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Conditional Probability is considered one of the most asked concept.
33 Questions around this concept.
It is given that the events are such that Then is:
A bag contains 8 balls, whose colours are either white or black. 4 balls are drawn at random without replacement and it was found that 2 balls are white and other 2 balls are black. The probability that the bag contains equal number of white and black balls is :
Conditional Probability
Conditional probability is a measure of the probability of an event given that another event has already occurred. If A and B are two events associated with the same sample space of a random experiment, the conditional probability of the event A given that B has already occurred is written as P(A|B), P(A/B) or .
The formula to calculate P(A|B) is
For example, suppose we toss one fair, six-sided die. The sample space S = {1, 2, 3, 4, 5, 6}. Let A = face is 2 or 3 and B = face is even number (2, 4, 6).
Here, P(A|B) means that the probability of occurrence of face 2 or 3 when an even number has occurred which means that one of 2, 4 and 6 has occurred.
To calculate P(A|B), we count the number of outcomes 2 or 3 in the modified sample space B = {2, 4, 6}: meaning the common part in A and B. Then we divide that by the number of outcomes in B (rather than S).
Properties of Conditional Probability
Let A and B are events of a sample space S of an experiment, then
Property 1 P(S|A) = P(A|A) = 1
Proof:
Property 2 If A and B are any two events of a sample space S and C is an event of S such that P(C) ≠ 0, then
P((A ∪ B)|C) = P(A|C) + P(B|C) – P((A ∩ B)|C)
In particular, if A and B are disjoint events, then
P((A∪B)|C) = P(A|C) + P(B|C)
Proof:
Property 3 P(A’|B) =1 - P(A|B) , if P(B) ≠ 0
Proof:
From Property 1, we know that P(S|B) = 1
⇒ P((A ∪ A’)|B) = 1 (as A ∪ A’ = S )
⇒ P(A|B) + P(A’|B) = 1 (as A and A’ are disjoint event)
⇒ P(A’|B) = 1 - P(A|B)
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