Probability Calculator

Based on the design of experiments, the outcome of events can be classified as independent, complement, mutual, non-mutual, union, intersection and conditional probability of events. The best example for the probability of events to occur is flipping a coin or throwing a dice.

For flipping a coin,
the sample space of total outcomes
S = {H,T} where H and T corresponds to Head and Tail of a coin respectively
S = {2} for a single toss of a coin

Complement of events A is the probability function which represents the subset of all members which are not present in A (event of not A) generally denoted by the symbol A^'. P(A^c) = 1 - P(A)

For example, the elements of sample space
S = {pen, pencil, eraser, sharpener, note, box}

If events of A = {pen, pencil, eraser, sharpener}
then the events of A^' = {note, box}

By using this probability calculator, users may find the possibility of complement of events A of sample space S in the statistical experiments.

Use this probability calculator to find the possibility of complement of events A in the sample space S.

In a sample space of statistical experiment, the mutually exclusive events A and B denoted by A ∩ B = null that represents that there is no elements of events A and B are in common.

In a sample space of statistical experiment, the union of events A and B denoted by A U B represents that there is the common elements of events A, B or both A and B.
The probability function of mutual events A and B
P(A U B) = P(A) + P(B) - P(A ∩ B)

For example, the elements of sample space
S = {1, 2, 3, 4, 5, 6}

Events of A = {1, 2, 3, 4}
Events of B = {3, 4, 5, 6}
(A ∩ B) = {3, 4}

A U B = {1, 2, 3, 4} + {3, 4, 5, 6} - {3, 4}
A U B = {1, 2, 3, 4, 5, 6}

By using this probability calculator to find the possibility of successful events A or B or both in the sample space S.

Users may refer the below solved example calculations to learn how to find what is the possibility of single or multiple independent, complement, mutual or non-mutual, union, intersection and condition probability of events to occur in statistical experiments.

Example Problem
What is the probability of getting 2 heads in 3 tosses of a coin?

Solution
Step by step workout :
For tossing a coin 3 times,
The elements of sample space S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
Expected element of Event A = {HHH, HHT, HTH, THH}
P(A) = ?

The total number of possible outcomes in a sample space for tossing a coin 3 times is 8.
The number of expected outcome from tossing a coin 3 times is 4.
Therefore,
P(A) = Number of Successful Events ÷ Total Events of Sample Space
= 2/8
= 0.5
P(A) = 0.5

0.5 is the probability of getting 2 heads in 3 tosses.

Example Problem
What is the probability for getting 5 when throwing a dice?

Solution
Step by step workout :
For a dice,
The total elements of sample space S = {1, 2, 3, 4, 5, 6}
Expected element of Event A = {5}
P(A) = ?

The total number of possible outcomes in a sample space for throwing a dice is 6.
Expected element of Event A is 1

Therefore,
P(A) = Number of Successful Events ÷ Total Events of Sample Space
= 1/6
= 0.16
P(A) = 0.16

0.16 is the probability of getting 5.

Example Problem
What is the probability for getting face card from the deck of cards?

Solution
Step by step workout :
For a getting a face card,
The elements of sample space S for a deck of cards = {spade (Ace, 2, 3, 4, 5, 6, 7, 8, 9, Jack, Queen, King), clubs (Ace, 2, 3, 4, 5, 6, 7, 8, 9, Jack, Queen, King), heart (Ace, 2, 3, 4, 5, 6, 7, 8, 9, Jack, Queen, King), diamond (Ace, 2, 3, 4, 5, 6, 7, 8, 9, Jack, Queen, King)}

S = 52

Expected elements or face cards of Event A = {spade (Jack, Queen, King), clubs (Jack, Queen, King), heart (Jack, Queen, King), diamond (Jack, Queen, King)}
The total group is 4. Three face cards in each groups.
A = 3 x 4 = 12
P(A) = ?

The total number of possible outcomes in a sample space for a deck of cards is 52.
Expected elements of Event A is 12

Therefore,
P(A) = Number of Successful Events ÷ Total Events of Sample Space
P(A) = 12/52
= 3/13
= 0.23
P(A) = 0.23

0.23 is the probability of getting face card from a deck of cards.

Example Problem
What is the probability for getting Ace from the deck of cards?
Solution
Step by step workout :
For a getting a Ace card,
The events of sample space S for a deck of cards = {spade (Ace, 2, 3, 4, 5, 6, 7, 8, 9, Jack, Queen, King), clubs (Ace, 2, 3, 4, 5, 6, 7, 8, 9, Jack, Queen, King), heart (Ace, 2, 3, 4, 5, 6, 7, 8, 9, Jack, Queen, King), diamond (Ace, 2, 3, 4, 5, 6, 7, 8, 9, Jack, Queen, King)}

S = 52
Expected elements or Ace of Events A = {spade Ace, clubs Ace, heart Ace, diamond Ace}
The total group is 4. One Ace card in each groups.
A = 1 x 4 = 4
P(A) = ?
For a single draw, the total number of possible outcomes or events in a sample space for a deck of cards is 52.
Expected events A is 4.

Therefore,
P(A) = Number of Successful Events ÷ Total Events of Sample Space
P(A) = 4/52
= 1/13
= 0.07
P(A) = 0.07

0.07 is the probability of getting Ace from a deck of 52 cards.

Example Problem
What is the probability of getting Ace, if the drawn Ace is not placed back in the deck.

Solution
Step by step workout :
If the drawn Ace is not placed back, the total number of possible outcomes or events in a sample space for a deck of cards is 51.

Expected events A is 3

Therefore,
P(A) = Number of Successful Events ÷ Total Events of Sample Space
P(A) = 3/51
= 1/17
= 0.058
P(A) = 0.058

0.058 is the probability of getting Ace from a deck of 51 cards.

Grade school students, teachers or professionals may use the above solved example calculations to learn how to solve the probability problems, and use this calculator to verify the answers or practice the worksheets or compute the different trials of statistical experiments to analyze the sample space and probability of events.