P(A/B) denotes the probability of event A happening, given that event B has happened. It is the conditional probability of A, given B. While calculating P(A/B), we assume that event B has occurred. It implies that the outcomes favourable to B become the total outcomes and hence outcomes favourable to P(A/B) are outcomes common to A and B.
P(A/B) = P(A given B)
= (Total no. of favourable cases)/(Total no. of cases)= n(A∩B)/n(B) = P(A∩B)/P(B).
Thus P(A/B) = P(A∩B)/P(B).
=> (P(A∩B) = P(B).(P(A/B) = P(A).(P(B/A)).
Note that if A and B are independent events then
P(A∩B) = P(A) P(B)
=> P(A/B) = P(A∩B)/P(B) P(A)P(B)/P(B) = P(A) which should be the case as occurrence of A does not depend upon B. Similarly P(B/A) = P(B).
More often than not the probability of one event is affected by occurrence of other events. Let a bag contains 3 red, 2 green and 4 yellow balls. A ball is drawn and found to be red. Consider the following cases:
(i) If the red ball is replaced then find the probability that the next ball to be
drawn is yellow.
(ii) What happens when red ball will not be replaced?
Let, R : red ball is first drawn
Y : yellow ball in next drawn.
In first case when the red ball is replaced.
P(Y) = (Total number of yellow balls)/(total number of balls)=4/9
In second case the red is ball is not replaced, so,
P(Y) = (Total number of yellow balls)/(Total number of balls left)=4/8
The probability in these two cases is different. In the first case when the first ball was red and it replaced it does not affect the probability of yellow ball in second draw but in second case when the ball was not replaced the probability of yellow ball is changed. It happened because in second case the sample space is changed. So from here we can draw a conclusion that when occurrence of any event changes the sample space, the probability of other event change. Hence the probability also changes.
It means that, occurrence of any event change the sample space. Then Probability of other events change, in other words probability of some event depend or the occurrence of other events. This is known as "Conditional Probability".
Let A and B be the two events. If A has already happened (i.e. given) then probability of B can be found by using the formula
P(B/A) = P(A∩B)/P(A)
If A and B are independent events, then
P(B/A) = P(B) i.e. P(A∩B) = P(A) P(B)
or, P(A/B) = P(A)
(i) n events A1, A2, A3, ........... An are said to be pair-wise independent tiff P
(Ai∩Aj) = P(Ai) P(Aj), where 1, I = 1, 2, 3, ........., n
(ii) These n events are said to be mutually independent iff
P(A1 ∩ A2 ∩......... ∩ An) = P(A1) P(A2) ....... P(An).
 Note : If the set of n events related to a sample space are pair-wise independents, they must be mutually independent, but vice versa is not always true.
To apply total probability formula we must check whether A1 and A2 fulfill the following three conditions or not.
A1 ∩ A2 = Φ
and A1 U A2 = S.
A1  S and A2 S.
Probability of an event which depends on more than one event
Suppose we have n events A1, A2, ........., An related to a sample space such that:
(i) They are mutually exclusive i.e.
Ai ∩ Aj = Φ, I, j = 1, 2 .......... n, I ≠ j.
(ii) They are exhaustive i.e.
A1 U A2 U A3 ......... U An = S.
(iii) They are proper subsets of sample space S i.e.
Ai S, i = 1, 2, .......... n.
Then probability of an event A which depends on the events
A1, A2, ........., An can be calculated as
P(A) = ∑n(i=1)
P(A/Ai) P(Ai)
= P(A/A1) P(A1) + P(A/A2) P(A2) +......+ P(A/An) P(An)
This is a formula for total probability.
Illustration:
Take the case of previous illustration in which a bag contains 3 Red, 2 Green and 4 Yellow balls. A ball is drawn and if it is Red or Yellow it is replaced otherwise not. Find the probability that second ball drawn is a yellow ball.
Solution:
First of all, we should be very clear about the event of which probability is to be found and the events on which it depends.
Let's A1 : First ball is either red or yellow
A2 : First ball is green
A : Second ball drawn is yellow
We have to calculate P(A) which definitely depends on A1 and A2.
Therefore, P(A) = P(A/A1) P(A1) + P(A/A2) P(A2)
P(A/A2) = Probability of A when A1 is already happen.
= 4/9 (i.e. probability of A given A1)
P(A1) = 7/9
P(A/A2) = Probability of A when A2 is already happened (Probability of A given A2). = 4/8
P(A2) = 2/9
Hence, P(A) = 4/9×7/9+4/8×2/9 = 37/81
Illustration:
If a dice is thrown, what is the probability of occurrence of a number greater than 1, if it is known that only odd numbers can come up.
Solution:
S=the sample space = {1, 2, 3, 4, 5, 6}
A=the event of occurrence of an odd number = {1, 3, 5}
B=the event of occurrence of a number greater than 1={2, 3, 4, 5, 6}
Here A∩B = {3, 5}
so that P(B/A) = P(A∩B)/(P(A))=(n(A∩B))/(n(A))=2/3.
Illustration:
In a college, 25% students failed in Mathematics, 155 students failed in Physics, and 10% failed in Mathematics and Physics. A student is selected at random:
(i) If he failed in Physics, then find the
chance of his failure in Mathematics,
(ii) If he failed in Mathematics, then find
the chance of his failure in Physics,
(iii) Find the chance of his failure in Mathematics or Physics.
Solution:
Let E1 and E2 be the events of failure in mathematics and physics respectively. Let the total number of students appearing in the examination be 100.
Since 25% students failed in Mathematics, n(E1) = 25.
=> P(E1) = n(E1)/(n(S))=25/100=1/4.
Since 15% students failed in Physics, n(E2) = 15
=> P(E2) = n(E2)/(n(S))=15/100=3/20.
Again 10% students failed in Physics and Mathematics both, so n(E1ÇE2) = 10
=> P(E1∩E2) = n(E1∩E2 )/(n(S))=10/100=1/10..
(i) The chance of failure in Mathematics
while he has failed in Physics is given by
P(E1/E2) = = (P(E1∩E2))/P(E2 ) =(1/10)/(3/20)=2/3.
(ii) The chance of failure in Physics while he has failed in Mathematics is given
by P(E2/E1) = (P(E1∩E2))/P(E2 ) =(1/10)/(1/4)=2/5..
(iii) The chance of failure in Mathematics or Physics is:
P(E1UE2) = P(E1) + P(E2) - P(E1∩E2)
[since both the events are not mutually exclusive]
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