Classical Definition of Probability and its Limitations - Quantitative Techniques for management

This definition, also known as the mathematical definition of probability, was given by J. Bernoulli. With the use of this definition, the probabilities associated with the occurrence of various events are determined by specifying the conditions of a random experiment. It is because of this that the classical definition is also known as 'a priori' definition of probability.


If n is the number of equally likely, mutually exclusive and exhaustive outcomes of a random experiment out of which m outcomes are favorable to the occurrence of an event A, then the probability that A occurs, denoted by P(A), is given by :

P(A) = Number of outcomes favorable to A/Number of exhaustive outcomes
= m/n

Various terms used in the above definition are explained below:

  1. Equally likely outcomes: The outcomes of random experiment are said to be equally likely or equally probable if the occurrence of none of them is expected in preference to others. For example, if an unbiased coin is tossed, the two possible outcomes, a head or a tail are equally likely.
  2. Mutually exclusive outcomes: Events that cannot occur together are called mutually exclusive events. Such events do not have any common outcomes. If two or more events are mutually exclusive, then at most one of them will occur every time we repeat the experiment. Thus the occurrence of one event excludes the occurrence of the other event or events. For any experiment, the final outcomes are always mutually exclusive because one and only one of these outcomes is expected to occur in one repetition of the experiment.
  3. Exhaustive outcomes: Exhaustive events contains all possible elementary events for an experiment. Thus, all sample spaces are collectively exhaustive lists. The sample space for an experiment can be described as a list of events that are mutually exclusive and collectively exhaustive. Sample space events do not overlap or intersect, and the list is complete.
  4. Event: The occurrence or non-occurrence of a phenomenon is called an event. For example, in the toss of two coins, there are four exhaustive outcomes, viz. (H, H), (H, T), (T, H), (T, T). The events associated with this experiment can be defined in a number of ways. For example, (i) the event of occurrence of head on both the coins, (ii) the event of occurrence of head on at least one of the two coins, (iii) the event of non-occurrence of head on the two coins, etc. An event can be simple or composite depending upon whether it corresponds to a single outcome of the experiment or not. In the example, given above, the event defined by (i) is simple, while those defined by (ii) and (iii) are composite events.

Example : What is the probability of obtaining a head in the toss of an unbiased coin?

Solution: This experiment has two possible outcomes, i.e., occurrence of a head or tail. These two outcomes are mutually exclusive and exhaustive. Since the coin is given to be unbiased, the two outcomes are equally likely. Thus, all the conditions of the classical definition are satisfied.

No. of cases favorable to the occurrence of head = 1
No. of exhaustive cases = 2
∴ Probability of obtaining head P(H) =1/2

Example : What is the probability of obtaining at least one head in the simultaneous toss of two unbiased coins?

Solution: The equally likely, mutually exclusive and exhaustive outcomes of the experiment are (H, H), (H, T), (T, H) and (T, T), where H denotes a head and T denotes a tail.
Thus, n = 4.

Let A be the event that at least one head occurs. This event corresponds the first three outcomes of the random experiment. Therefore, m = 3.
Hence, probability that A occurs, i.e., P (A)=3/4

Example : Find the probability of obtaining an odd number in the roll of an unbiased die.

Solution: The number of equally likely, mutually exclusive and exhaustive outcomes, i.e., n = 6. There are three odd numbers out of the numbers 1, 2, 3, 4, 5 and 6. Therefore, m = 3.

Thus, probability of occurrence of an odd number= 3/6 =1/2

Example : What is the chance of drawing a face card in a draw from a pack of 52 well-shuffled cards?

Solution: Total possible outcomes n = 52.

Since the pack is well-shuffled, these outcomes are equally likely. Further, since only one card is to be drawn, the outcomes are mutually exclusive.

There are 12 face cards, ∴ m = 12.

Thus, probability of drawing a face card = 12/52 = 3/13.

Example : What is the probability that a leap year selected at random will contain 53 Sundays?

Solution: A leap year has 366 days. It contains 52 complete weeks, i.e, 52 Sundays. The remaining two days of the year could be anyone of the following pairs:

(Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday), (Sunday, Monday). Thus, there are seven possibilities out of which last two are favorable to the occurrence of 53rd Sunday.

Hence, the required probability = 2/7.

Example : Find the probability of throwing a total of six in a single throw with two unbiased dice.

Solution: The number of exhaustive cases n = 36, because with two dice all the possible outcomes are:

(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6).

Out of these outcomes the number of cases favorable to the event A of getting 6 are:
(1, 5), (2, 4), (3, 3), (4, 2), (5, 1). Thus, we have m = 5.

∴ P (A) = 5/36

Example : A bag contains 15 tickets marked with numbers 1 to 15. One ticket is drawn at random. Find the probability that:

(i) the number on it is greater than 10,
(ii) the number on it is even,
(iii) the number on it is a multiple of 2 or 5.

Solution: Number of exhaustive cases n = 15

  1. Tickets with number greater than 10 are 11, 12, 13, 14 and 15. Therefore, m = 5 and hence the required probability = 5/15 = 1/3
  2. Number of even numbered tickets m = 7
    ∴ Required probability = 7/15
  3. The multiple of 2 are: 2, 4, 6, 8, 10, 12, 14 and the multiple of 5 are: 5, 10, 15.
    m = 9 (note that 10 is repeated in both multiples will be counted only once).
    Thus, the required probability = 9/15 = 3/5.

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