
Probability Calculator
Easily calculate the likelihood of single or multiple events, normal distributions, and winning odds with our free, accurate Probability Calculator.
| Result | ||
|---|---|---|
| Probability of A NOT occuring: P(A') | 0.5 | |
| Probability of B NOT occuring: P(B') | 0.6 | |
| Probability of A and B both occuring: P(A∩B) | 0.2 | |
| Probability that A or B or both occur: P(A∪B) | 0.7 | |
| Probability that A or B occurs but NOT both: P(AΔB) | 0.5 | |
| Probability of neither A nor B occuring: P((A∪B)') | 0.3 | |
| Probability of A occuring but NOT B: | 0.3 | |
| Probability of B occuring but NOT A: | 0.2 | |
Probability
Probability of A: P(A) = 0.5
Probability of B: P(B) = 0.4
Probability of A NOT occuring: P(A') = 1 - P(A) = 0.5
Probability of B NOT occuring: P(B') = 1 - P(B) = 0.6
Probability of A and B both occuring: P(A∩B) = P(A) × P(B) = 0.2
Probability that A or B or both occur: P(A∪B) = P(A) + P(B) - P(A∩B) = 0.7
Probability that A or B occurs but NOT both: P(AΔB) = P(A) + P(B) - 2P(A∩B) = 0.5
Probability of neither A nor B occuring: P((A∪B)') = 1 - P(A∪B) = 0.3
Probability of A occuring but NOT B: P(A) × (1 - P(B)) = 0.3
Probability of B occuring but NOT A: (1 - P(A)) × P(B) = 0.2
Probability
Probability of A occuring 5 time(s) = 0.65 = 0.07776
Probability of A NOT occuring = (1-0.6)5 = 0.01024
Probability of A occuring = 1-(1-0.6)5 = 0.98976
Probability of B occuring 3 time(s) = 0.33 = 0.027
Probability of B NOT occuring = (1-0.3)3 = 0.343
Probability of B occuring = 1-(1-0.3)3 = 0.657
Probability of A occuring 5 time(s) and B occuring 3 time(s) = 0.65 × 0.33 = 0.00209952
Probability of neither A nor B occuring = (1-0.6)5 × (1-0.3)3 = 0.00351232
Probability of both A and B occuring = (1-(1-0.6)5) × (1-(1-0.3)3) = 0.65027232
Probability of A occuring 5 times but not B = 0.65 × (1-0.3)3 = 0.02667168
Probability of B occuring 3 times but not A = (1-0.6)5 × 0.33 = 2.7648e-4
Probability of A occuring but not B = (1-(1-0.6)5) × (1-0.3)3 = 0.33948768
Probability of B occuring but not A = (1-0.6)5 × (1-(1-0.3)3) = 0.00672768
Probability
The probability between -1 and 1 is 0.68268
The probability outside of -1 and 1 is 0.31732
The probability of -1 or less (≤-1) is 0.15866
The probability of 1 or more (≥1) is 0.15866
| CONFIDENCE INTERVALS TABLE | ||
|---|---|---|
| CONFIDENCE | RANGE | N |
| 0.6828 | -1.00000 – 1.00000 | 1 |
| 0.8 | -1.28155 – 1.28155 | 1.281551565545 |
| 0.9 | -1.64485 – 1.64485 | 1.644853626951 |
| 0.95 | -1.95996 – 1.95996 | 1.959963984540 |
| 0.98 | -2.32635 – 2.32635 | 2.326347874041 |
| 0.99 | -2.57583 – 2.57583 | 2.575829303549 |
| 0.995 | -2.80703 – 2.80703 | 2.807033768344 |
| 0.998 | -3.09023 – 3.09023 | 3.090232306168 |
| 0.999 | -3.29053 – 3.29053 | 3.290526731492 |
| 0.9999 | -3.89059 – 3.89059 | 3.890591886413 |
| 0.99999 | -4.41717 – 4.41717 | 4.417173413469 |
There was an error with your calculation.
Last updated: June 26, 2026
Table of Contents
- Probability of Two Events Calculator
- Probability Solver for Two Events
- Probability of a Series of Independent Events
- Probability of a Normal Distribution
- Introduction to Probability
- Rules of Event Operations
- Example: Event Operations
- Complement of an Event
- Intersection of Events
- Independent Events
- Union of Events
- Normal Distribution
- Calculating the Probability of a Normal Distribution
- Example: Normal Distribution
Probability of Two Events Calculator
When you know the probabilities of two independent events, you can use the Probability of Two Events Calculator to determine the likelihood of them occurring together. Simply enter the probabilities of your two independent events (Probability of A and Probability of B) into the tool. The calculator will instantly generate the union, intersection, and other related probabilities, complete with visual Venn diagrams to help you understand the results.
Probability Solver for Two Events
The Probability Solver for Two Events allows you to compute various probabilities for two independent events as long as you have any two input values. This is incredibly useful when the initial probabilities of one or both events are unknown. Not only does this tool provide the final answer, but it also displays the complete, step-by-step calculations for your reference.
Probability of a Series of Independent Events
You can use the Probability of a Series of Independent Events Calculator to evaluate experiments where independent events happen one after another. To find the probability of these consecutive occurrences, simply input the required probabilities and set the number of times the event takes place.
Probability of a Normal Distribution
Our Normal Distribution Probability Calculator is an excellent tool for determining the probability under a normal curve. Simply input the mean μ, standard deviation σ, and your desired boundaries. This normal probability calculator will quickly compute the probability for the specified boundaries and provide confidence intervals across a range of confidence levels.
Introduction to Probability
Probability is the statistical likelihood that a specific event will occur. When an event is absolutely certain to happen, its probability is 1. Conversely, when an event is impossible, its probability is 0. Consequently, any given event's probability always falls between 0 and 1. Using a dedicated probability calculator makes evaluating these chances incredibly simple and accurate.
Rules of Event Operations
In statistics, any specific grouping of an experiment's outcomes is referred to as an event. Essentially, an event is any subset of a sample space. The core operations used to analyze these events are the complement, intersection, and union. Let's explore each of these rules using a practical example.
Example: Event Operations
Suppose your college has various departments, including a Business Faculty. International students are also enrolled at the college. As part of a project, you need to conduct interviews with students, and you decide to start with the first person who walks through the gate. You are aware of the following probabilities:
A = The first student is from the Business Faculty.
B = The first student is an international student.
P(A) = 0.6
P(B) = 0.3
Complement of an Event
The complement of an event includes all the outcomes in a sample space that are not part of that specific event.
For example, the complement of event A means the first student selected is from a department other than the Business Faculty. This can be denoted by $A\prime$ or Aᶜ.
Let's visualize the complement of event A using a Venn diagram.

