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Probability Calculator

Probability of Two Events

To find out the union, intersection, and other related probabilities of two independent events.

Please input values between 0 and 1.

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Probability Solver for Two Events

Please provide any 2 values below to calculate the rest probabilities of two independent events.

Please input values between 0 and 1.

Probability of a Series of Independent Events

ProbabilityRepeat Times Event A Event B

Probability of a Normal Distribution

Use the calculator below to find the area P shown in the normal distribution, as well as the confidence intervals for a range of confidence levels.

P
For negative infinite, use -inf For positive infinite, use inf

Probability of Two Events

Probability is the measure of the likelihood of an event occurring. It is quantified as a number between 0 and 1, with 1 signifying certainty, and 0 signifying that the event cannot occur. It follows that the higher the probability of an event, the more certain it is that the event will occur. In its most general case, probability can be defined numerically as the number of desired outcomes divided by the total number of outcomes. This is further affected by whether the events being studied are independent, mutually exclusive, or conditional, among other things. The calculator provided computes the probability that an event A or B does not occur, the probability A and/or B occur when they are not mutually exclusive, the probability that both event A and B occur, and the probability that either event A or event B occurs, but not both.

Complement of A and B

Given a probability A, denoted by P(A), it is simple to calculate the complement, or the probability that the event described by P(A) does not occur, P(A'). If, for example, P(A) = 0.65 represents the probability that Bob does not do his homework, his teacher Sally can predict the probability that Bob does his homework as follows:

P(A') = 1 - P(A) = 1 - 0.65 = 0.35

Given this scenario, there is, therefore, a 35% chance that Bob does his homework. Any P(B') would be calculated in the same manner, and it is worth noting that in the calculator above, A and B can be independent; i.e. if P(A) = 0.65, P(B) does not necessarily have to equal 0.35, and can equal 0.30 or some other number.

Intersection of A and B

The intersection of events A and B, written as P(A ∩ B), is the joint probability of the two events occurring together. In the case where A and B are mutually exclusive events, P(A ∩ B) = 0. Consider the probability of rolling a 4 and 6 on a single roll of a die; it is not possible. These events would therefore be considered mutually exclusive. Computing P(A ∩ B) is simple if the events are independent: the probabilities of events A and B are multiplied. To find the probability that two separate rolls of a die result in 6 each time:

P(A ∩ B) = P(A) × P(B) = 16 × 16 = 0.0278

AB

The calculator provided considers the case where the probabilities are independent. Calculating the probability is slightly more involved when the events are dependent, and involves an understanding of conditional probability, or the probability of event A given that event B has occurred, P(A|B). Take the example of a bag of 10 marbles, 7 of which are black, and 3 of which are blue. The probability of drawing a blue marble is P(A) = 3/10, and of drawing a black marble P(B) = 7/10. If a blue marble has been withdrawn without replacement, the probability of then drawing a black marble is P(B|A) = 7/9, so the probability of drawing a blue and then a black marble is:

P(A ∩ B) = P(A) × P(B|A) = (3/10) × (7/9) = 0.2333

Union of A and B

In probability, the union of events, P(A ∪ B), essentially involves the condition where any or all of the events being considered occur. Note that P(A ∪ B) can also be written as P(A OR B). In this case, the "inclusive OR" is being used. This means that while at least one of the conditions within the union must hold true, all conditions can be simultaneously true. There are two cases for the union of events; the events are either mutually exclusive, or the events are not mutually exclusive. In the case where the events are mutually exclusive, the calculation of the probability is simpler:

P(A ∪ B) = P(A) + P(B)

A basic example of mutually exclusive events would be the rolling of a die, where event A is the probability that an even number is rolled, and event B is the probability that an odd number is rolled. It is clear in this case that the events are mutually exclusive since a number cannot be both even and odd, so P(A ∪ B) would be 3/6 + 3/6 = 1, since a standard die only has odd and even numbers.

