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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\\n

Commented byEDWIN88 last updated on 04/Mar/21

Answer available (a) 0.5 (b) 0.4 (c) 0.3 (d)0.2 (e) 0.1

$$\mathrm{Answer}\:\mathrm{available}\: \\ $$ $$\left(\mathrm{a}\right)\:\mathrm{0}.\mathrm{5}\:\:\:\:\:\left(\mathrm{b}\right)\:\mathrm{0}.\mathrm{4}\:\:\:\:\:\:\left(\mathrm{c}\right)\:\mathrm{0}.\mathrm{3}\:\:\:\:\:\left(\mathrm{d}\right)\mathrm{0}.\mathrm{2}\:\:\:\:\:\left(\mathrm{e}\right)\:\mathrm{0}.\mathrm{1} \\ $$

Commented bybenjo_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

$$\mathrm{case}\:\left(\mathrm{1}\right)\: \\ $$ $$\mathrm{P}\left(\mathrm{A}_{\mathrm{1}} \right)=\begin{pmatrix}{\mathrm{3}}\\{\mathrm{2}}\end{pmatrix}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{1}} ×\begin{pmatrix}{\mathrm{5}}\\{\mathrm{0}}\end{pmatrix}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{5}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{0}} \\ $$ $$\:=\:\frac{\mathrm{9}}{\mathrm{2}^{\mathrm{6}} }×\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{5}} }\:=\:\frac{\mathrm{9}}{\mathrm{2}^{\mathrm{11}} } \\ $$ $$\mathrm{\color{mathblue}{c}\color{mathblue}{a}\color{mathblue}{s}\color{mathblue}{e}}\color{mathblue}{\left(}\mathrm{\color{mathblue}{2}}\color{mathblue}{\right)} \\ $$ $$\mathrm{\color{mathblue}{P}}\color{mathblue}{\left(}\mathrm{\color{mathblue}{A}}_{\mathrm{\color{mathblue}{2}}} \color{mathblue}{\right)}\color{mathblue}{=}\begin{pmatrix}{\mathrm{\color{mathblue}{3}}}\\{\mathrm{\color{mathblue}{1}}}\end{pmatrix}\color{mathblue}{\left(}\frac{\mathrm{\color{mathblue}{1}}}{\mathrm{\color{mathblue}{4}}}\color{mathblue}{\right)}^{\mathrm{\color{mathblue}{1}}} \color{mathblue}{\left(}\frac{\mathrm{\color{mathblue}{3}}}{\mathrm{\color{mathblue}{4}}}\color{mathblue}{\right)}^{\mathrm{\color{mathblue}{2}}} \color{mathblue}{×}\begin{pmatrix}{\mathrm{\color{mathblue}{5}}}\\{\mathrm{\color{mathblue}{1}}}\end{pmatrix}\color{mathblue}{\left(}\frac{\mathrm{\color{mathblue}{1}}}{\mathrm{\color{mathblue}{2}}}\color{mathblue}{\right)}^{\mathrm{\color{mathblue}{1}}} \color{mathblue}{\left(}\frac{\mathrm{\color{mathblue}{1}}}{\mathrm{\color{mathblue}{2}}}\color{mathblue}{\right)}^{\mathrm{\color{mathblue}{4}}} \\ $$ $$\color{mathblue}{=}\color{mathblue}{\:}\frac{\mathrm{\color{mathblue}{2}\color{mathblue}{7}}}{\mathrm{\color{mathblue}{4}}^{\mathrm{\color{mathblue}{3}}} }\color{mathblue}{\:}\color{mathblue}{×}\color{mathblue}{\:}\frac{\mathrm{\color{mathblue}{5}}}{\mathrm{\color{mathblue}{2}}^{\mathrm{\color{mathblue}{5}}} }\color{mathblue}{\:}\color{mathblue}{=}\color{mathblue}{\:}\frac{\mathrm{\color{mathblue}{1}\color{mathblue}{3}\color{mathblue}{5}}}{\mathrm{\color{mathblue}{2}}^{\mathrm{\color{mathblue}{1}\color{mathblue}{1}}} } \\ $$ $$\mathrm{\color{mathred}{c}\color{mathred}{a}\color{mathred}{s}\color{mathred}{e}}\color{mathred}{\left(}\mathrm{\color{mathred}{3}}\color{mathred}{\right)} \\ $$ $$\mathrm{\color{mathred}{P}}\color{mathred}{\left(}\mathrm{\color{mathred}{A}}_{\mathrm{\color{mathred}{3}}} \color{mathred}{\right)}\color{mathred}{=}\color{mathred}{\:}\begin{pmatrix}{\mathrm{\color{mathred}{3}}}\\{\mathrm{\color{mathred}{0}}}\end{pmatrix}\color{mathred}{\left(}\frac{\mathrm{\color{mathred}{1}}}{\mathrm{\color{mathred}{4}}}\color{mathred}{\right)}^{\mathrm{\color{mathred}{0}}} \color{mathred}{\left(}\frac{\mathrm{\color{mathred}{3}}}{\mathrm{\color{mathred}{4}}}\color{mathred}{\right)}^{\mathrm{\color{mathred}{3}}} \color{mathred}{×}\begin{pmatrix}{\mathrm{\color{mathred}{5}}}\\{\mathrm{\color{mathred}{2}}}\end{pmatrix}\color{mathred}{\left(}\frac{\mathrm{\color{mathred}{1}}}{\mathrm{\color{mathred}{2}}}\color{mathred}{\right)}^{\mathrm{\color{mathred}{2}}} \color{mathred}{\left(}\frac{\mathrm{\color{mathred}{1}}}{\mathrm{\color{mathred}{2}}}\color{mathred}{\right)}^{\mathrm{\color{mathred}{3}}} \\ $$ $$\color{mathred}{=}\color{mathred}{\:}\frac{\mathrm{\color{mathred}{2}\color{mathred}{7}}}{\mathrm{\color{mathred}{2}}^{\mathrm{\color{mathred}{6}}} }\color{mathred}{\:}\color{mathred}{×}\color{mathred}{\:}\frac{\mathrm{\color{mathred}{1}\color{mathred}{0}}}{\mathrm{\color{mathred}{2}}^{\mathrm{\color{mathred}{5}}} }\color{mathred}{\:}\color{mathred}{=}\color{mathred}{\:}\frac{\mathrm{\color{mathred}{2}\color{mathred}{7}\color{mathred}{0}}}{\mathrm{\color{mathred}{2}}^{\mathrm{\color{mathred}{1}\color{mathred}{1}}} } \\ $$ $$\mathrm{totally}\:\Rightarrow\mathrm{P}\left(\mathrm{A}\right)=\:\frac{\mathrm{9}+\mathrm{135}+\mathrm{270}}{\mathrm{2}^{\mathrm{11}} }\approx\mathrm{0}.\mathrm{202148} \\ $$ $$ \\ $$

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