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.
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.
- Step 1: All outcomes: 6 for the first die × 6 for the second die = 36 outcomes (multiplication rule).
- 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.
- Step 3: Probability = 6 ÷ 36 = 1/6.
- 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.
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.
- Step 1: The complement: tails all 3 times. Its probability is 1/2 × 1/2 × 1/2 = 1/8.
- Step 2: 1 − 1/8 = 7/8.
- Check: If you list all 8 outcomes (HHH, HHT, …, TTT), the only one with no heads is TTT, so 7/8 is correct.
- Step 1: The probability of not rolling a 6 on one roll is 5/6.
- Step 2: The probability of no 6 in all 4 rolls: (5/6)⁴ = 625/1296.
- Step 3: 1 − 625/1296 = 671/1296 ≈ 0.518.
- 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.
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.
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.
- Step 1: Coin tosses are independent of each other, so the probability of heads on the 5th toss is 1/2.
- Step 2: The probability of tails all 5 times from the start: (1/2)⁵ = 1/32.
- 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.
📌 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
🔁 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
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🤖 Try asking AI like this
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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.
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Storage is unavailable in this browser, so this lasts only for this page.📐 Math Basics
- 1Numbers and Operations: Fractions and Decimals Revisited
- 2Ratios and Rates: Reading Percentages Correctly
- 3Equations: A Balance for Finding Unknown Numbers
- 4Functions and Graphs: An Eye for Change
- 5Exponents and Logarithms: A World That Grows by Multiplying
- 6Geometry Basics: Area and Pythagoras
- 7Probability: Putting Numbers on Uncertainty
- 8Statistics: Mean, Median, and Variance
- 9Reading Data: The Traps in Graphs
- 10The Math for Understanding AI