$$\\$$Badu took a test consisting of 3 multiple choice questions with 4 answer choices and 5 true-false type questions. If Badu answers all the questions by guessing randomly, the chances of him answering correctly are only 2 questions
$$\mathrm{Answer}\:\mathrm{available}\: \\$$$$\left(\mathrm{a}\right)\:\mathrm{0}.\mathrm{5}\:\:\:\:\:\left(\mathrm{b}\right)\:\mathrm{0}.\mathrm{4}\:\:\:\:\:\:\left(\mathrm{c}\right)\:\mathrm{0}.\mathrm{3}\:\:\:\:\:\left(\mathrm{d}\right)\mathrm{0}.\mathrm{2}\:\:\:\:\:\left(\mathrm{e}\right)\:\mathrm{0}.\mathrm{1} \\$$
$$\mathrm{case}\:\left(\mathrm{1}\right)\: \\$$$$\mathrm{P}\left(\mathrm{A}_{\mathrm{1}} \right)=\begin{pmatrix}{\mathrm{3}}\\{\mathrm{2}}\end{pmatrix}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{1}} ×\begin{pmatrix}{\mathrm{5}}\\{\mathrm{0}}\end{pmatrix}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{5}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{0}} \\$$$$\:=\:\frac{\mathrm{9}}{\mathrm{2}^{\mathrm{6}} }×\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{5}} }\:=\:\frac{\mathrm{9}}{\mathrm{2}^{\mathrm{11}} } \\$$$$\mathrm{case}\left(\mathrm{2}\right) \\$$$$\mathrm{P}\left(\mathrm{A}_{\mathrm{2}} \right)=\begin{pmatrix}{\mathrm{3}}\\{\mathrm{1}}\end{pmatrix}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} ×\begin{pmatrix}{\mathrm{5}}\\{\mathrm{1}}\end{pmatrix}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{4}} \\$$$$=\:\frac{\mathrm{27}}{\mathrm{4}^{\mathrm{3}} }\:×\:\frac{\mathrm{5}}{\mathrm{2}^{\mathrm{5}} }\:=\:\frac{\mathrm{135}}{\mathrm{2}^{\mathrm{11}} } \\$$$$\mathrm{case}\left(\mathrm{3}\right) \\$$$$\mathrm{P}\left(\mathrm{A}_{\mathrm{3}} \right)=\:\begin{pmatrix}{\mathrm{3}}\\{\mathrm{0}}\end{pmatrix}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{0}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{3}} ×\begin{pmatrix}{\mathrm{5}}\\{\mathrm{2}}\end{pmatrix}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} \\$$$$=\:\frac{\mathrm{27}}{\mathrm{2}^{\mathrm{6}} }\:×\:\frac{\mathrm{10}}{\mathrm{2}^{\mathrm{5}} }\:=\:\frac{\mathrm{270}}{\mathrm{2}^{\mathrm{11}} } \\$$$$\mathrm{totally}\:\Rightarrow\mathrm{P}\left(\mathrm{A}\right)=\:\frac{\mathrm{9}+\mathrm{135}+\mathrm{270}}{\mathrm{2}^{\mathrm{11}} }\approx\mathrm{0}.\mathrm{202148} \\$$$$\\$$