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
Various terms used in the above definition are explained below:
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
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