(a)OR(b)

In answering a question on a multiple choice test, a student either knows the answer or guesses. Let $\displaystyle \frac{3}{5}$ be the probability that he knows the answer and $\displaystyle \frac{2}{5}$ be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability $\displaystyle \frac{1}{3}$. What is the probability that the student knows the answer, given that he answered it correctly?

Let events $\displaystyle \mathrm{A}, \mathrm{B}$ and E be defined as: A : Student knows the answer B : Student guesses the answer E : student answered correctly\mathrm{P}(\mathrm{~A})=\frac{3}{5}, \mathrm{P}(\mathrm{~B})=\frac{2}{5}Here, $\displaystyle \mathrm{P}\left(\frac{\mathrm{E}}{\mathrm{A}}\right)=1$ and $\displaystyle \mathrm{P}\left(\frac{\mathrm{E}}{\mathrm{B}}\right)=\frac{1}{3}$ By Bayes' Theorem\mathrm{P}\left(\frac{\mathrm{~A}}{\mathrm{E}}\right)=\frac{\mathrm{P}(\mathrm{~A}) \cdot \mathrm{P}\left(\dfrac{\mathrm{E}}{\mathrm{~A}}\right)}{\mathrm{P}(\mathrm{~A}) \cdot \mathrm{P}\left(\dfrac{\mathrm{E}}{\mathrm{~A}}\right)+\mathrm{P}(\mathrm{~B}) \cdot \mathrm{P}\left(\dfrac{\mathrm{E}}{\mathrm{~B}}\right)}Let X denote the prize value. Here X can take values of $\displaystyle 8,4$ and 2.\mathrm{P}(\mathrm{X}=8)=\frac{2}{10}, \text { or } \frac{1}{5} \mathrm{P}(\mathrm{X}=4)=\frac{5}{10}, \text { or } \frac{1}{2} \mathrm{P}(\mathrm{X}=2)=\frac{3}{10}

X	$\displaystyle 8$	$\displaystyle 4$	$\displaystyle 2$
$\displaystyle \mathrm{P}(\mathrm{X})$	$\displaystyle \frac{1}{5}$	$\displaystyle \frac{1}{2}$	$\displaystyle \frac{3}{10}$
$\displaystyle \mathrm{XP}(\mathrm{X})$	$\displaystyle \frac{8}{5}$	$\displaystyle \frac{4}{2}$	$\displaystyle \frac{6}{10}$

Hence, Mean value of $\displaystyle \mathrm{X}=\sum \mathrm{X} \mathrm{P}(\mathrm{X})=\frac{8}{5}+2+\frac{6}{10}$