1) The probabilities of events A and E given the other are calculated from the information provided about the sample space and events. 2) Given probabilities of people reading English and Hindi newspapers, the probability of Hindi reading persons given English reading is calculated. 3) The probability of selecting a Hindi knowing woman teacher from a list is calculated using the information about number of men/women, Hindi knowing/not knowing people, and teachers in the list. 4) The probability of selecting a red ball from the second box after transferring a ball randomly from the first box to the second box is calculated.

1. Conditional Probability
RAJU DAVID.C, MSc, MBA, PGDCA, BCC
Faculty, Department of Psychology
Rajagiri College of Social Science
Kalamassery, Ernakulam

2. Conditional probability - Definition
Conditional probability of an event B, given that the event A has already occurred, is defined as
the ratio of the probability of the joint occurrence of both the events to the probability of
already occurred event A, and is denoted by P(B|A).
i.e., P(B|A) = P(AՈB) / P(A) ……….(1) [P(B|A) is read as “Probability B, given A”]
Similarly, P(A|B) = P(AՈB) / P(B) ……….(2)
Here the probability of an event is determined after imposing a condition that the other event
has already been occurred.
(Note: Here the probability of the event already occurred should not be 0 (zero).)
Comparing the above two equations,
P(B|A) = P(AՈB) / P(A) ……(1), implies P(A) x P(B|A) = P(AՈB) ……(3)
P(A|B) = P(AՈB) / P(B) ……(2), implies P(B) x P(A|B) = P(AՈB) ……(4)
So, from (3) & (4) we see that, P(AՈB) = P(A) x P(B|A) = P(B) x P(A|B)

3. MULTIPLICATION THEOREM of Probability
Two events are said to be independent, if the probability of their joint
occurrence is the same as the product of their individual probabilities.
i.e., P(AՈB) = P(A) x P(B) …..(1)
In conditional probability/(Dependent)
P(AՈB) = P(A) x P(B|A) = P(B) x P(A|B) ……(2)
Comparing (1) & (2), it is clear that if A & B are independent, then, P(B|A)
should be equal to P(B). Similarly, P(A|B) should be equal to P(A)

4. A husband and wife appears in an interview for 2 vacancies in the same post. The prob of husband’s
selection is 1|7 and that of wife is 1|5. What is the prob that
1) both of them will be selected? 2) only one of them will be selected?
3) none of them will be selected?
Let A & E denote the events of selection of husband and wife. It is given that P(A) = 1|7 & P(E) = 1|5
Prob of non-selection of husband P(Ā) = 1 - P(A) = 1 – (1/7) = 6/7, Prob of non-selection of wife P(Ē) = 1 – (1/5) = 4/5
The selection|non-selection of husband and wife are independent events
(because the selection or non-selection of husband or wife is not at all affected the selection of the other).
If 2 events are independent, then the prob of their joint occurrence is the product of their individual probs.
Ans. 1) Both of them selected –
Prob. of joint occurrence of A & E = P(AE) = P(A) x P(E) = (1/7) x (1/5) = 1/35
Ans. 2) Only one of them be selected –
That means, when A selected, B not selected or when A not selected, B selected.
It is the P(AĒ or ĀE) = P(AĒ U ĀE) = P(AĒ) + P(ĀE) - (the events AĒ & ĀE are mutually exclusive)
P(AĒ) = P(A) x P(Ē)…….(1) & P(ĀE) = P(Ā) x P(E)……..(2)
So, Prob of only one of them is selected = P(AĒ U ĀE) = P(AĒ) + P(ĀE)
= P(A) x P(Ē) + P(Ā) x P(E)
= (1/7) x (4/5) + (1/5 x 6/7) = (4+6)/35 = 10/35
Ans. 3) Prob that none of them selected – P(ĀĒ) = P(Ā) x P(Ē) = (6/7) x (4/5) = 24/35

5. Calculate P(B|A) & P(A|B), if P(A) = 4/15, P(B) = 11/24
and P(AUB) = 61/120
Ans:
By definition P(B|A) = P(AՈB)/P(A) or P(AB)/P(A) and
P(A|B) = P(AՈB)/P(B) or P(AB)/P(B)
So want P(AB), as it is not given directly.
We know that P(AUB) = P(A) + P(B) - P(AB) or P(AB) = P(A) + P(B) – P(AUB)
P(A) = 4/15, P(B) = 11/24, P(AUB) = 61/120
So, P(AB) = 4/15 + 11/24 - 61/120 = 32/120 + 55/120 - 61/120
= (32+55-61)/120 = 26/120
Then, P(B|A) = P(AB)/P(A) = (26/120) / (4/15) = (26 x 15)/(120 x 4) = 13/16
And P(A|B) = P(AB)/P(B) = (26/120) / (11/24) = 26/55

6. Q1. Calculate P(B|A), if P(A) = 0.75, P(B) = 0.60 and P(A|B) = 0.901.
We have
P(A) x P(B|A) = P(B) x P(A|B) = P(AՈB)
0.75 X P(B|A) = 0.6 X 0.9
P(B|A) = (0.6 X 0.9)/0.75 = 0.54/0.75 = 0.72
Q2. What is the prob. that the total of two dice greater than 9, given that the first die is 5?
Let A denotes the event first die is 5 and event B is the total of two dice is greater than 9
A= {5} P(A) = 1/6, B = {(4,6),(5,5), (5,6), (6,4), (6,5), (6,6)}
AB = {(5,5),(5,6)} P(AB) = 2/36
Therefore, P(B/A) = P(AB) / P(A) = (2/36) / (1/6) = 1/3

7. EXERCISES
1. Given S = {2,3,5,7,11,13,17,19,23,29,31}, A = {3,5,7,17,19,11,31,23},
E = {19,7,13,23,17,3,29}. Find P(A|E), P(E|A), P(Ā|E), P(E|Ā).
Also establish that P(A)P(E|A) = P(E)P(A|E) = P(AE)
2. In a city 70% of the population read English news paper, 65% read Hindi newspaper. If 10%
people do not read any newspaper, find the prob of Hindi reading persons, given that English
reading persons.
3. A University has to select an examiner from a list of 50 persons. 20 of them are women and 30
are men. 10 of them know Hindi and 40 do not know. 15 of them are teachers. What is the prob of
selecting a Hindi knowing woman teacher?
4. One box contains 6 red 4 green balls and a second box contains 7 red and 3 green balls. A ball is
randomly selected from the first box and placed in the second box. Then a ball is selected from
the second box. What is the prob that it is a red ball?