Probability puts a number on how likely something is. It runs from 0 (it cannot happen) to 1 (it always happens), and it is written as a fraction, a decimal or a percent. A fair coin lands heads with probability $latex \frac{1}{2}$, 0.5 or 50%, and all three say the same thing.
This page builds the ideas in order: the sample space, events, the basic formula, the complement, the addition rule and independent events, using dice, coins and playing cards. Two interactive tools let you count outcomes on a grid and watch random trials settle on the theoretical value.
Experiments, outcomes and the sample space
A random experiment is a process whose result cannot be predicted with certainty, although all the possible results are known. Each possible result is an outcome. The set of all outcomes is the sample space, written $latex S$.
- One coin: $latex S=\{H,T\}$, 2 outcomes.
- One die: $latex S=\{1,2,3,4,5,6\}$, 6 outcomes.
- Two coins: $latex S=\{HH,HT,TH,TT\}$, 4 outcomes.
- Two dice: 36 ordered pairs, from $latex (1,1)$ to $latex (6,6)$.
- A deck of cards: 52 cards, 4 suits (hearts and diamonds are red, clubs and spades are black) with 13 ranks each (A, 2 to 10, J, Q, K). The J, Q and K are the 12 face cards.
Counting the sample space correctly is half of any probability problem. When two dice are told apart, $latex (2,5)$ and $latex (5,2)$ are different outcomes.
Events and the basic formula
An event is a set of outcomes we care about. "The die shows an even number" is the event $latex A=\{2,4,6\}$. When all outcomes are equally likely:
$latex P(A)=\frac{\text{number of favorable outcomes}}{\text{number of possible outcomes}}=\frac{n(A)}{n(S)}$
- $latex 0 \le P(A) \le 1$ always. An impossible event has probability 0 and a certain event has probability 1.
- The probabilities of all the outcomes of the experiment add up to 1.
For the even number on a die, $latex P(A)=\frac{3}{6}=\frac{1}{2}$.
The phrase "equally likely" matters. With two dice, a sum of 7 is far more likely than a sum of 2, because 7 can be made in six ways and 2 in only one. That is why probability problems count outcomes (pairs), not sums.
See the two-dice sample space
Each cell is one outcome and shows the sum. Choose an event and count the highlighted cells: the probability is that count divided by 36. Tick the box to see the complement.
Sample space of two dice
Each cell is one outcome. Pick an event and count the highlighted cells.
The complement: "not A"
The complement of an event $latex A$ is every outcome that is not in $latex A$. Since an outcome is either in $latex A$ or not, the two probabilities add up to 1:
$latex P(A') = 1 - P(A)$
It saves work for "at least one" questions, because "none" is a single case. The chance of at least one head in 3 flips is 1 minus the chance of all tails: $latex 1-\frac{1}{8}=\frac{7}{8}$.
The addition rule: "A or B"
$latex P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$
Subtracting the overlap avoids counting the outcomes that are in both events twice. If the events are mutually exclusive (they cannot happen together), the overlap is 0 and you just add.
Independent events: "A and B"
Two events are independent when one does not change the probability of the other. Then:
$latex P(A \text{ and } B) = P(A)\cdot P(B)$
A coin and a die do not affect each other, so the chance of heads and a 5 is $latex \frac{1}{2}\cdot\frac{1}{6}=\frac{1}{12}$. A tree diagram shows the same multiplication along each path, here for two coin flips:
Events are dependent when the first changes the second. Drawing cards without putting them back is the classic case: after one ace is gone, the deck has 51 cards and 3 aces left, so $latex P(\text{second ace}\mid\text{first ace})=\frac{3}{51}$.
Watch the frequency settle on the probability
The relative frequency is successes divided by trials. With few trials it jumps around. With thousands it settles near the theoretical probability, a fact called the law of large numbers. It does not mean the next trial is due to balance things out: each flip starts fresh.
Simulator: watch the frequency settle on the probability
Pick an experiment, run trials and watch the running frequency approach the theoretical probability.
Run trials
Worked examples
EXAMPLE 1
A fair die is rolled. Find the probability of a number greater than 4.
The favorable outcomes are 5 and 6, so $latex P=\frac{2}{6}=\frac{1}{3}$, about 33.33%.
EXAMPLE 2
Two dice are rolled. Find the probability that the sum is 7, and that it is at least 10.
The sum 7 comes from (1,6), (2,5), (3,4), (4,3), (5,2), (6,1): 6 of 36, so $latex P=\frac{6}{36}=\frac{1}{6}$.
A sum of at least 10 means 10 (3 ways), 11 (2 ways) or 12 (1 way): 6 of 36, so $latex P=\frac{6}{36}=\frac{1}{6}$ as well.
