Formulas for Conditional Probability – Multiplication Theorem

Here, you will learn formulas for conditional probability and multiplication theorem with examples.

Let’s begin –

Formulas for Conditional Probability

Let A and B be two events such that P(A) > 0. Then P(B|A) denotes the Conditional Probability of B given that A has occurred. Since A is known to have occurred, it becomes the new sample space replacing the original S. From this we led to definition

P(B|A) = \({P(A\cap B)}\over P(A)\) = which is called Conditional Probability of B given A.

Formula for Multiplication Theorem

P(\(A\cap B\)) = P(A).P(B|A)

It is called compound probability or multiplication theorem. It says the probability that both A and B occur is equal to the probability that A occur times the probability that B occurs given that A has occurred.

Note :

For any three events \(A_1\), \(A_2\), \(A_3\) we have

P(\(A_1\cap A_2\cap A_3\)) = P(\(A_1\)).P(\(A_2|A_1\)).P(\(A_3|(A_1\cap A_2\)))

Example : A bag contains 3 red, 6 white and 7 blue balls. Two balls are drawn one by one. What is the probability that first ball is white and second ball is blue when first drawn ball is not replaced in the bag?

Solution : Let A be the event of drawing first ball white and B be the event of drawing second ball blue.
Here A and B are dependent events.

P(A) = \(6\over 16\), P(B|A) = \(7\over 15\)

P(AB) = P(A).P(B|A) = \(6\over 16\)x\(7\over 15\) = \(7\over 40\)

DE MORGAN’S LAW :

If A & B are two subsets of a universal set U, then

(i)  \(A\cup B)^c\) = \(A^c\cap B^c\)

(ii)  (\(A\cap B)^c\) = \(A^c\cup B^c\)

Note :

(i)  (\(A\cup B\cup C)^c\) = \(A^c\cap B^c\cap C^c\) & (\(A\cap B\cap C)^c\) = \(A^c\cup B^c\cup C^c\)

(ii)  (ii) \(A\cup (B\cap C)\) = (\(A\cup B)\cap (A\cup C)\) & \(A\cap (B\cup C)\) = (\(A\cap B)\cup (A\cap C)\)

Example : If A and B are two events such that P(\(A\cup B)\) = \(3\over 4\), P(\(A\cap B)\) = \(1\over 4\) and \(P(A^c)\) = \(2\over 3\). Then find-(i) P(A) (ii) P(B) (iii) \(P(A\cap B^c)\) (iv) \(P({A^c}\cap B)\)

Solution : P(A) = 1 – P(\(A^c\)) = 1 – \(2\over 3\) = \(1\over 3\)

P(B) = P(\(A\cup B\)) + P(\(A\cap B\)) – P(A) = \(3\over 4\) + \(1\over 4\) – \(1\over 3\) = \(2\over 3\)

P(\(A\cap B^c\)) = P(A) – P(\(A\cap B\)) = \(1\over 3\) – \(1\over 4\) = \(1\over 12\)

P(\({A^c}\cap B\)) = P(B) – P(\(A\cap B\)) = \(2\over 3\) – \(1\over 4\) = \(5\over 12\)

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