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Conditional probability, math, Marketing Management
Two events A and B are said to be dependent when
B
can occur only when A is known to have occurred (or vice versa). The probability attached to such an event is called the conditional probability and is denoted by
P(A/B)
or, in other words probability of A given that B has occurred.
If two events A and B are dependent, then the conditional probability of B given A is:
P(B/A) = P(AB)/P(A)
Proof: suppose
a
_{1}
is the number of cases for the simultaneous happening of A and B out of
a
_{1}
+ a
_{2}
cases in which A can happen with or without happening of
B.
∴ P(B/A) = a
_{1}
/(a
_{1}
+ a
_{2}
) = (a
_{1}
/n)/(a
_{1}
+ a
_{2}
)/n = P(AB)/P(A)
Similarly it can be shown that
P(A/B) = P[(AB)/P(B)]
The general rule of multiplication in its modified form in terms of conditional probability becomes:
P(A and B) = P(B) × P(A/B)
Or, P(A and B) = P(A) × P(B/A)
For three events A, B and C we have
P(ABC) = P(A) × P(B/A) × P(C/AB)
i.e. the probability of occurrence of A, B and C is equal to the probability of A, times of the probability of B given that A has occurred, times the probability of C given that both A and B have occurred.
Illustration:
a bag contains 5 white and 3 black balls. Two balls are drawn at random one after the other without replacement. Find the probability that both balls drawn are black.
Solution: probability of drawing a black ball in the first attempt is
P(A) = 3/(5 + 3) = 3/8
Probability of drawing the second black ball given that the first ball drawn is black
P(B/A) = 2/(5 + 2) = 2/7
∴
The probability that both balls drawn are black is given by
P(AB) = P(A) × P(B/A) = 3/8 × 2/7 = 3/28
Posted Date: 2/14/2012 1:39:25 PM  Location : United States
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