In the Venn diagram above, the colored area represents the complement of event A.
The total area of the rectangle represents the overall probability of the sample space, which is exactly 1. The space outside of circle A illustrates the probability of the complement of event A. This visual representation allows us to establish the following relationship:
$$P\left(A\right)+P\left(A^\prime\right)=1$$
Therefore,
$$P\left(A^\prime\right)=1-P\left(A\right)$$
Now, let's calculate the corresponding probabilities.
The probability that the first student selected for the interview is not from the Business Faculty is:
$$P\left(A^\prime\right)=1-P\left(A\right)=1-0.6=0.4$$
The probability that the first student selected is not an international student is:
$$P\left(B^\prime\right)=1-P\left(B\right)=1-0.3=0.7$$
Intersection of Events
The intersection of two events, A and B, consists of the set of all common elements shared by both events. The word "AND" is frequently used to indicate this intersection.
In our example, the intersection of event A and event B means selecting a student who is an international student AND belongs to the Business Faculty. This is denoted mathematically as follows:
$$A\cap B$$
Let's look at the intersection of events A and B through a Venn diagram.

In the Venn diagram above, the shaded area highlights the intersection of events A and B.
Now, let's introduce event C: selecting a local student for the interview. We can display events A and C in a new Venn diagram.

Since a student cannot be both local and international simultaneously, selecting an international student inherently excludes the possibility of selecting a local student. Because they cannot occur at the same time, events A and C are considered mutually exclusive.
Mutually exclusive events share no common elements. As a result, their intersection is empty:
$$A\cap C=φ$$
You can calculate the probability of an intersection using a few different methods depending on the known variables. The intersection for events A and B can be written using these formulas:
$$P\left(A\cap B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cup B\right)$$
$$P\left(A\cap B\right)=P(A)× P(B/A)$$
$$P\left(A\cap B\right)=P(B)× P(A/B)$$
Independent Events
Independent events are events whose outcomes do not influence one another. Returning to our example, selecting a student from the Business Faculty has no effect on whether that student is international or local. Therefore, event A and event B are independent events.
When events are completely independent, the probability of one happening does not rely on the occurrence of the other. Thus, the mathematical relationship is expressed as:
$$P(B/A)=P(B)\ and\ P(A/B)=P(A)$$
You can substitute these into our previous equations to simplify finding the probability of intersecting independent events:
$$P\left(A\cap B\right)=P\left(A\right)× P\left(\mathrm{B/A}\right)=P(A)× P(B)$$
$$P\left(A\cap B\right)=P\left(B\right)× P\left(\mathrm{A/B}\right)=P(B)× P(A)$$
This means you can easily find the intersection of two independent events simply by multiplying their individual probabilities:
$$P\left(A\cap B\right)=P\left(A\right)× P\left(B\right)=P(B)× P(A)$$
Given that events A and B are independent, let's determine the probability that the first student chosen for the interview will be both from the Business Faculty and an international student:
$$P\left(A\cap B\right)=P\left(A\right)× P\left(B\right)=0.6× 0.3=0.18$$
Union of Events
The union of two events results in a broader event that contains all elements from either or both original events. The word "OR" is commonly used to describe this type of relationship.
In our running example, the union of events A and B means selecting a student who is either international OR from the Business Faculty (or both). This is denoted as:
$$A\cup B$$
Let's visualize the union of events A and B with a Venn diagram.

