# Multiplication Theorem Probability

Here you will learn formula for multiplication theorem probability with examples.

Let’s begin –

## Multiplication Theorem Probability

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.

If A and B are two events associated with the given random experiment, then

P($$A\cap B$$) = P(A).P(B/A), if P(A) $$\ne$$ 0

or, P($$A\cap B$$) = P(B).P(A/B), if P(B) $$\ne$$ 0

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($$A\cap B$$) = $$2\over 3$$ – $$1\over 4$$ = $$5\over 12$$