P(A ⋂ B) Formula

Before going onto the formula for the probability of A intersection B, let’s recall what is meant by A intersection B. In set theory, A intersection B is the set containing elements common in both A and B. The intersection of A and B can be represented as A ⋂ B, and it is read as “A intersection B”. Sometimes A intersection B can also be pronounced as A and B. Learn how to find the intersection of sets here, for a better understanding.

In this article, you will learn the meaning and formula for the probability of A and B, i.e. A intersection B along with examples.

P(A ⋂ B) Meaning

P(A ∩ B) indicates the probability of A and B, or, the probability of A intersection B means the likelihood of two events simultaneously, i.e. the probability of happening two events at the same time. There exist different formulas based on the events given, whether they are dependent events or independent events.

As we know, if A and B are two events, then the set A ∩ B denotes the event ‘A and B’. Thus, A ∩ B = {x : x ∈ A and x ∈ B}

Based on the above expression, we can find the probability of A intersection B.

P(A and B) Formula

Consider two events A and B of sample space S, such that their intersection is A ⋂ B.

Now,

n(S) = Total number of elements in the sample space

n(A) = Number of elements in set A

n(B) = Number of elements in set B

n(A ⋂ B) = Number of elements in set A ⋂ B

P(A) = n(A)/n(S)

P(B) = n(B)/n(S)

P(A ⋂ B) = n(A ⋂ B)/n(S)

Where, n(A∩B) = n(A) + n(B) – n(A U B)

Let’s have a look at the formula of the probability of A and B in the case of dependent and independent events.

P(A ⋂ B) Formula for Independent Events

If A and B are independent events, then the probability of A intersection B is given by:

P(A ⋂ B) = P(A) P(B)

Here,

P(A ∩ B) = Probability of both independent events A and B happen together

P(A) = Probability of an event A

P(B) = Probability of an event B

Learn about the independent events of probability here.

Go through the example given below to understand how to find the probability of A intersection B in this case.

Example:

Consider an experiment of throwing a pair of dice.

Sample space = S = {(1, 1), (1, 2),…., (6, 5), (6, 6)}

n(S) = 6

Let A be the event ‘score on the first die is six’.

A = {(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

n(A) = 6

B is the event ‘sum of two scores on two dice is at least 11’.

B = {(5,6), (6,5), (6,6)}

n(B) = 3

Thus, A ∩ B = {(6,5), (6,6)}

Here, the set A ∩ B = {(6,5), (6,6)} may represent the event that the score on the first die is six and the sum of the scores is at least 11.

Thus, the occurrence of A does not affect the occurrence of B.

So, A and B are independent events.

Now,

P(A) = n(A)/n(S) = 6/36

P(B) = n(B)/n(S) = 3/36

P(A ∩ B) = P(A) P(B)

= (6/36) × (3/36)

= 1/72

Therefore, the probability of getting the score on the first die is six and the sum of the scores is at least 11 = 1/72.

P(A ⋂ B) Formula for Mutually Exclusive Events

Two events, say A and B, are called mutually exclusive events if the occurrence of any one of them omits the occurrence of the other event, i.e., if they can not occur simultaneously. In this case, sets A and B are called disjoint. That means the intersection of these two events is an empty set.

i.e. A ∩ B = φ

Thus, P(A ∩ B) = 0

Click here to understand more about mutually exclusive events.

P(A ⋂ B) Formula for Dependent Events

P(A∩B) formula for dependent events can be given based on the concept of conditional probability. In this case, the probability of A intersection B formulas will be:

P(A∩B) = P(A|B)P(B) or P(A∩B) = P(B|A)P(A)

Here,

P(A|B) = P(A given B)

P(B|A) = P(B given A)

Solved Problem

Question:

The probability that the baseball team wins the game is 1/30, and the probability that the football team wins the game is 1/32. What is the probability that both teams win their respective games?

Solution:

Given,

The probability that the baseball team wins the game = P(A) = 1/30

The probability that the football team wins the game = P(B) = 1/32

Here, the probability of each event occurring is independent of the other.

So, P(A ∩ B) = P(A) P(B)

= (1/30) (1/32)

= 1/960

= 0.00104

Therefore, the probability that both teams win their respective games is 0.00104.

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