# Conditional probability regarding multiple choice

Q: A multiple choice exam has 4 choices for each question. A student has studied enough so that the probability they will know the answer to a question is 0.5, the probability that they will be able to eliminate one choice is 0.25, otherwise all 4 choices seem equally plausible.

If they know the answer they will get the question right. If not they have to guess from the 3 or 4 choices. As the teacher you want the test to measure what the student knows. If the student answers a question correctly what’s the probability they knew the answer?

A: Stuck in the process of trying to answer this.

The way I thought about it was using Baye's rule and treating it as a conditional probability.

In that, $$P(answer correct | knew answer) = \frac{P(know answer | answer correct) \cdot P(answer correct)}{P(knew answer)}$$

The probability that the student knows the answer is 0.5 so that gives us the denominator. Figuring out probability for the answer being correct and knowing the answer given a correct answer is a bit more confusing for me. Would appreciate guidance.

Let A represent the event that the question is answered correctly

Let X represent the event that the student knows the correct answer

Let Y represent the event that the student is able to eliminate one choice

Let Z represent the event that the student is not able to eliminate any choice

I interpret the statement ....

the probability they will know the answer to a question is 0.5, the probability that they will be able to eliminate one choice is 0.25, otherwise all 4 choices seem equally plausible.

... to mean ...

$$P(X)=0.5; P(Y)=.25;p(Z)=.25$$

So ...

$$\begin{eqnarray*} P(X|A) &=& \frac{P(A \cap X)}{P(A)} \\&=& \frac{P(A \cap X)}{P(A\cap X)+P(A\cap Y)+P(A\cap Z)} \\&=& \frac{P(A |X)P(X)}{P(A |X)P(X)+P(A |Y)P(Y)+P(A |Z)P(Z)} \\&=& \frac{1 \cdot \frac 12}{1 \cdot \frac 12+\frac 14 \cdot \frac 13+\frac 14 \cdot \frac 14} \\&=& \frac {\frac 12} {\frac{24}{48}+\frac4{48}+\frac3{48}} =\frac{24}{31} \end{eqnarray*}$$

• Ah, yes, I interpreted the problem statement differently in my Answer, but I bet you have the correct interpretation! +1 – Bram28 Jul 12 '17 at 19:36

With

$A$: students answers correctly

$B$: student knew answer

what you want is $P(B|A)$, rather than $P(A|B)$

Now, we are given that:

$$P(B)= \frac{1}{2}$$

and hence

$$P(B^C)=1-\frac{1}{2}=\frac{1}{2}$$

Also:

$P(A|B)=1$ (if they knew the answer they give the correct answer)

and

$$P(A|B^C)=\frac{1}{4}\cdot \frac{1}{3}+\frac{3}{4}\cdot \frac{1}{4}=\frac{1}{12}+\frac{3}{16}=\frac{4}{48}+\frac{9}{48}=\frac{13}{48}$$

(When they don't know the answer there is a $\frac{1}{4}$ probability they can eliminate one of the answers and thus have a $\frac{1}{3}$ probability of guessing correctly between the remaining 3, and there is a $\frac{3}{4}$ probability they can't eliminate any one answer in which case they have a $\frac{1}{4}$ probability of guessing correctly)

Next, we have that:

$$P(A \cap B)=P(A|B)\cdot P(B)=1\cdot \frac{1}{2}=\frac{1}{2}$$

and

$$P(A \cap B^C)=P(A|B^C)\cdot P(B^C)=\frac{13}{48}\cdot \frac{1}{2}=\frac{13}{96}$$

and thus:

$$P(A)=P(A \cap B)+P(A \cap B^C)=\frac{1}{2}+\frac{13}{96}=\frac{61}{96}$$

So finally:

$$P(B|A)=\frac{P(A\cap B)}{P(A)}=\frac{\frac{1}{2}}{\frac{61}{96}}=\frac{48}{61}$$