B Binance · The world's largest crypto exchangeBinance Sign up → AD OKX OKX · A leading global crypto exchangeOKX Sign up → AD
📐 Math Basics · Lesson 7 / 10

Probability: Putting Numbers on Uncertainty

Probability is "the share of desired outcomes among all possible outcomes." Counting the outcomes with nothing missed and nothing counted twice is almost all there is to calculating probability.

⏱ About 17 min ✍️ 4 practice questions 🔁 Unlimited drills Updated 2026-10-08
🎯 By the end of this lesson you can
  • Count outcomes with the multiplication rule and the addition rule
  • Calculate the probability of "at least once" using the complement
  • Find the probability that independent events happen together by multiplying
  • Explain why the gambler's fallacy is a fallacy

1.What Probability Means

When all outcomes are equally likely, the probability of an event is "the number of outcomes in which the event happens ÷ the number of all possible outcomes." The probability of rolling an even number with one die is the 3 even outcomes (2, 4, 6) divided by all 6 outcomes: 3/6 = 1/2.

Probability is always between 0 and 1. 0 means it never happens, and 1 means it is certain to happen. A probability of 1/2 is not a promise that "it will happen once in every two tries"; it means that if you repeat it very many times, the share of times it happens gets close to 1/2. Getting 7 heads in 10 coin tosses is not unusual, but with 10,000 tosses the share of heads usually clusters around 0.5.

P(event) = number of outcomes in which the event happens ÷ number of all outcomes
0 ≤ P ≤ 1

2.Counting Outcomes: The Multiplication and Addition Rules

How many outfits can you make from 3 shirts and 4 pairs of pants? For each shirt you can choose from 4 pairs of pants, so 3 × 4 = 12 outfits. When choices are linked by "and" like this, you multiply the numbers of outcomes (the multiplication rule). A 4-digit numeric passcode has 10 options for each digit, so there are 10 × 10 × 10 × 10 = 10,000 possibilities.

On the other hand, when outcomes are linked by "or" and the two events cannot happen at the same time, you add the numbers of outcomes (the addition rule). On a die, "2 or less, or 5 or more" is 2 outcomes + 2 outcomes = 4 outcomes. But if the two events can overlap, you have to subtract the overlap once. "Even or a multiple of 3" is 3 even outcomes + 2 multiples of 3, minus the 6 that is both, counted once: 4 outcomes (2, 3, 4, 6).

You also need to consider whether order matters. There are 5 × 4 = 20 ways to choose a president and a vice president from 5 people, but to choose 2 representatives without order, (A, B) and (B, A) must count as the same, so there are 20 ÷ 2 = 10 ways.

ExampleFind the probability that the sum of two dice is 7.
  1. Step 1: All outcomes: 6 for the first die × 6 for the second die = 36 outcomes (multiplication rule).
  2. Step 2: List every outcome with a sum of 7, missing none: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). That is 6 outcomes.
  3. Step 3: Probability = 6 ÷ 36 = 1/6.
  4. Check: Whatever the first die shows, there is exactly one face of the second die that makes the sum 7, so the probability must be 1/6. Same conclusion.
Answer1/6

3.Complements: Count "At Least Once" Backward

Counting "the probability that it happens at least once" directly is complicated, because you have to consider every case: happening once, twice, three times, and so on. In that situation, it's simpler to find the opposite, "the probability that it never happens," and subtract it from 1. The event that a given event does not happen is called its complement.

P(at least once) = 1 − P(never)
ExampleFind the probability of getting heads at least once when you toss a coin 3 times.
  1. Step 1: The complement: tails all 3 times. Its probability is 1/2 × 1/2 × 1/2 = 1/8.
  2. Step 2: 1 − 1/8 = 7/8.
  3. Check: If you list all 8 outcomes (HHH, HHT, …, TTT), the only one with no heads is TTT, so 7/8 is correct.
Answer7/8
ExampleFind the probability of rolling a 6 at least once when you roll a die 4 times.
  1. Step 1: The probability of not rolling a 6 on one roll is 5/6.
  2. Step 2: The probability of no 6 in all 4 rolls: (5/6)⁴ = 625/1296.
  3. Step 3: 1 − 625/1296 = 671/1296 ≈ 0.518.
  4. Check: 6⁴ = 1296 and 5⁴ = 625, and 1296 − 625 = 671. The result is a little more than 1/2; note that it is smaller than the intuition "with 4 rolls it should be 4/6" (about 0.667). Probabilities do not simply add up.
Answer671/1296 (about 51.8%)

4.Independent Events: Multiply the Probabilities

Two events are independent when the outcome of one does not affect the probability of the other. A coin toss and a die roll don't affect each other, so the probability of getting heads and also rolling a 6 is 1/2 × 1/6 = 1/12.

When events are not independent, the conditions change before you multiply. If you draw balls from a bag with 3 red balls and 2 blue balls without putting them back, then once the first ball is red, the probability that the second is also red is not 3/5 but 2/4. So the probability that both balls are red is 3/5 × 2/4 = 6/20 = 3/10. Real-world events often look independent but are actually tangled up with each other, like the weather in the same area on the same day.

If independent: P(A and B) = P(A) × P(B)
The birthday problem: in a group of 23 people, the probability that at least one pair shares a birthday is about 50.7%, more than half (assuming 1 year has 365 days and every day is equally likely). It is found by multiplying out the complement, "everyone has a different birthday," and it shows how far intuition and calculation can diverge.