The calculator above computes the other case, where the events A and B are not mutually exclusive. In this case:

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

Using the example of rolling a die again, find the probability that an even number or a number that is a multiple of 3 is rolled. The sample space is S = {1, 2, 3, 4, 5, 6}; the probability of an even number is P(A) = {2, 4, 6} = 3/6, of a multiple of 3 is P(B) = {3, 6} = 2/6, and the intersection is P(A ∩ B) = {6} = 1/6, so P(A ∪ B) = 3/6 + 2/6 - 1/6 = 2/3.

Exclusive OR of A and B

Another possible scenario that the calculator above computes is P(A XOR B): the event that A or B occurs, but not simultaneously. The equation is as follows:

P(A XOR B) = P(A) + P(B) - 2 × P(A ∩ B)

AB

As an example, imagine it is Halloween, and two buckets of candy are set outside the house, one containing Snickers, and the other containing Reese's. Multiple flashing neon signs insist that each trick-or-treater only takes one Snickers OR Reese's but not both! Given a probability of Reese's being chosen as P(A) = 0.65, or Snickers being chosen with P(B) = 0.349, calculate the probability that Snickers or Reese's is chosen, but not both:

0.65 + 0.349 - 2 × 0.65 × 0.349 = 0.999 - 0.4537 = 0.5453

Therefore, there is a 54.53% chance that Snickers or Reese's is chosen, but not both.

Normal Distribution

The normal distribution or Gaussian distribution is a continuous probability distribution that follows the function:

f(x) = 1σ√(2π) e-(x - μ)22σ2

where μ is the mean and σ2 is the variance. Note that standard deviation is typically denoted as σ. Also, in the special case where μ = 0 and σ = 1, the distribution is referred to as a standard normal distribution.

The normal distribution is often used to describe and approximate any variable that tends to cluster around the mean, for example, the heights of male students in a college, the leaf sizes on a tree, the scores of a test, etc. Use the "Normal Distribution" calculator above to determine the probability of an event with a normal distribution lying between two given values. Finding P involves standardizing the two desired values to a z-score by subtracting the given mean and dividing by the standard deviation, as well as using a Z-table to find probabilities for Z. If, for example, it is desired to find the probability that a student at a university has a height between 60 inches and 72 inches tall given a mean of 68 inches tall with a standard deviation of 4 inches, 60 and 72 inches would be standardized as such: (60 - 68) / 4 = -2 and (72 - 68) / 4 = 1.

To determine the probability of a value between 0 and 2 using the table below, find 2 in the first column, since this table by definition provides probabilities between the mean (which is 0 in the standard normal distribution) and the value in question. If, instead, the value in question were 2.11, the 2.1 row would be matched with the 0.01 column and the value would be 0.48257. Also, note that even though the actual value of interest is -2 on the graph, the table only provides positive values. Since the normal distribution is symmetrical, only the displacement is important. Thus, the probability of a value falling between 0 and 2 is 0.47725, while a value between 0 and 1 has a probability of 0.34134. Since the desired area is between -2 and 1, the probabilities are added to yield 0.81859, or approximately 81.859%. Returning to the example, this means that there is an 81.859% chance that a male student at the given university has a height between 60 and 72 inches.

The calculator also provides a table of confidence intervals for various confidence levels. Refer to the Sample Size Calculator for a more detailed explanation of confidence intervals and levels. Briefly, a confidence interval is a way of estimating a population parameter that provides an interval of the parameter rather than a single value. A confidence interval is always qualified by a confidence level, usually expressed as a percentage such as 95%. It is an indicator of the reliability of the estimate.

