Question Number 134472 by EDWIN88 last updated on 04/Mar/21
![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](https://www.tinkutara.com/question/Q134472.png)
$$ \\ $$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
Commented by EDWIN88 last updated on 04/Mar/21
![Answer available (a) 0.5 (b) 0.4 (c) 0.3 (d)0.2 (e) 0.1](https://www.tinkutara.com/question/Q134473.png)
$$\mathrm{Answer}\:\mathrm{available}\: \\ $$$$\left(\mathrm{a}\right)\:\mathrm{0}.\mathrm{5}\:\:\:\:\:\left(\mathrm{b}\right)\:\mathrm{0}.\mathrm{4}\:\:\:\:\:\:\left(\mathrm{c}\right)\:\mathrm{0}.\mathrm{3}\:\:\:\:\:\left(\mathrm{d}\right)\mathrm{0}.\mathrm{2}\:\:\:\:\:\left(\mathrm{e}\right)\:\mathrm{0}.\mathrm{1} \\ $$
Commented by benjo_mathlover last updated on 04/Mar/21
![case (1) P(A_1 )= ((3),(2) )((1/4))^2 ((3/4))^1 × ((5),(0) )((1/2))^5 ((1/2))^0 = (9/2^6 )×(1/2^5 ) = (9/2^(11) ) case(2) P(A_2 )= ((3),(1) )((1/4))^1 ((3/4))^2 × ((5),(1) )((1/2))^1 ((1/2))^4 = ((27)/4^3 ) × (5/2^5 ) = ((135)/2^(11) ) case(3) P(A_3 )= ((3),(0) )((1/4))^0 ((3/4))^3 × ((5),(2) )((1/2))^2 ((1/2))^3 = ((27)/2^6 ) × ((10)/2^5 ) = ((270)/2^(11) ) totally ⇒P(A)= ((9+135+270)/2^(11) )≈0.202148](https://www.tinkutara.com/question/Q134476.png)
$$\mathrm{case}\:\left(\mathrm{1}\right)\: \\ $$$$\mathrm{P}\left(\mathrm{A}_{\mathrm{1}} \right)=\begin{pmatrix}{\mathrm{3}}\\{\mathrm{2}}\end{pmatrix}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{1}} ×\begin{pmatrix}{\mathrm{5}}\\{\mathrm{0}}\end{pmatrix}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{5}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{0}} \\ $$$$\:=\:\frac{\mathrm{9}}{\mathrm{2}^{\mathrm{6}} }×\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{5}} }\:=\:\frac{\mathrm{9}}{\mathrm{2}^{\mathrm{11}} } \\ $$$$\mathrm{case}\left(\mathrm{2}\right) \\ $$$$\mathrm{P}\left(\mathrm{A}_{\mathrm{2}} \right)=\begin{pmatrix}{\mathrm{3}}\\{\mathrm{1}}\end{pmatrix}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} ×\begin{pmatrix}{\mathrm{5}}\\{\mathrm{1}}\end{pmatrix}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{4}} \\ $$$$=\:\frac{\mathrm{27}}{\mathrm{4}^{\mathrm{3}} }\:×\:\frac{\mathrm{5}}{\mathrm{2}^{\mathrm{5}} }\:=\:\frac{\mathrm{135}}{\mathrm{2}^{\mathrm{11}} } \\ $$$$\mathrm{case}\left(\mathrm{3}\right) \\ $$$$\mathrm{P}\left(\mathrm{A}_{\mathrm{3}} \right)=\:\begin{pmatrix}{\mathrm{3}}\\{\mathrm{0}}\end{pmatrix}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{0}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{3}} ×\begin{pmatrix}{\mathrm{5}}\\{\mathrm{2}}\end{pmatrix}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} \\ $$$$=\:\frac{\mathrm{27}}{\mathrm{2}^{\mathrm{6}} }\:×\:\frac{\mathrm{10}}{\mathrm{2}^{\mathrm{5}} }\:=\:\frac{\mathrm{270}}{\mathrm{2}^{\mathrm{11}} } \\ $$$$\mathrm{totally}\:\Rightarrow\mathrm{P}\left(\mathrm{A}\right)=\:\frac{\mathrm{9}+\mathrm{135}+\mathrm{270}}{\mathrm{2}^{\mathrm{11}} }\approx\mathrm{0}.\mathrm{202148} \\ $$$$ \\ $$