Question

An exam has two parts. Raul has probability .9 of answering a question from part A correctly and .7 of answering a question from part B correctly. What is the probability that Raul's grade is greater than 75%?

Part A contains 25 questions and part B contains 36 questions. Assume that answering each question correctly are independent events.

Answer 1

I got roughly 76.3%

Explanation:

Let's first of all see that we have binomial probabilities with answering Part A and Part B of the exam. The general form of a binomial probability is:

`sum_(k=0)^(n)C_(n,k)(p)^k((~p)^(n-k))`

For Part A it's `sum_(k=0)^(25)C_(25,k)(.9)^k((.1)^(25-k))`

and for Part B: `sum_(k=0)^(36)C_(36,k)(.7)^k((.3)^(25-k))`

When we sum up the summations, we end up with 1 for both Part A and Part B - in essence, the entirety of the probability results.

What we're looking for are combinations of results where Raul ends up with greater than a 75% mark. Each question appears to be worth the same, and so he needs to get:

`0.75=x/(25+36)=>x=45.75 -> 46`

and so he needs to get 46 or more right to achieve the needed score. And so we will multiply results from Part A and Part B that will get us to 46 or more answers right.

This means that if Raul gets all of Part A right (25 questions), he can get at least 21 Part B questions right and be ok:

`(("Part A right","Part B at least right","multipications"),(25,21,16),(24,22,15),(vdots,vdots,vdots),(10,36,1))`

for a total of 136 multiplications. Which I'll do on a separate spreadsheet.

I got roughly 76.3%