5.The Gambler's Fallacy

You toss a coin and get heads 5 times in a row. Is tails due next? No. A coin has no memory. If it is a fair coin, the probability of heads on the sixth toss is still 1/2. Believing that "the other side is due now" is called the gambler's fallacy.

It's confusing because two questions get mixed up. "Before tossing, the probability of getting 6 heads in a row" is small: (1/2)⁶ = 1/64. But "given that 5 heads have already come up, the probability that the next toss is heads" is 1/2. What has already happened is no longer a matter of probability.

Lottery numbers, "numbers that are due" in a lotto draw, and the idea that "it's my turn to win" after a losing streak are all the same fallacy. On the other hand, when outcomes do affect each other (as with the bag above, where balls are not put back), earlier outcomes change the next probability. The key is to first ask whether the events are independent.

ExampleYou tossed a fair coin 4 times and got tails every time. What is the probability of heads on the 5th toss? And before the first toss, what was the probability of getting "tails all 5 times"?
  1. Step 1: Coin tosses are independent of each other, so the probability of heads on the 5th toss is 1/2.
  2. Step 2: The probability of tails all 5 times from the start: (1/2)⁵ = 1/32.
  3. Check: There are 2⁵ = 32 possible outcomes for 5 tosses, and "TTTTT" is 1 of them, so 1/32 is correct. But once the first 4 are already fixed as tails, only 2 outcomes remain, "TTTTH" and "TTTTT," and they are equally likely, so the answer is 1/2.
Answer1/2 for the next toss; 1/32 for 5 tails in a row from the start

📌 Key points

  • Probability = number of desired outcomes ÷ number of all outcomes (when all outcomes are equally likely)
  • Multiply for "and"; add for "or" when the events can't happen together, and subtract the overlap once if they can
  • Calculate "at least once" as 1 − (probability that it never happens)
  • The probability that independent events happen together is the product of their probabilities
  • Independent trials have no memory; a streak of outcomes does not change the next probability

✍️ Practice questions

Answer first, then open "Answer and explanation".

Q1. When you toss two coins, what is the probability that both land heads?

⭕ Correct

❌ Not quite — see the explanation

Answer and explanation
Answer ③ 1/4

There are 4 outcomes in all (HH, HT, TH, TT), and 1 of them is both heads, so it is 1/4. Since the tosses are independent, you can also get it as 1/2 × 1/2 = 1/4. 1/3 comes from wrongly counting HT and TH as one, giving 3 outcomes.

Q2. Find the probability of getting heads at least once when you toss a coin 2 times.

Answer and explanation
Answer 3/4

The probability of the complement, "tails both times," is 1/4, so the answer is 1 − 1/4 = 3/4.

Q3. Red has come up 8 times in a row on a roulette wheel. If the wheel is fair, which statement about the next spin is correct?

⭕ Correct

❌ Not quite — see the explanation

Answer and explanation
Answer ③ The probability for the next spin is unchanged, regardless of earlier results

Each spin of a fair roulette wheel is independent. Earlier results don't change the probability of the next spin. Believing black is "due" is the gambler's fallacy.

Q4. What is the probability that the sum of two dice is 12?

⭕ Correct

❌ Not quite — see the explanation

Answer and explanation
Answer ③ 1/36

The only outcome with a sum of 12 is (6, 6), and there are 36 outcomes in all, so it is 1/36.

🔁 Unlimited practice

Problems are generated endlessly. Type your answer and press "Check" to have it graded right away, or press "Show solution" to see a step-by-step solution in the same order as the lesson. You can choose the difficulty, and your streak of correct answers is counted.

  • Probabilities with one or two dice
  • Counting outcomes (the multiplication rule, choosing with and without order)
  • Getting heads k times in n coin tosses
  • Complements: "at least once"
  • Multiplying independent events
  • Hard: drawing twice without putting the first back

These drills are generated in your browser with JavaScript, which is not running right now. Use the examples and practice questions above, then reopen this page with JavaScript turned on.

🤖 Try asking AI like this

Copy a prompt and replace the [ ] parts with your own situation. Don't take the answer on trust — check it against this lesson.

When checking whether you missed any outcomes in a probability solution

For this probability problem, first show me a table listing every possible outcome, with none missing. Then mark the desired outcomes, count them, and find the probability. Also check whether the answer from the formula matches the answer from listing and counting. Problem: [problem]

When you want to test your probability intuition

Give me 5 probability problems that people often get wrong by intuition. First ask for my intuitive answer, and when I answer, explain with an exact calculation why the intuition misses.

When verifying a probability an AI gave you

Tell me whether you assumed the events were independent in the probability calculation you just did. If they aren't independent, show how the calculation changes, and assess whether the independence assumption is realistic in this situation.
References
  • Standard middle and high school math textbook content (counting, probability)

Reached every goal above? Mark the lesson complete.

📐 Math Basics

  1. 1Numbers and Operations: Fractions and Decimals Revisited
  2. 2Ratios and Rates: Reading Percentages Correctly
  3. 3Equations: A Balance for Finding Unknown Numbers
  4. 4Functions and Graphs: An Eye for Change
  5. 5Exponents and Logarithms: A World That Grows by Multiplying
  6. 6Geometry Basics: Area and Pythagoras
  7. 7Probability: Putting Numbers on Uncertainty
  8. 8Statistics: Mean, Median, and Variance
  9. 9Reading Data: The Traps in Graphs
  10. 10The Math for Understanding AI
📚 Worth reading
🧠What Generative AI Does and Where It Fails→ ✍️How to Write a Good Prompt→ 🔍Checking AI Answers→ 📚Using AI for Study and Work Without Plagiarism→
← Foundations for the AI Era