Conditional Probability: Overview, Questions, Preparation

When the occurrence of one event is related to the occurrence of another event, you can find its probability with conditional probability. If event A occurs only if event B occurs first, it is called conditional probability. It is represented as P(A|B).

The Formula of Conditional Probability

Conditional probability P(A|B) can be expressed mathematically as:

P(A|B) = N(A ∩ B)/N(B),

where N(A ∩ B) represents the elements that are shared by both A and B and N(B) are the elements that are in B. N(B) can never be 0.

If N is the number of elements in the sample space, then the below statement is also true:

P(A|B) = N(A ∩ B)/N / N(B)/N ---- (1)

The probability of any event is the ratio obtained by dividing the number of expected outcomes with the total outcomes. 

Therefore, N(A ∩ B)/N = P (A ∩ B) and N(B)/N = P(B) ----- (2)

From statements 1 and 2, we get:

P(A|B) = P (A ∩ B)/P(B)

Suppose E and F be the events that belong to the sample space S. 

1^st Property: 

P(S|F) = P (S ∩ F)/P(F) = P(F)/P(F) = 1

and P(F|F) = P (F ∩ F)/P(F) = P(F)/P(F) = 1

Therefore, P(S|F) and P(F|F) = 1.

2^nd Property:

If the sample space comprises events A and B, then the below statement holds true:

P((A ∪ B)|F) = P(A|F) + P(B|F) – P((A ∩ B)|F).

3^rd Property: 

If M and M′ are mutually exclusive or disjoint events, then the following statement is true:

P(M′|F) = 1 - P(M|F)

You will learn conditional probability in detail in Class XII in ‘Probability’. The chapter has a weightage of around 10 marks, which means that conditional probability will have a maximum weightage of around 4 to 5 marks. 

1. Find the value of P(A|B) if P(A) = ⅚ and P(B) = 7/6 and P (A ∩ B) = ⅙ 

Solution. P(A|B) = P(A ∩ B)/P(B) = ⅙ / 7/6 = 1/7 

2. A dice is thrown twice, and the numbers appearing are observed. If the sum of the numbers appearing on the dice is 5, what is the probability of number 2 appearing at least once?

Solution. Suppose that F is the event when the sum of the numbers is 5 

F = { (1,4), (2,3), (3,2), (4,1)} 

Therefore, P(F) = 4/36 = 1/9

Now, let E be the event that the number 2 appears at least once. 

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

P(E) = 11/36

E ∩ F = {(2.3), (3.2)}

P(E ∩ F) = 2/36 = 1/18 

Therefore, 

P(E|F) = P(E ∩ F)/P(F) = 1/18 /  1/9 = ½.

Q: What are disjoint events?

Q: Who found the concept of conditional probability?

Q: What is the difference between joint probability and conditional probability?

Q: What is the Bayes Theorem?

Q: Give the formula for Bayes Theorem.