Question

State and prove Multiplication Theorem of Probability.

Accepted Answer

According to the multiplication rule of probability, the probability of occurrence of both the events $A$ and $B$ is equal to the product of the probability of $B$ occurring and the conditional probability that event $A$ occurring given that event $B$ occurs.

If $A$ and $B$ are dependent events, then the probability of both events occurring simultaneously is given by:

$P (A \cap B) = P (B) \times P (\frac{A}{B}) = P (A) \times P (\frac{B}{A})$ , where $P (A) \neq 0, P (B) \neq 0$ .

If $A$ and $B$ are two independent events in an experiment, then the probability of both events occurring simultaneously is given by:

$P (A \cap B) = P (B) \times P (A)$

Proof:

We know that the conditional probability of event $A$ given that $B$ has occurred is denoted by $P (\frac{A}{B})$ and is given by:

$P (\frac{A}{B}) = \frac{P (A \cap B)}{P (B)}$ , where $P (B) \neq 0$ .

$P (A \cap B) = P (B) \times P (\frac{A}{B}) \dots \dots \dots \dots . . (1)$

And, $P (\frac{B}{A}) = \frac{P (B \cap A)}{P (A)}$ , where $P (A) \neq 0$ .

$P (B \cap A) = P (A) \times P (\frac{B}{A})$

Since, $P (A \cap B) = P (B \cap A)$

$P (A \cap B) = P (A) \times P (\frac{B}{A}) \dots \dots \dots \dots \dots . . (2)$

From $(1)$ and $(2)$ we get,

$P (A \cap B) = P (B) \times P (\frac{A}{B}) = P (A) \times P (\frac{B}{A})$ , where $P (A) \neq 0, P (B) \neq 0$

The above result is known as the multiplication rule of probability.

For independent events $A$ and $B$ , $P (\frac{B}{A}) = P (B)$ . The equation $(2)$ can be modified into,

$P (A \cap B) = P (B) \times P (A)$ .

State and prove Multiplication Theorem of Probability.

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