Probability answers: ”How likely is this to happen?” We write it as a number between 0 and 1. A probability of 0 means impossible. A probability of 1 means it definitely happens. Most things are somewhere in between — and we usually write them as fractions.

This works when every outcome has the same chance. A fair die, a fair coin, or reaching into a bag without looking — all equally likely.

Example 1: You roll one fair die. What’s the chance of rolling a 4? Only 1 side shows a 4, and there are 6 sides total. P(rolling a 4) = 1/6.

Example 2: A bag has 3 red candies and 2 blue candies. You grab one without looking. P(picking red) = 3/5, because 3 out of 5 are red.

Always check: does each outcome have the same chance? If yes, you’re good to use this formula.

Sometimes it’s way easier to count what you don’t want than what you do. This is called the complement.

If we call the chance that event A happens P(A), then the chance it doesn’t happen is:

This is especially powerful for ”at least one” problems. Rather than listing every way you could succeed, find the one way you’d get nothing — then subtract.

Example: Flip 3 coins. What’s the chance of getting at least one heads? The only way you get zero heads is all three tails: P(all tails) = ½ × ½ × ½ = 1/8. So P(at least one heads) = 1 − 1/8 = 7/8.

Shortcut: whenever you see “at least one” in a problem, think complement first — find the “none” case and subtract from 1.

What if you want two things to both happen? Like drawing a red card and then rolling a 6? When two events don’t affect each other, you multiply their probabilities.

With replacement (independent)

You draw a marble from a bag, write down the color, and put it back before drawing again. Because the bag is the same both times, the two draws don’t affect each other — they’re independent.

P(red on 1st and red on 2nd) = P(red) × P(red)

Without replacement (dependent)

This time you keep the first marble out. Now the bag has one fewer marble, so the probabilities change for the second draw. The events are now dependent.

Example: A standard deck has 52 cards and 4 aces. You draw an ace and set it aside. Now there are 3 aces left in 51 cards.

P(both aces, no replacement) = (4/52) × (3/51)

P(both aces, with replacement) = (4/52) × (4/52)

Ask yourself: does the first pick go back in? If yes, same odds each time. If no, update the numbers for the second pick.

What if you want either event A or event B to happen? Whether you add or adjust depends on one question: can they both happen at the same time?

When they can't both happen (no overlap)

Rolling a 2 or a 5 on one die — a single roll can never be both. Add the probabilities directly.

Example: P(rolling 2 or 5) = 1/6 + 1/6 = 2/6 = 1/3.

When they can both happen at once (overlap)

Drawing a heart or a face card — the King of Hearts is both! Adding the probabilities directly would count those overlap cards twice. Subtract the overlap once to correct for this:

Example: P(heart or face card) = 13/52 + 12/52 − 3/52 = 22/52. (There are 3 face cards that are also hearts: J♥ Q♥ K♥.)

Spot the word "or" in a problem — then check if the two events can happen together. If yes, subtract the overlap to avoid counting it twice.

Imagine you could play the same carnival game 100 times. Would you win more money than you spend, or slowly lose it? Expected value answers this — it’s the long-run average outcome if you repeated something over and over.

To calculate it: multiply each possible result by how likely it is, then add everything up.

Example 1 — rolling a die: Each face (1 through 6) has a 1/6 chance. E = (1 + 2 + 3 + 4 + 5 + 6) ÷ 6 = 21/6 = 3.5. Roll it many times and the average will be close to 3.5.

Example 2 — carnival game: You win $10 with probability 1/5, or lose $2 with probability 4/5. E = 10 × (1/5) + (−2) × (4/5) = 2 − 1.6 = $0.40. On average you’d make 40 cents per game.

Expected value is about the long run. You might win or lose on any single try — but average it out over many tries and you’ll get close to E.

• Basic probability: Probability = favorable ÷ total. We write this as P(event). Works when all outcomes are equally likely.
• Complement trick: P(not A) = 1 − P(A). For “at least one,” find P(none) and subtract from 1.
• Two events in a row: Multiply probabilities. With replacement → same odds each time. Without replacement → odds shift for the second pick.
• Expected value: Add up (each outcome × its probability). This is your long-run average.

Bonus facts to know: Two dice have 36 total outcomes. The most common sum is 7 — so P(sum is 7) = 6/36 = 1/6. Flip n coins and there are 2ⁿ total outcomes.

Ready? Head to the quiz and test yourself!

Pick an answer for each question, then click Submit answers to see how you did — every question includes an explanation!