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The probability of occurrence of the event A as well as B is given by the product of (unconditional) probability of A and conditional probability of B, assuming that A has actually occurred, i.e.
Probability of (A and B) = Probability of A X Conditional Probability of B, assuming A
In symbolic form, P(AB) = P(A).P(B|A)
This is also known as Multiplication Theorem.
Proof: Suppose that a random experiment has n mutually exclusive, exhaustive and equally likely outcomes, among which m cases are favourable to an event A. So, the unconditional probability of A is
P(A) = m / n
Out of these m cases, let m1 cases be favourable to another event B also; i.e. the number of cases favourable to A as well as B is m1. Hence, by the classical definition of probability,
P(AB) = m1 / n
It may be noticed that once event A is known to have actually occurred, the occurrence of B as well is limited to only m1 cases out of m (in which A occurs). So, the conditional probability of B, assuming that A has already occurred, is
P(B|A) = m1 / m
We find that, m1/n = (m/n).(m1/m)
i.e. P(AB) = P(A).P(B|A)
Probability of (A and B) = Probability of A X Conditional Probability of B, assuming A
In symbolic form, P(AB) = P(A).P(B|A)
This is also known as Multiplication Theorem.
Proof: Suppose that a random experiment has n mutually exclusive, exhaustive and equally likely outcomes, among which m cases are favourable to an event A. So, the unconditional probability of A is
P(A) = m / n
Out of these m cases, let m1 cases be favourable to another event B also; i.e. the number of cases favourable to A as well as B is m1. Hence, by the classical definition of probability,
P(AB) = m1 / n
It may be noticed that once event A is known to have actually occurred, the occurrence of B as well is limited to only m1 cases out of m (in which A occurs). So, the conditional probability of B, assuming that A has already occurred, is
P(B|A) = m1 / m
We find that, m1/n = (m/n).(m1/m)
i.e. P(AB) = P(A).P(B|A)
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