Question

In answering on a multiple choice test, a student either know the answer or guesses. Let p be the probability that the students knows the answer and 1-p be the probability that the student guesses. Assume that a student who guesses at the answer will be correct with probability 1/m, where m is the number of multiple choice alternatives. What is the conditional probability that a student knew the answer to a question when he or she answered it correctly?

Answer 1

Answer:

$P(A_{1}|B ) =\frac{mp}{1+p(m-1)}$

Step-by-step explanation:

For mutually exclusive events as A1, A2, A3, etc, Bayes' theorem states:

$P(A|B)= \frac{P(B|A)P(A)}{P(B)}$

P(A|B) is a conditional probability: the likelihood of event A occurring given that B is true.

P(B|A) is a conditional probability: the likelihood of event B occurring given that A is true.

P(A) is the probability that A occurs

P(B) is the probability that B occurs

For this problem:

A1 is the probability that the student knows the answer

A2 is the probability that the student guesses the answer

B is the probability that the student answer correctly

$P(A_{1})=p \\P(A_{2})=1-p \\P(B|A_{1})=1 \\P(B|A_{2})=\frac{1}{m} \\P(B)= P(A_{1})P(B|A_{1}) + P(A_{2})P(B|A_{2})= p+\frac{1-p}{m}$

P(B|A₁) means the probability that the answer is correct when he knew the answer

P(B|A₂) means the probability that the answer is correct when he guessed the answer

P(A₁|B) means the probability that he knew the answer when the answer was correct

Replacing everything in the Bayes' theorem you get:

$P(A_{1}|B)= \frac{P(B|A_{1})P(A_{1})}{P(B)}=\frac{(1)(p)}{p+\frac{1-p}{m}} =\frac{mp}{mp+1-p} =\frac{mp}{1+p(m-1)}$