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Unit 3: Conditional Probability and Independence Baye’s Theorem

Therefore, P(B | A) = (1/4) 1 (1/2) = 1/2. Notes

So we have P(B | A) > P(B).
On the other hand, if C= { 1,2,3 } and D = {1,2,4}, then P(C) = P(D) = 3/4 and P(C  D) = 112.
Thus,

1/2
P(D | C) =  2/3, and in this case,
3/4
P(D | C) < P(D)
This example illustrates that additional information (about the occurrence of an event) can
increase or decrease the probability of occurrence of another event: We would be interested in
those situations which correspond to the cases when P(B | A) = P(B), as in the following example.

Example 19: We continue with the previous example. But now define H = {1,2} and
K = {l , 3}. Then
P(H) = 1/2, P(K) = 1/2 and P(H  K) = 114.
Hence
In this example, knowledge of the occurrence of H does not alter the probability of occurrence
of K. We call such events, independent events.
Thus, two events A and B are independent, if
P(B | A) = P(B). ...(13)
However, in this definition, we need to have P(A) > 0. Using the definition of P(B | A), we can
rewrite (13) as
P(A  B) = P(A) P(B) ...(14)
which does not require that P(A) or P(B) be positive. We shall now use (14) to define independence
of two events.
Definition 4 : Let A and B be two events associated with the same random experiment. They are
said to be stochastically independent or simply independent if
P(A  B) – P(A) P(B)

So the events A and B in Example 18 are not independent. Similarly, events C and D are also not
independent. But events K and H in Example 19 are independent.
See if you can apply Definition 4 and solve this exercise.

Task Two unbiased dice are rolled. Let
A be the event “odd face with the first die”
1
A be the event “odd face with the second die”
2
B be the event that the score on the first die is 1
1
B be the event that the total score is at most 3.
2
Check the independence of the events
(a) A and A
1 2
(b) B and B
1 2

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