“Professor, I am nervous about the test.”

“Professor, how should we study for the test?”

“Professor, what questions can we expect on the test?”

“Professor, what formulas will you provide on the test?”

“Professor, I am getting stressed out.”

I want to respond.

“**I know how you feel.**”

This response will not help because if I was taking a test, and I was nervous, I don’t need confirmation from my professor that he knows how I feel.

“**I have been in your shoes.**” or “**Been there, done that.**”

Really… besides not helping, this response will make them angry. I am not in their shoes now, and who cares if I have done it before.

“**You are on your own. Go figure it out.**”

No, I am on their side, and we are taking down the data analysis beast together.

Since the responses are not working, I provide a way to think probabilistically.

Let us say there are 15 questions that you need to study for the test. These 15 questions are our sample space.

The probability of the sample space is 1 — **Probability Axiom 1**

The likelihood of being tested from these 15 questions is 1. You cannot escape this unless you decide to skip the test.

Let us now group these questions into concepts. Suppose there are three concepts, Concept A, Concept B and Concept C. Go through each question and identify to which concept it belongs. For instance,

We see that **six** questions belong to concept A, **seven** questions belong to concept B, and **five** questions belong to concept C. **Three** questions have both concept A and concept B. Switch to a visual mode – you will see this.

By now, you should know that the probability of getting a question from concept **A is 6/15 = 0.4**,** **the probability of getting a question from concept **B is 7/15 = 0.466**, and the probability of getting a question from concept **C is 5/15 = 0.333**.

Underlying this number is the rule that the probability of any event in the sample space is between 0 and 1 — **Probability axiom 2**.

In other words, if there were no questions that belonged to concept A, then, the probability would have been **0** (0/15). If all the 15 questions were from concept A, the probability would have been **1** (15/15).

Let us now think about the probability of getting a question from concept A or concept C. **11 questions** in total belong to **concept A or concept C**; six from A and five from C. Hence, the probability of getting a question from concept A or concept C is

In this example, there are no questions that have both concept A and concept C. They are disjoint sets. For **disjoint or mutually exclusive events**, the probability that one or the other of the two events occurs is the sum of their individual probabilities — **Probability axiom 3**.

Let us extend this understanding to the probability of getting a question from concept A or concept B. if you look at the Venn diagram above; you will see that there are **3 + 3 + 4** questions that are either concept A or concept B or both. Hence, the probability of getting a question from concept A or concept B is

Since **three **questions belong to both** A and B **(i.e., the intersection of A and B), it will be sufficient to count them once. In the equation, you are adding the probability of A (**6 questions out of 15**), to the probability of B (**7 questions out of 15**), and subtracting the intersection (**3 common questions out of 15**) to avoid double counting.

Now that the ground rules (**Axioms of Probability**) are set, all you need to do is calculate the probability of getting exactly one question from binomial distribution in a total of 5 questions and hope your prediction will come true! * That was for my friends in the class*.

Oh, I love this game. They anticipate my behavior and predict the probability of getting any question. I anticipate that they do this, hence, try to outsmart them.

I get a sense that my folks figured out the pattern here. **Beavers are Strivers**. Wikipedia also tells me that beavers build dams, canals, and lodges.

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I’ll wrap my comments into words: 1) thank, 2)you.