Probability Calculator

This calculator can calculate the probability of two events, as well as that of a normal distribution. Also, learn more about different types of probabilities below. You can also add this tool to your favourite tools list.

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The Probability of Two Events

Probability is a measure of how likely something is to happen. It is expressed as a number between 0 and 1, with 1 denoting certainty and 0 denoting the impossibility of the event. It follows that the likelihood of an event increasing increases with probability, making its occurrence more likely. Probability can be expressed mathematically in the most generic sense as the ratio of desired outcomes to total outcomes. This is further influenced, among other things, by whether the events being investigated are independent, mutually exclusive, or conditional. With the given calculator, you can determine the likelihood that an event A or B will not happen, the likelihood that A or B will happen when they are not mutually exclusive, the likelihood that both A and B will happen, and the likelihood that either A or B will happen but not both.

Complement of A and B

It is easy to determine the complement, or the likelihood that the event indicated by P(A) does not occur, P(A'), given a probability A, denoted by P(A). If, for instance, P(A) = 0.65 denotes the likelihood that Bob will not complete his schoolwork, his teacher Sally can forecast the likelihood that Bob will do it as follows:

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

Hence, there is a 35% likelihood that Bob completes his assignment in light of this circumstance. Any P(B') would be computed in the same way. It's important to note that in the calculator above, P(B') can be independent, meaning that if P(A) = 0.65, P(B') doesn't necessarily have to equal 0.35 but might also equal 0.30 or another value.

Intersection of A and B

The combined probability of at least two occurrences is represented below in a Venn diagram as the intersection of events A and B, also written as P(A B) or P(A AND B). P(A B) = 0 when A and B are occurrences that cannot both occur at the same time. Take into account the impossibility of rolling both a 4 and a 6 on the same die. Hence, these occurrences would be thought to be mutually exclusive. If the occurrences are independent, computing P(A B) is easy. The likelihood of events A and B in this situation are doubled.

The calculator takes into account the scenario in which the probability are unrelated. When two events are interdependent, calculating the likelihood requires knowledge of conditional probability, often known as the probability of event A given that event B has occurred, or P(A|B). Consider a bag of ten marbles, seven of which are black and three of which are blue. Determine the likelihood of drawing a black marble in the event that a blue marble has been removed from the bag without being replaced (lowering the total number of marbles in the bag):

The likelihood of drawing a blue marble

P(A) = 3/10

The probability of drawing a black marble

P(B) = 7/10

Probability of drawing a black marble after drawing a blue marble:

P(B|A) = 7/9

As can be observed, any previous instance in which a black or blue marble was drawn without a replacement has an impact on the likelihood that a black marble will be drawn. In order to ascertain the likelihood of taking a blue and then a black marble from the bag:

With the probabilities determined above, what are the odds of drawing a blue marble first, then a black marble?

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

Union of A and B

The Venn diagram below illustrates how the condition where any or all of the events being studied occur is what is meant by the union of events, or P(A U B). Keep in mind that P(A U B) is another way to write P. (A OR B). The "inclusive OR" is applied in this situation. This means that all of the conditions in the union may be true at the same time, provided that at least one of them does. There are two scenarios for the union of events: either the occurrences are mutually exclusive or they are not. The calculation of the probability is easier when the events are mutually exclusive:

The rolling of a die, where event A represents the likelihood that an even number will be rolled and event B represents the likelihood that an odd number will be rolled, serves as a simple illustration of mutually exclusive occurrences. As a number cannot be both even and odd in this situation, it is obvious that the occurrences are mutually exclusive. Hence, P(A U B) would be 3/6 + 3/6 = 1, since a conventional die only produces odd and even numbers.

The alternative case, in which occurrences A and B are not mutually exclusive, is computed using the aforementioned calculator. In this instance

P(A U B) is equal to P(A) + P(B) - P(A B).

