Rachit Kumar SaxenaManager-Editorial
What is Conditional Probability?
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)
Properties of Conditional Probability
Suppose E and F be the events that belong to the sample space S.
1st 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.
2nd 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).
3rd Property:
If M and M′ are mutually exclusive or disjoint events, then the following statement is true:
P(M′|F) = 1 - P(M|F)
Weightage of Conditional Probability
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.
Illustrative Examples on Conditional Probability
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.
FAQs on Conditional Probability
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.
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