Module 2. Theory of probability
Lesson 8
MULTIPLICATION THEOREM OF PROBABILITY
8.1
Introduction
In the previous lesson we have studied the addition theorem of probability for mutually exclusive events as well as for those events which are not mutually exclusive. In many situations we want to find the probability of simultaneous occurrence of two or more events. Sometimes the information is available that an event A has occurred and one is required to find the probability of occurrence of another event B utilizing the information about event A. Such a probability is known as conditional probability. In this lesson we shall discuss the important concept of conditional probability of an event which will be helpful in understanding the concept of multiplication theorem of probability as well as independence of events.
8.2 Multiplication Theorem for Independent Events
Statement: This theorem states that if two events A and B are independent
then the probability that both of them will occur is equal to the product of
their individual probabilities.
Symbolically P (AB) = P (A∩B) = A (A
and B) = P (A). P (B)
Proof
If an event A can happen in
n1 ways out of which a1 are favorable and the event B can
happen in n2 ways out of which a2 are favorable, we can
combine each favorable event in the first with each favorable event in the
second case. Thus, the total number of favorable cases is a1 × a2.
Similarly, the total number of possible cases is n1 × n2.
Then by definition the probability of happening of both the independent events
is
Similarly we can extend the theorem to three events
P (A∩B∩C) = P (A) × P (B) × P(C)
The following examples illustrate the application of this theorem:
Example 1. From a pack of 52 cards, two cards are drawn at random one after the other with replacement. What is the probability that both cards are kings?
Solution:
The probability of drawing a king P (A) = 
The probability of drawing again the king after
replacement P (B) =
Since the two events are independent, the probability of drawing two kings is:
Example 2. A bag contains 4 red balls, 3 white balls and 5 black balls. Two balls are drawn one after the other with replacement. Find the probability that first is red and the second is black.
Solution:
Probability of red ball in the first draw = 
The probability of a black ball in the second draw = 
Since the events are independent, the probability that first is red and the second are black will be:
8.3 Conditional Probability
In many situations we have the information about the occurrence of an event A and are required to find out the probability of the occurrence of another event B. Two events A and B are said to be dependent when event A can occur only when event B 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. For example, if we want to find the probability of an ace of spade if we know that card drawn from a pack of cards is black. Let us consider another problem relating to dairy plant. There are two lots of full cream pouches A and B, each containing some defective pouches. A coin is tossed and if it turns up with its head upside lot A is selected and if it turns with tail up, lot B is selected. In this problem we are interested to know the probability of the event that a milk pouch selected from the lot obtained in this manner is defective.
Definition: If two events A and B are dependent, then the conditional probability of B given that event A has occurred is defined as
Let us consider the experiment of throwing of a die once. The sample space of this experiment is {1, 2, 3, 4, 5, and 6}.
Let E1: ‘an even number shows up’ and E2: ‘multiple of 3 show up’.
Then E1: {2,
4, 6} and E2: {3, 6}. Hence, P (E1) = 3/6 =
˝ and P (E1) = 2/6 =1/3
In order to find the probability of occurrence of E2 when it is given that E1 has occurred we know that in a single throw of die ‘2’or ‘4’or ‘6’has come up. Out of these only ‘6’ is favorable to E2. So the probability of occurrence of E2 when it is given that E1 has occurred is equal to 1/3. This probability of E2 when E1 has occurred is written as P (E2|E1). Here we find that P (E2|E1) =P (E2). Let us consider the event E3: ‘a number greater than 3 shows up’ then E3:{4,5,6} and P(E3)=3/6=1/2 Out of 2,4 and 6, two numbers namely 4 and 6 are favorable to E3. Therefore, P (E3|E1) =2/3. The events of the type E1 and E2 are called independent events as the occurrence or non-occurrence of E1 does not affect the probability of occurrence or non-occurrence of E2. The events E1and E3 are not independent.
8.4 Multiplication Theorem of Probability for Dependent Events
Statement: The probability of simultaneous happening of two events A and B is given by:
Where P (B|A) is the conditional probability of happening of B under the condition that A has happened and P (A|B) is the conditional probability of happening of A under the condition that B has happened.
Proof:
Let
A and B be the events associated with the sample space S of a random experiment
with exhaustive number of outcomes (sample points) N, i.e., n(S) = N. Then by
definition
………(Eq. 1)
For the conditional event
A|B (i.e., the happening of A under the condition that B has happened), the
favorable outcomes (sample points) must be out of the sample points of B. In
other words, for the event A|B, the sample space is B and hence
Similarly, we have
On multiplying and dividing equation (1) by n (A), we get
Also
Generalization
The multiplication theorem of probability can be extended to more than two events. Thus, for three events A1, A2 and A3 we have
For
n events A1, A2,…, An we have
P(A1 ∩ A2 ∩ … ∩ An) = P(A1) P(A2|A1) P(A3|A1 ∩ A2)× … × P(An|A1∩ A2 ∩ … ∩ An-1)
In particular, if A1, A2, …, An are independent events then
i.e., the probability of the simultaneous happening of n independent events is equal to the product of their individual probabilities.
The following examples illustrate the application of this theorem
Example 3. A bag contains 5 white and 8 red balls. Two successive drawings of 3 balls are made such that (a) the balls are replaced before the second drawing, and (b) the balls are not replaced before the second draw. Find the probability that the first drawing will give 3 white and the second 3 red balls in each case.
Solution:
(a)
When balls are replaced.