Find the likelihood of rolling an even number or a number that is a multiple of three using the dice-rolling example once more. The six values of the dice are used to represent the set in this instance and are written as follows: S = {1,2,3,4,5,6}
Odds of getting an even number:
P(A) = {2,4,6} = 3/6
Likelihood of a 3+ number:
P(B) = {3,6} = 2/6
The point where A and B meet:
P(A ∩ B) = {6} = 1/6
P(A U B) = 3/6 + 2/6 -1/6 = 2/3
the only OR of A and B
The calculator above can also calculate P(A XOR B), which is depicted in the Venn diagram below. The event that A or B happens, but not both, is known as the "Exclusive OR" operation. The calculation looks like this:

Consider, for illustration, that two candy buckets, one containing Snickers and the other carrying Reese's, are placed outside the house on Halloween. The buckets of candy are surrounded by numerous neon flashing signs warning trick-or-treaters to only take one Snickers OR one Reese's, but not both! Nonetheless, it is unlikely that every child will abide by the flashing neon signs. Calculate the likelihood that Snickers or Reese's will be chosen, but not both, given that the probability of a child exercising restraint while taking into account the negative effects of a potential future cavity is P(unlikely) = 0.001 and P(A) = 0.65 for Reese's and P(B) = 0.349 for Snickers, respectively.

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

Snickers or Reese's have a 54.53% probability of being picked, but not both.

Normal Distribution

A continuous probability distribution that fits the following function is the normal distribution, often known as the Gaussian distribution:

where the mean is and the variance is 2. Keep in mind that standard deviation is usually written as. Moreover, the distribution is referred to as a standard normal distribution in the exceptional case when = 0 and = 1. A typical normal distribution curve is shown in the diagram above along with the calculator.

Any variable that tends to cluster around the mean, like the heights of male college students, the sizes of leaves on a tree, test results, etc., is frequently described and approximated by the normal distribution. Using the "Normal Distribution" calculator above to calculate the likelihood that an event with a normal distribution will fall between two specified values (i.e., P in the graphic above); for instance, the likelihood that a male student's height in a college will range between 5 and 6 feet. According to the above diagram, in order to calculate P, the two desired values must be standardised to a z-score by subtracting the specified mean and dividing by the standard deviation. Moreover, a Z-table must be used to determine the probabilities for Z. If, for instance, it were intended to determine the likelihood that a university student would be between 60 and 72 inches tall given a mean height of 68 inches and a standard deviation of 4 inches, then 60 and 72 inches would be standardized as follows:

Given μ = 68; σ = 4 (60 - 68)/4 = -8/4 = -2 \s(72 - 68)/4 = 4/4 = 1

The region of interest in the normal distribution is shown in the graph above. Use the standard normal Z-table that is provided at the bottom of the page to compute the probability represented by the area of the graph that is shaded. It should be noted that there are various kinds of standard normal Z-tables. The likelihood that a statistic is between 0 and Z, where 0 is the mean in the ordinary normal distribution, is shown in the table below. The desired probability can be determined by subtracting the pertinent values from Z-tables that offer probabilities to the left or right of Z.

As this table by definition provides probabilities between the mean (which is 0 in the standard normal distribution) and the number of options, in this case 2, locate 2 in the first column of the table to get the likelihood of a value between 0 and 2. While the value in question is 2, it should be noted that the table can be read by aligning the 2 row with the 0 column and then reading the value there. If the amount in question were 2.11 instead, the 2.1 row and the 0.01 column would match, and the result would be 0.48257. Furthermore take note that the table only displays positive values, despite the fact that the real value of interest is -2 on the graph. Only the displacement matters because the normal distribution is symmetrical, and a displacement of 0 to -2 or 0 to 2 will have the same area under the curve. Hence, a number between 0 and 2 has a probability of 0.47725, while a value between 0 and 1 has a probability of 0.34134. As the intended range is between -2 and 1, adding the probabilities results in 0.81859, or around 81.859% of the total. Back to the example, this indicates that there is an 81.859% chance that a male student at the specified university is between 60 and 72 inches tall in this instance.

A table of confidence intervals for various confidence levels is also provided by the calculator. For a more thorough discussion of confidence intervals and levels, see the Sample Size Calculator for Proportions. A confidence interval is an approach to population parameter estimation that, rather than providing a single value, provides an interval of the parameter. A confidence level, typically stated as a percentage like 95%, is always used to qualify a confidence interval. It serves as a gauge of how trustworthy the estimate is.