5.5 RULES OF PROBABILITY AND ALGEBRA OF EVENTS
5.5.1 Rules of Addition
The addition rules are helpful when we have two events and are interested in knowing the probability that at least one of the events occurs.
Mutually Exclusive Events
The rule of addition for mutually exclusive (disjoint), exhaustive, and equally likely events states thatIf two events A and B are mutually exclusive, exhaustive, and equiprobable, then the probability of either event A or B or both occurring is equal to the sum of their individual probabilities.
This rule is expressed in the following formula
P(A or B) = P(A B) (A B) (A) (B)
(S) (S) n n n n n ∪ + ∪ = = = (A) (B) P(A) P(B) (S) (S) n n n + n = + (5-1)
where A ∪ B (read as ‘A union B’) denotes the union of two events A and B and it is the set of all sample points belonging to A or B or both. This rule can also be illustrated by the Venn diagram shown in Fig. 5.3. Here two circles contain all the sample points in events A and B. The overlap of the circles indicates that some sample points are contained in both A and B.
Illustration: Consider the pattern of arrival of customers at a service counter during
the first hour it is open along with its probability:
No. of persons : 0 1 2 3 4 or more
Probability : 0.1 0.2 0.3 0.3 0.1
To understand the probability that either 2 or 3 persons will be there during the first hour, we have
P(2 or 3) = P(2) + P(3) = 0.3 + 0.3 = 0.6
The formula (5-1) can be expanded to include more than two events. In particular, if there are n events in a sample space that are mutually exclusive, then the probability of the union of these events is given by
P(A1 ∪ A2 ∪ . . . ∪ An) = P(A1) + P(A2) + . . . + P(An) (5-2)
For example, if we are interested in knowing the probability that there will be two or more persons during the first hour, then using formula (5-2), we have
P(2 or more) = P(2, 3, 4 or more) = P(2) + P(3) + P(4) = 0.3 + 0.3 + 0.1 = 0.7
Marginal probability: The
unconditional probability of an event occurring.
Venn diagram: A pictorial
representation for showing the sample space and operations involving events. The sample space is represented by a rectangle and events as circles.
Figure 5.3
An important special case of formula (5-1) is for complementary events. Let A be any event and A be the complement of A. Obviously A and A are mutually exclusive and exhaustive events. Thus, either A occurs or it does not, is given by
P(A or A ) = P(A) + P(A ) = P(A) + {1 – P(A)} = 1
or P(A) = 1 – P(A ) (5-3)
For example, if a dice is rolled, then the probability whether an odd number of spots occurs or does not.
Partially Overlapping (or Joint) Events
If events A and B are not mutually exclusive, it is possible for both events to occur simultaneously? This means these events have some sample points in common. Such events are also called joint (or overlapping) events. The sample points in common (belong to both events) represent the joint event A ∩ B (read as: A intersection B). The addition rule in this case is stated as:If two events A and B are not mutually exclusive, then the probability of either A or B or both occurring is equal to the sum of their individual probabilities minus the probability of A and B occurring together.
This rule is expressed in the following formula: P(A or B) = P(A) + P(B) – P(A and B)
or P(A ∪ B) = P(A) + P(B) – P(A ∩ B) (5-4)
This addition rule can also be illustrated by the Venn-diagram shown in Fig. 5.4.
Illustration: Suppose 70 per cent of all tourists who come to India will visit Agra while
60 per cent will visit Goa and 50 per cent of them will visit both Agra and Goa. The probability that a tourist will visit either Goa or Agra or both is obtained by applying formula (5-4) as follows:
P(Agra or Goa) = P(Agra) + P(Goa) – P(both Agra and Goa) = 0.70 + 0.60 – 0.50 = 0.8
Consequently, the probability that a tourist will visit neither Agra nor Goa is calculated by
P(neither Agra nor Goa) = 1 – P(Agra or Goa) = 1 – 0.80 = 0.20 The formula (5-4) can be expanded to include more than two events. In particular, if there are three events that are not mutually exclusive, then the probability of the union of these events is given by
P(A ∪ B ∪ C) =P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C)
– P(C ∩ A) + P(A ∩ B ∩ C) (5-5)
Remark: The rules of addition are applicable for calculating probability of events in
case of simultaneous trails.
Example 5.4: What is the probability that a randomly chosen card from a deck of cards
will be either a king or a heart.
Solution: Let event A and B be the king and heart in a deck of 52 cards, respectively.
