Question

In a test examine either guesses or copies or knows the answer to a multiple choice question with $m$ choices out of which exactly one is correct. The probability that he makes a guess is $\dfrac{1}{3}$ and probability that he copies the answer is $\dfrac{1}{6}$ .The probability that his answer is correct given that the copied it, is $\dfrac{1}{8}$ . If the probability that he knew the answer to the question given that he correctly answered it is 120/141. Find $m$ .

Answer 1

This question is of Bayes theorem. Under given conditions, probabilities of different conditions are given. To solve this question, we need to write all given conditions first then check what else anything extra is needed to satisfy the conditions and then put them in Bayes theorem ‘s formula. By doing so you will get the answer.

Complete step-by-step answer:
Consider the following events:
$X1$ = he guesses the answer
$X2$ = copies the answer
$X3$ = he knows the answer
$X$ his answer is correct
Now $P(X1) = \dfrac{1}{3}$
$P(X2) = \dfrac{1}{6}$
So we don’t know the probability of $X3$ so will calculate it by subtracting the given probability from 1.
$P(X3) = 1 - P(X1) - P(X2) = 1 - \dfrac{1}{3} - \dfrac{1}{6} = \dfrac{1}{2}$
Formula of Bayes Theorem:
$P(\dfrac{A}{B}) = \dfrac{{P(\dfrac{B}{A}).P(A)}}{{P(B)}}$
Not, according to conditions:
$P(\dfrac{X}{{X1}}) = \dfrac{1}{m}$ probability when he guess correct answer
$P(\dfrac{X}{{X2}}) = \dfrac{1}{8}$ probability when he copies correct answer
$P(\dfrac{X}{{X3}}) = 1$ probability when know the answer
$P(\dfrac{{X3}}{x}) = \dfrac{{120}}{{141}}$ probability that his answer is correct
Therefore:
$P(\dfrac{{X3}}{x}) = \dfrac{{P(\dfrac{X}{{X3}}).P(X3)}}{{P(\dfrac{X}{{X1}}).P(X1) + P(\dfrac{X}{{X2}}).P(X2) + P(\dfrac{X}{{X3}}).P(X3)}}$
After putting values:
$P(\dfrac{{X3}}{x}) = \dfrac{{1 \times \dfrac{1}{2}}}{{\dfrac{1}{m} \times \dfrac{1}{3} + \dfrac{1}{8} \times \dfrac{1}{6} + 1 \times \dfrac{1}{2}}} = \dfrac{{120}}{{141}}$
$P(\dfrac{{X3}}{x}) = \dfrac{{\dfrac{1}{2}}}{{\dfrac{1}{{3m}} + \dfrac{1}{{48}} + \dfrac{1}{2}}} = \dfrac{{\dfrac{1}{2}}}{{\dfrac{{24m + m + 16}}{{48m}}}}$
$P(\dfrac{{X3}}{x}) = \dfrac{1}{{\dfrac{{25m + 16}}{{24m}}}} = \dfrac{{24m}}{{25m + 16}} = \dfrac{{120}}{{141}}$
$\dfrac{{24m}}{{25m + 16}} = \dfrac{{120}}{{141}}$
$141m = 125m + 80$
$16m = 80$
$m = 5$

Thus, the number of options is 5.

Note: In such questions, sometimes all quantities are not given. Same thing happened in this question, the probability that “he knows the answer” is not so we calculated it by subtracting the given probability from 1. Remember such cases when something is missing try to find relation between given quantities and then proceed it will be easy.