In the Venn diagram above, the entire colored area represents the union of events A and B.
To calculate the probability of event A or event B occurring, we add the probabilities of both individual events together and then subtract the probability of their intersection.
The formula for the probability of a union between events A and B is written as follows:
$$P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)$$
We can also modify this to create a specific formula for the union of two independent events. This is especially helpful when the intersection probability is unknown.
Because the events are independent:
$$P\left(A\cap B\right)=P(A)× P(B)$$
Therefore, the union formula becomes:
$$P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P(A)× P(B)$$
Let's calculate the probability of the union of events A and B. In other words, what is the chance of selecting a student who is a business major, an international student, or both at the same time?
$$P(A\cup B)=P(A)+P(B)-P(A\cap B)=0.6+0.3-0.18=0.72$$
Thanks to our Probability of Two Events Calculator and Probability Solver for Two Events, you can perform all these calculations instantly. These tools are perfect for verifying manual math, as they display comprehensive step-by-step calculations alongside the final answers.
Normal Distribution
A normal distribution is a symmetrical, bell-shaped curve. In a perfect normal distribution, the mean, median, and mode are all identical. Exactly 50% of the data falls above the mean, and the other 50% falls below it. As the curve extends away from the mean in both directions, it approaches—but never actually touches—the X-axis. The total area under this curve is always equal to 1.

If a random variable X follows a normal distribution with parameters μ (mean) and σ² (variance), it is written as X ~ N(μ, σ²).
Calculating the Probability of a Normal Distribution
The probability density function of a normal distribution is expressed as:
$$f\left(x\right)=\frac{1}{\sqrt{2π\sigma^2}}× e^\frac{-{(x-\mu)}^2}{2\sigma^2}$$
In this function:
- μ is the mean of the distribution;
- σ² is the variance of the distribution;
- π is approximately 3.14;
- e is approximately 2.7182.
Because there are an infinite number of normal curves, it is impossible to create a single probability table for every possible combination of means and standard deviations. To solve this, statisticians use the standard normal distribution. This is a special case of the normal distribution where the mean is 0 and the standard deviation is 1.
To manually calculate the probability of a normal distribution, you must first transform your specific distribution into a standard normal distribution using a z-score. Once converted, you can use a z-table to find the probability. Alternatively, our Normal Probability Calculator seamlessly functions as a standard normal probability calculator, instantly computing probabilities and confidence intervals without manual chart lookups.
The formula for the z-score is:
$$Z=\frac{X-\mu}{\sigma}$$
The standard normal distribution curve is a powerful tool for solving real-world statistical problems. It is specifically used to determine the probability of continuous variables. A continuous variable can take on an infinite number of values, including decimals—such as height, weight, and temperature.
Let's look at a practical example to understand how to find the probability within a normal distribution.
Example: Normal Distribution
Suppose the results of your statistics course are normally distributed, featuring a mean score of 65 and a standard deviation of 10. If a student is selected at random, determine the probability of the following scenarios:
- The student's score is equal to or above 70.
- The student's score is strictly less than 70.
- The student's score falls between 50 and 70.
Solution
$$P\left(X≥70\right)=P\left(Z≥\frac{70-65}{10}\right)=P\left(Z≥0.5\right)=1-0.6915=0.3085$$
$$P\left(X<70\right)=P\left(Z<\frac{70-65}{10}\right)=P\left(Z<0.5\right)=0.6915$$
$$P\left(50<X<70\right)=P\left(\frac{50-65}{10}<Z<\frac{70-65}{10}\right)=P\left(-1.5<Z<0.5\right)=0.4332+0.1915=0.6247$$
Computing the probability of a normal curve manually involves multiple complex steps and requires reading z-tables. Fortunately, our Normal Distribution Probability Calculator allows you to bypass the hassle. Simply enter four values—the mean, the standard deviation, and the left and right boundaries—and the calculator will compute the precise probability for you instantly.



