\\ & =\mathrm{P}\left(F\right)+\mathrm{P}\left(B\right)-\mathrm{P}(F\cap B)\\ & =\frac{140}{300}+\frac{112}{300}-\frac{40}{300}\\ & =\frac{53}{75}∎\end{array}$$

$$\begin{array}{cc}& \mathrm{P}(M\cap A)\\ & =\frac{54}{300}\\ & =\frac{9}{50}∎\end{array}$$

$$\begin{array}{cc}& \mathrm{P}\left(M\right)\mathrm{P}\left(A\right)\\ & =\frac{160}{300}\left(\frac{90}{300}\right)\\ & =\frac{4}{25}\end{array}$$

$$\begin{array}{cc}& \mathrm{P}\left(\text{exactly 2 out of 3 study Art}\right)\\ & =\mathrm{P}\left(AA{A}^{\prime }\right)+\mathrm{P}\left(A{A}^{\prime }A\right)+\mathrm{P}\left(AA{A}^{\prime }\right)\\ & =\frac{90}{300}\left(\frac{89}{299}\right)\left(\frac{210}{298}\right)+\frac{90}{300}\left(\frac{89}{299}\right)\left(\frac{210}{298}\right)+\frac{90}{300}\left(\frac{89}{299}\right)\left(\frac{210}{298}\right)\\ & =0.189∎\end{array}$$