Question

A student answers a multiple-choice question with five alternatives, of which exactly one is correct. The probability that he knows the correct answer is $p,0 < p < 1$ . If he does not know the correct answer, he randomly ticks one answer. Given that he has answered the question correctly, the probability that he did not tick the answer randomly, is,
A. $\dfrac{{3p}}{{4p + 3}}$
B. $\dfrac{{5p}}{{3p + 2}}$
C. $\dfrac{{5p}}{{4p + 1}}$
D. $\dfrac{{4p}}{{3p + 1}}$

Answer 1

Hint : For solving this particular question , we have to use Bayes theorem that is if we have ‘n’ mutually exclusive ( events that cannot happen at same time) and exhaustive (one of the event must occur) $E$ is any event such that $P({E_i}) > 0$ for $0 \leqslant i \leqslant n,$ then
for $0 \leqslant k \leqslant n$ ,
$P({E_i}) = \dfrac{{P({E_i})P\left( {\dfrac{E}{{{E_i}}}} \right)}}{{\sum\limits_{k = 1}^n {P({E_k})P\left( {\dfrac{E}{{{E_k}}}} \right)} }}$ .

Complete step-by-step answer :
Let us consider events ,
Let ${E_1}$ be the student does not know the correct option ,
${E_2}$ be the event that student knows the correct option ,
And $E$ be the event that student answered correctly.
Also probability that the student knows the correct option, $P\left( {{E_2}} \right) = p$
$\therefore$ probability that the student does not knows the correct option, $P\left( {{E_1}} \right) = 1 - p$
Since it is also given in the question, that there are five alternatives for each question, so the probability of student ticking the correct option knowing the correct one and probability of ticking correct one when not knowing the answer will be given respectively,
$P\left( {\dfrac{E}{{{E_2}}}} \right) = 1\;{\text{and}}\;P\left( {\dfrac{E}{{{E_1}}}} \right) = \dfrac{1}{5}$
Therefore, the probability that student did not tick the answer randomly is equals to the probability that student tick the answer correctly,

$$= \dfrac{{P({E_2})P\left( {\dfrac{E}{{{E_2}}}} \right)}}{{P({E_1})P\left( {\dfrac{E}{{{E_1}}}} \right) + P({E_2})P\left( {\dfrac{E}{{{E_2}}}} \right)}} \\\ = \dfrac{{p \times 1}}{{(1 - p)\dfrac{1}{5} + p \times 1}} \\\ = \dfrac{p}{{\dfrac{{1 - p}}{5} + p}} \\\ = \dfrac{p}{{\dfrac{{1 - p + 5p}}{5}}} \\\ = \dfrac{{5p}}{{1 + 4p}} \;$$

Hence, we get the required result.
So, the correct answer is “Option C”.

Note : Questions similar in nature as that of above can be approached in a similar manner and we can solve it easily. We used Bayes theorem that is if we have ‘n’ mutually exclusive ( events that cannot happen at same time) and exhaustive (one of the event must occur) $E$ is any event such that $P({E_i}) > 0$ for $0 \leqslant i \leqslant n,$ then
for $0 \leqslant k \leqslant n$ ,
$P({E_i}) = \dfrac{{P({E_i})P\left( {\dfrac{E}{{{E_i}}}} \right)}}{{\sum\limits_{k = 1}^n {P({E_k})P\left( {\dfrac{E}{{{E_k}}}} \right)} }}$ .