EXAMPLE 3
A die is rolled 4 times. Find the probability of getting at least one 6.
Use the complement. The chance of no 6 in one roll is $latex \frac{5}{6}$, and the rolls are independent, so $latex P(\text{no 6})=\left(\frac{5}{6}\right)^4=\frac{625}{1296}$.
$latex P(\text{at least one 6})=1-\frac{625}{1296}=\frac{671}{1296}\approx 0.5177$, about 51.8%.
EXAMPLE 4
One card is drawn from a standard deck. Find the probability that it is a heart or a king.
There are 13 hearts and 4 kings, and the king of hearts is in both groups: $latex P=\frac{13}{52}+\frac{4}{52}-\frac{1}{52}=\frac{16}{52}=\frac{4}{13}$, about 30.77%.
EXAMPLE 5
Two cards are drawn from a standard deck. Find the probability that both are aces, (a) putting the first card back, (b) keeping it out.
(a) Independent: $latex \frac{4}{52}\cdot\frac{4}{52}=\frac{1}{169}\approx 0.00592$.
(b) Dependent: $latex \frac{4}{52}\cdot\frac{3}{51}=\frac{1}{221}\approx 0.00452$. Without replacement the chance is smaller.
Common mistakes
- Assuming outcomes are equally likely when they are not. With two coins, "0, 1 or 2 heads" are not three equal cases: one head happens in 2 of the 4 outcomes.
- Counting sums instead of pairs. A sum of 7 is six outcomes of 36, not one outcome of 11.
- Adding when you should multiply. "A and B" for independent events multiplies. "A or B" adds, minus the overlap.
- Ignoring the overlap in "or". Heart or king is 16 cards, not 17.
- Forgetting that drawing without replacement changes the odds. Update the numerator and denominator after each draw.
- Believing in "due" results. After five heads in a row, the next flip is still 50% heads.
Solved exercises
EXERCISE 1
A die is rolled. What is the probability of a prime number?
Solution
The primes on a die are 2, 3 and 5, so $latex P=\frac{3}{6}=\frac{1}{2}$.
EXERCISE 2
Two coins are flipped. What is the probability of exactly one head?
Solution
The outcomes with one head are HT and TH, 2 of 4: $latex P=\frac{2}{4}=\frac{1}{2}$.
EXERCISE 3
One card is drawn from a standard deck. What is the probability that it is a red face card?
Solution
Face cards are J, Q, K in each suit. The red suits have 3 each: 6 cards. $latex P=\frac{6}{52}=\frac{3}{26}\approx 0.1154$.
EXERCISE 4
Two dice are rolled. What is the probability that the sum is 9?
Solution
The pairs are (3,6), (4,5), (5,4), (6,3): 4 of 36. $latex P=\frac{4}{36}=\frac{1}{9}$.
EXERCISE 5
A die is rolled twice. What is the probability of at least one 6?
Solution
Complement: no 6 in two rolls is $latex \frac{5}{6}\cdot\frac{5}{6}=\frac{25}{36}$. So $latex P=1-\frac{25}{36}=\frac{11}{36}\approx 0.3056$.
EXERCISE 6
A bag holds 5 red and 3 blue marbles. Two marbles are drawn without replacement. What is the probability that both are blue?
Solution
$latex P=\frac{3}{8}\cdot\frac{2}{7}=\frac{6}{56}=\frac{3}{28}\approx 0.1071$.
Frequently asked questions
What is the difference between probability and odds?
Probability compares successes with all outcomes. Odds compare successes with failures. For a six on a die, the probability is 1 in 6 and the odds are 1 to 5.
Can a probability be greater than 1 or negative?
No. If you get one, a count is wrong. Check the sample space and the favorable outcomes.
After five heads in a row, is tails more likely?
No. Coin flips are independent, so every flip is 50% heads. The simulator above shows that the frequency settles by dilution, not by correction.
How do I turn a probability into a percent?
Multiply by 100. For example, $latex \frac{1}{6}\approx 0.1667$ is about 16.67%.
Related calculators
- Dice and coin probability simulator
- Combinations calculator (nCr)
- Permutations calculator (nPr)
- Percentage calculator
More tools are in the calculators hub.
See also
- Permutations and Combinations - Definitions, Formulas and Examples
- Difference Between Permutations and Combinations
- Factorials - Example and Practice Problems
- Combinations - Example and Practice Problems
- Applications of Pascal's Triangle
- Permutations - Example and Practice Problems
- Frequency tables and histograms: how to build and read them
- Variance and standard deviation step by step