Then, it is given that Figure 5.4
Card Probability Reason
King P(A) = 4/52 4 kings in a deck of 52 cards Heart P(B) = 13/52 13 hearts in a deck of 52 cards King of heart P(A and (B) = 1/52 1 King of heart in a deck of 52 cards
Using the formula (5-4), we get
P(A or B) = P(A) + P(B) – P(A and B) = 52 52 524 +13− 1 = 1652 = 0.3077
Example 5.5: Of 1000 assembled components, 10 have a working defect and 20 have a
structural defect. There is a good reason to assume that no component has both defects. What is the probability that randomly chosen component will have either type of defect? [Delhi Univ., MBA, 2003]
Solution: Let the event A and B be the component which has working defect and has
structural defect, respectively. Then it is given that
P(A) = 10/1000 = 0.01, P(B) = 20/1000 = 0.02 and P(A and B) = 0
The probability that a randomly chosen component will have either type of defect is given by
P(A or B) = P(A) + P(B) – P(A and B) = 0.01 + 0.02 – 0.0 = 0.03
Example 5.6: A survey of 200 retail grocery shops revealed following monthly income
pattern:
Monthly Income (Rs) Number of Shops
Under Rs 20,000 102
20,000 to 30,000 061
30,000 and above 037
(a) What is the probability that a particular shop has monthly income under Rs 20,000
(b) What is the probability that a shop selected at random has either an income between Rs 20,000 and Rs 30,000 or an income of Rs. 30,000 and more?
Solution: Let the events A, B and C represent the income under three categories,
respectively.
(a) Probability that a particular shop has monthly income under Rs 20,000 is P(A) = 102/200= 0.51.
(b) Probability that shop selected at random has income between Rs 20,000 and Rs 30,000 or Rs 30,000 and more is given by
P(A or B) = P(A) + P(B)
= 61 37
200 200+ = 0.305 + 0.185 = 0.49
Example 5.7: From a sales force of 150 persons, one will be selected to attend a special
sales meeting. If 52 of them are unmarried, 72 are college graduates, and 3/4 of the 52 that are unmarried are college graduates, find the probability that the sales person selected at random will be neither single nor a college graduate.
Solution: Let A and B be the events that a sales person selected is married and that he is
a college graduate, respectively. Then, it is given that
P(A) = 52/150, P(B) = 72/150; P(A and B) = (3/4) (52/150) = 39/150
The probability that a salesperson selected at random will be neither single nor a college graduate is:
P(A∩B) = 1 – P(A ∪ B) = 1 – {P(A) + P(B) – P(A ∩ B)} = 1 – 52 72 39 13 150 150 150 30 + − =
Example 5.8: From a computer tally based on employer records, the personnel manager
of a large manufacturing firm finds that 15 per cent of the firm’s employees are supervisors and 25 per cent of the firm’s employees are college graduates. He also discovers that 5 per cent are both supervisors and college graduates. Suppose an employee is selected at random from the firm’s personnel records, what is the probability of:
(a) selecting a person who is both a college graduate and a supervisor? (b) selecting a person who is neither a supervisor nor a college graduate?
Solution: Let A and B be the events that the person selected is a supervisor and that he
is a college graduate, respectively. Given that
P(A) = 15/100; P(B) = 25/100; P(A and B) = 5/100
(a) Probability of selecting a person who is both a college graduate and a supervisor is: P (A and B) = 5/100 = 0.05
(b) Probability of selecting a person who is neither a supervisor nor a college graduate is: P (A and B) =1 – P(A or B) = 1 – [P(A) + P(B) – P(A and B)]
=1 – 15 25 5 100 100 100 + − = 65 100 = 0.65
Example 5.9: The probability that a contractor will get a plumbing contract is 2/3 and
the probability that he will not get an electrical contract is 5/9. If the probability of getting at least one contract is 4/5, what is the probability that he will get both?
Solution: Let A and B denote the events that the contractor will get a plumbing and
electrical contract, respectively. Given that
P(A) = 2/3; P(B) = 1 – (5/9) = 4/9; P(A ∪ B) = 4/5 P(A ∩ B) = P(A) + P(B) – P(A ∪ B) = 2 4 4
3 9+ − = 5 14
45 = 0.31 Thus, the probability that the contractor will get both the contracts is 0.31.
Example 5.10: An MBA applies for a job in two firms X and Y. The probability of his
being selected in firm X is 0.7 and being rejected at Y is 0.5. The probability of at least one of his applications being rejected is 0.6. What is the probability that he will be selected by one of the firms?
Solution: Let A and B denote the event that an MBA will be selected in firm X and will
be rejected in firm Y, respectively. Then, given that P(A) = 0.7, P(A) = 1 – 0.7 = 0.3
P(B) = 0.5, P( B ) = 1 – 0.5 = 0.5, P( A ∪ B ) = 0.6 Since P(A ∩ B) = 1 – P(A∪B) = 1 – 0.6 = 0.4
therefore, probability that he will be selected by one of the firms is given by P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 0.7 + 0.5 – 0.4 = 0.8
Thus, the probability of an MBA being selected by one of the firms is 0.8.