Probability of guessing the likely number of question that he was sure on a multiple choice test.

Question

Ian completed a multiple-choice test with four options for each question. He was 90% sure of his answer on some of the questions. But on questions in which he wasn't 90% sure, he randomly guessed an answer. If there were 20 questions, and he got 11 right, what was the most likely number of questions that he was 90% sure on?

Note: choices are independent of each

P.S. - This a question written by our TA and may not be exactly logical.

My attempt: The form mutually exclusive and exhaustive set of events. So P(knows,correct)+P(guess,correct)+P(guess,incorrect)=1 and it is given P(knows,correct)=0.9 and his probability in case he doesn't know the answer and guess out of 4 option is 0.25 .

                        So finally-> (1 x .09) + (.25 x .1) 

Answer provided:9

Answer 1

Let $q$ be the number of questions that Ian was confident on. Notice that because there are 20 questions, and Ian always must have been either confident or guessing on every question, he guessed on $20 - q$ questions.

So, the expected number of correct answers is $E(q) = 0.9(q) + 0.25(20-q) = 5 + 0.65q$.

Let's set $E$ equal to 11: $0.65q + 5 = 11 \to 0.65q = 6 \to q = 9.23$, which is closest to $9$.

Answer 2

Easiest approach here is (in the mud) trial and error.

Let $q = $ number of questions that he was 90% sure on.
Let $e(q) = $ expected number of right answers, based on $q$.

Try different values of $q$.

$e(8) = (0.9 \times 8) + (0.25 \times 12) = 10.2.$

Note that as $q$ increases by 1, $e(q)$ increases by $(0.65)$.

Therefore, $e(9) = 10.85$ and $e(10) = 11.5$.

$q = 9$ gives $e(q)$ closest to 11.

Edit
Possible criticism of my answer.

I have made the (intuitive) assumption that the most likely value of $q$ is the one that yields $e(q)$ closest to 11. In fact, my knowledge of probability theory is insufficient to formally prove this quickly; however the assumption does seem reasonable.