Question

A student answers a multiple question 5 alternatives of exactly one is correct the probability that he knows the 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 of the answers. 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 for the event must occur)\[E\] is any event such that \[P({{E}_{i}})>0\]for \[0\le i\le n\]then for\[0\le k\le n\], \[P({{E}_{i}})=\dfrac{P({{E}_{i}})P\left( \dfrac{E}{{{E}_{i}}} \right)}{\sum\limits_{k=1}^{n}{P({{E}_{k}})P\left( \dfrac{E}{{{E}_{k}}} \right)}}\]

Complete step by step answer:
Consider the Event,
Let \[{{E}_{1}}\] be the student does not know the correct option,
Let \[{{E}_{2}}\] be the event that student knows the correct option,
And \[E\]be the event that the student answered correctly.
Also probability that the student knows the correct option, \[P({{E}_{2}})=p\]
\[\therefore \]Probability that the student does not knows the correct option, \[P({{E}_{1}})=1-p\]
Since it is also given in the question, that there are five alternatives for each question, so the probability of student ticking correct one when not knowing the answer will be given respectively,
\[P\left( \dfrac{E}{{{E}_{2}}} \right)=1\,\] 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 solve it easily. We used Bayes theorem that is if we have ‘n’ mutually exclusive (events that cannot happen at the same time) and exhaustive (one of the event must occur) \[E\] is any event such that \[P({{E}_{i}})>0\] for \[0\le i\le n\], then for \[0\le k\le n\] by using the formula and substitute in the formula is important for this type of question.