Probability rules and counting (Statistics) · lesson

The addition rule for "or"

💡 P(A or B) = P(A) + P(B) − P(A and B), and the overlap is 0 when A and B can't both happen.

Adding P(A) and P(B) counts the overlap twice. Subtract it once to fix that.

If A and B can't happen together, they are mutually exclusive. The overlap is 0, so just add.

🧠 Let's work it out

P(A) = 0.45, P(B) = 0.35, and P(A and B) = 0.2. Find P(A or B).

Add: 0.45 + 0.35 = 0.8.
Subtract the overlap: 0.8 − 0.2 = 0.6.

What do you think comes next?

mutually exclusive: can't both happen at the same time
overlap: the outcomes where both events happen

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Complements

💡 The chance something does NOT happen is 1 minus the chance it does.

An event and its complement cover every outcome, so their chances add to 1.

This is often the quick way: find the chance of what you don't want and subtract from 1.

🧠 Let's work it out

A team's chance of winning is 0.28. What is the chance it does not win?

Not winning is the complement.
1 − 0.28 = 0.72.

What do you think comes next?

complement: everything that is not the event

The counting principle

💡 If one choice can be made m ways and another n ways, together they can be made m × n ways.

For a meal with a bread, a meat, and a cheese, every bread can go with every meat and every cheese.

So multiply the number of options at each step.

🧠 Let's work it out

A sandwich uses one of 5 breads, one of 3 meats, and one of 2 cheeses. How many sandwiches can you make?

5 × 3 = 15 bread-and-meat pairs.
Times the cheeses: 15 × 2 = 30.

What do you think comes next?

outcome: one possible result

Arrangements where order matters

💡 When order matters, multiply down: n × (n − 1) × … for each spot you fill.

Putting things in order is a permutation. Each item used is gone for later spots, so the choices shrink by one each time.

Arranging all n items gives n!, read "n factorial": n × (n − 1) × … × 1. Jobs like president and secretary are different, so order matters there too.

🧠 Let's work it out

Seven runners race. How many ways can 1st, 2nd, and 3rd place be filled?

7 choices for 1st.
Then 6 for 2nd and 5 for 3rd.
7 × 6 × 5 = 210.

What do you think comes next?

permutation: an arrangement where order matters
factorial: n! = n × (n − 1) × … × 1

Groups where order doesn't matter

💡 When order doesn't matter, count the ordered ways and divide by the ways to arrange the group: nCr.

A committee of 2 from Ana and Ben is the same as Ben and Ana. So the ordered count is too big.

nCr = (n × (n − 1) × … for r terms) ÷ r!. For a probability, divide the groups you want by all possible groups.

🧠 Let's work it out

A bag has 4 red and 3 blue marbles. You grab 2. What is the chance both are red?

Red pairs: 4C2 = (4 × 3) ÷ 2 = 6.
All pairs: 7C2 = (7 × 6) ÷ 2 = 21.
6/21 = 2/7.

What do you think comes next?

combination: a group where order doesn't matter
nCr: the number of ways to choose r things from n

The multiplication rule for "and"

💡 P(A and B) = P(A) × P(B | A); for independent events, that is just P(A) × P(B).

Independent events don't affect each other, like a coin and a die. Multiply their chances.

Without replacement, the first pick changes what's left. Multiply the first chance by the new chance for the second pick.

Ask: does knowing the first result change the chance of the second? If yes, the events are dependent.

🧠 Let's work it out

A bag has 5 red and 3 blue marbles. Draw two without putting the first back. What is the chance both are red?

P(first red) = 5/8.
Now 4 red are left out of 7, so P(second red) = 4/7.
5/8 × 4/7 = 20/56 = 5/14.

What do you think comes next?

without replacement: the first item is not put back before the next pick
dependent: one event changes the chance of the other