# 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.

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 ${\sum}_{k = 0}^{25} {C}_{25 , k} {\left(.9\right)}^{k} \left({\left(.1\right)}^{25 - k}\right)$

and for Part B: ${\sum}_{k = 0}^{36} {C}_{36 , k} {\left(.7\right)}^{k} \left({\left(.3\right)}^{25 - k}\right)$

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:

$0.75 = \frac{x}{25 + 36} \implies x = 45.75 \to 46$

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:

$\left(\begin{matrix}\text{Part A right" & "Part B at least right" & "multipications} \\ 25 & 21 & 16 \\ 24 & 22 & 15 \\ \vdots & \vdots & \vdots \\ 10 & 36 & 1\end{matrix}\right)$

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

I got roughly 76.3%