# Probability. 100 questions with 4 answers.

I'm currently trying to solve this probability question and I'm quite unsure if my answer is correct or not.

On a multiple-choice exam, there are $100$ questions each with $4$ possible choices.

A student is certain of the correct answer to each question with probability $0.6$ and guesses randomly among the $4$ choices otherwise.

i. What is the probability that the student correctly answers question #$1$?

ii. What is the probability that the student was certain of the answer to question #$1$ given that they got it correct?

My Solutions:

i. $(100\cdot4)\cdot(0.6/100)=400\cdot0.006=2.4$

ii. $2.4/100=0.024$

I think I might be a bit off.

Can anyone tell me if my calculations are correct?

• "certain of the correct answer" = "knows the correct answer" or "thinks he knows the correct answer"??? – barak manos Nov 24 '15 at 20:12
• That is what is said in my exam sample questions. I told my lecturer that part .ii is ambiguous and he just laughed at me. – User199932 Nov 24 '15 at 20:14
• The answer to the first question is most definitely wrong, since probability is by definition a value between $0$ and $1$. – barak manos Nov 24 '15 at 20:18
• The first question is about probability. The answer has to be between 0 and 1. – callculus Nov 24 '15 at 20:18
• As far as I can tell, the fact that there are 100 questions only serves to confuse you. It is not necessary for the solution, or relevant to the calculation of the probability of getting one answer correct. – The Chaz 2.0 Nov 24 '15 at 20:46

Let's assume that "is certain of the correct answer" = "knows the correct answer".

Without assuming that, this "certainty" would be equivalent to "guessing randomly".

What is the probability that the student correctly answers question #$1$?

Split it into disjoint events, and add up their probabilities:

$$0.6+(1-0.6)\cdot\frac14=0.7$$

What is the probability that the student was certain of the answer to question #$1$ given that they got it correct?

Let $A$ denote the event of the student being certain of the answer to question #$1$.

Let $B$ denote the event of the student correctly answering question #$1$.

Then:

$$P(A|B)=\frac{P(A \cap B)}{P(B)}=\frac{0.6}{0.7}=\frac67$$