Total balls in the bag = 8 + 5 = 13
3
balls can be drawn out of total of 13 balls in 13C3 ways.
3 white balls can be drawn out of 5 white balls in 5C3 ways.
Probability
of 3 white balls = 
Since
the balls are replaced after the first draw so again there are 13 balls in the
bag 3 red balls can be drawn out of 8 red balls in 8C3 ways.
Probability
of 3 red balls = 
Since the events are independent, the required probability is:
(b)
When the balls are not replaced before second draw
Total balls in the bag = 8 + 5 = 13
3
balls can be drawn out of 13 balls in 13C3 ways.
3
white balls can be drawn out of 5 white balls in 5C3 ways.
The
probability of drawing 3 white balls = 
After the first draw, balls left are 10, 3 balls can be drawn out of 10 balls in 10C3 ways.
3
red balls can be drawn out of 8 balls in 8C3 ways. Probability of drawing 3 red
balls = 
.
Since both the events are dependent, the required probability is:
Example 4. A bag contains 5 white and 3 red balls and four balls are successively drawn and are not replaced. What is the chance that (i) white and red balls appear alternatively and (ii) red and white balls appear alternatively?
Solution (i) The probability of drawing a white ball = 5/8
The probability of drawing a red ball = 3/7
The probability of drawing a white ball = 4/6 and the probability of drawing a red ball = 2/5
Since the events are dependent, therefore the required probability is:
(ii) The probability of drawing a red ball = 3/8 and the probability of drawing a white ball = 5/7
The probability of drawing a red ball = 2/6 and the probability of drawing a white ball = 4/5
Since the events are dependent, therefore the required probability is:
Example 5. A coin is tossed once. If it shows head, it is tossed again and if it shows tail, then a dice is tossed. Let E1 be the event: ‘the first throw of coin shows tail’ and E2 be the event: ‘the dice shows a number greater than 4’. Find P (E2|E1)
Solution: In this problem the random experiment was carried out in two stages
a) A coin is tossed b) If the first stage shows a head, coin is tossed again and if it shows a tail, a dice is thrown. The sample space is {HH, HT, T1, T2, T3, T4, T5, T6}.
P (HH) = P(HT) = 1/2x1/2=1/4
and P(T1) = P(T2) = P(T3) = P(T4) = P(T5) = P(T6) = 1/6X1/2 = 1/12
E1: the event the first throw of coin shows tail = {T1,T2,T3,T4,T5,T6}.
E2 : the event the dice shows a number greater than 4 = { T5,T6}
P(E1)=6x1/12=1/2 and P(E2)=2x1/12=1/6
Also 
8.5 Combined use of Addition and Multiplication Theorem
In some problems in probability both addition and multiplication theorems are used simultaneously. The following examples illustrate the combined use of addition and multiplication theorems.
Example 6. A bag contains 5 white and 4 black balls. A ball is drawn from this bag and is replaced and then second draw of a ball is made. What is the probability that two balls are of different colors.
Solution: There are two possibilities
i) First ball is white and the second ball drawn is black.
ii) First ball is black and the second ball drawn is white.
Since the events are independent, so by using multiplication theorem we have
i)
Probability
of drawing First ball white and the second ball black = 
ii)
Probability
of drawing First ball black and the second ball white = 
Since these probabilities are mutually exclusive, by using addition theorem
Probability that two balls are of
different colors = 
Example 7. A can hit a target 4 times in 5 shots. B 3 times in 4 shots and C twice in 3 shots. They fire a volley. What is the probability that
a) Two shots hit the target
b) At least two shots hit the target.
Solution:
a) There are three possibilities that two shots hit the target
i)
A
and B hit the target and C could not hit i.e. 
ii)
B
and C hit the target and A could not hit i.e. 
iii)
A
and C hit the target and B could not hit i.e. 
Since the events are independent, so using multiplication theorem
i)
ii)
iii)
Since the above three possibilities are mutually exclusive, so by using addition theorem, we have
b) There are two possibilities that at least two shots hit the target
i) Two could hit the target
ii)
Three
could hit the target i.e. A and B and C could hit i.e. 
Since the events are independent, so by using multiplication theorem
i)
P(Two
could hit the target) = 
ii) 
Since
the above two possibilities are mutually exclusive, so by addition theorem
P (At least two
shots hit the target) = 
Example 8. Three small sized Herds A, B and C consist of 3 cows and 1 buffalo, 2 cows and 2 buffaloes, 1 cow and 3 buffaloes, respectively. Find the probability of selecting one cow and two buffaloes from three Herds.
Solution: This particular example illustrates the combined application of Additive and Multiplicative theorem of probability. There are three Herds whose composition is given as under
|
Herd |
Type of animals |
|
|
Cow |
Buffalo |
|
|
A |
3 |
1 |
|
B |
2 |
2 |
|
C |
1 |
3 |
Probability of selection of one cow and two buffaloes from Herds A, B and C can take place in following three possible ways.
(i) One cow from Herd A and one buffalo each from Herd B and C respectively
(ii) One cow from Herd B and one buffalo each from Herd A and C respectively
(iii) One cow from Herd C and one buffalo each from Herd A and B respectively
The
probability for Situation (i) is 
The
probability for Situation (ii) is 
The
probability for Situation (iii) is 
Since either of the above situations can occur, therefore probability of getting selection either through situation (i) or situation (ii) or situation (iii) will be sum of the probabilities of three different situations which is equal to