Z Table from Mean (0 to Z)

z00.010.020.030.040.050.060.070.080.09
000.003990.007980.011970.015950.019940.023920.02790.031880.03586
0.10.039830.04380.047760.051720.055670.059620.063560.067490.071420.07535
0.20.079260.083170.087060.090950.094830.098710.102570.106420.110260.11409
0.30.117910.121720.125520.12930.133070.136830.140580.144310.148030.15173
0.40.155420.15910.162760.16640.170030.173640.177240.180820.184390.18793
0.50.191460.194970.198470.201940.20540.208840.212260.215660.219040.2224
0.60.225750.229070.232370.235650.238910.242150.245370.248570.251750.2549
0.70.258040.261150.264240.26730.270350.273370.276370.279350.28230.28524
0.80.288140.291030.293890.296730.299550.302340.305110.307850.310570.31327
0.90.315940.318590.321210.323810.326390.328940.331470.333980.336460.33891
10.341340.343750.346140.348490.350830.353140.355430.357690.359930.36214
1.10.364330.36650.368640.370760.372860.374930.376980.3790.3810.38298
1.20.384930.386860.388770.390650.392510.394350.396170.397960.399730.40147
1.30.40320.40490.406580.408240.409880.411490.413090.414660.416210.41774
1.40.419240.420730.42220.423640.425070.426470.427850.429220.430560.43189
1.50.433190.434480.435740.436990.438220.439430.440620.441790.442950.44408
1.60.44520.44630.447380.448450.44950.450530.451540.452540.453520.45449
1.70.455430.456370.457280.458180.459070.459940.46080.461640.462460.46327
1.80.464070.464850.465620.466380.467120.467840.468560.469260.469950.47062
1.90.471280.471930.472570.47320.473810.474410.4750.475580.476150.4767
20.477250.477780.478310.478820.479320.479820.48030.480770.481240.48169
2.10.482140.482570.4830.483410.483820.484220.484610.4850.485370.48574
2.20.48610.486450.486790.487130.487450.487780.488090.48840.48870.48899
2.30.489280.489560.489830.49010.490360.490610.490860.491110.491340.49158
2.40.49180.492020.492240.492450.492660.492860.493050.493240.493430.49361
2.50.493790.493960.494130.49430.494460.494610.494770.494920.495060.4952
2.60.495340.495470.49560.495730.495850.495980.496090.496210.496320.49643
2.70.496530.496640.496740.496830.496930.497020.497110.49720.497280.49736
2.80.497440.497520.49760.497670.497740.497810.497880.497950.498010.49807
2.90.498130.498190.498250.498310.498360.498410.498460.498510.498560.49861
30.498650.498690.498740.498780.498820.498860.498890.498930.498960.499
3.10.499030.499060.49910.499130.499160.499180.499210.499240.499260.49929
3.20.499310.499340.499360.499380.49940.499420.499440.499460.499480.4995
3.30.499520.499530.499550.499570.499580.49960.499610.499620.499640.49965
3.40.499660.499680.499690.49970.499710.499720.499730.499740.499750.49976
3.50.499770.499780.499780.499790.49980.499810.499810.499820.499830.49983
3.60.499840.499850.499850.499860.499860.499870.499870.499880.499880.49989
3.70.499890.49990.49990.49990.499910.499910.499920.499920.499920.49992
3.80.499930.499930.499930.499940.499940.499940.499940.499950.499950.49995
3.90.499950.499950.499960.499960.499960.499960.499960.499960.499970.49997
40.499970.499970.499970.499970.499970.499970.499980.499980.499980.49998

Frequently Asked Questions

How do you find the probability of A or B?

For events that are not mutually exclusive, P(A or B) = P(A) + P(B) - P(A and B). For independent events, P(A and B) = P(A) × P(B), so with P(A) = 0.5 and P(B) = 0.4 the union is 0.5 + 0.4 - 0.2 = 0.7.

What does the two-event solver do?

Give it any two of the eight related probabilities of two independent events - P(A), P(B), their complements, the intersection, the union, the exclusive or, and the probability of neither - and it works out the rest. Some pairs, such as the union and the intersection, have two mirror-image solutions, and both are shown.

How is a normal distribution probability calculated?

Standardize each bound to a z-score, z = (x - mean) / standard deviation, then take the difference of the standard normal cumulative probabilities. For a mean of 68 and a standard deviation of 4, the probability between 60 and 72 is the area between z = -2 and z = 1, which is 0.81859.