After reading this article you will learn about the meaning and theories of probability.
Meaning of Probability:
The word probability literally denotes chance and the theory of probability deals with laws governing the chances of occurrence of phenomena which are unpredictable in nature.
Probability is, therefore, defined as the likelihood of occurrence of a particular event, i.e., something to occur or not to occur.
If ‘a’ stands for event to happen and ‘b’ stands for the failure, then the probability of its happening p can be written as:
Theories of Probability:
1. Addition Theory:
When events, let us say A, B, C, D are mutually exclusive (that the events cannot occur simultaneously that is either A or B or C or D can occur), the chance of occurrence of either event can be calculated from the formula:
P(A or B or C or D) = P(A) + P(B) + P(C) + P(D).
Proof:
Let us consider that in a random experiment n possible outcomes are there which are mutually exclusive. If x of these cases are favourable to event A and y favourable to event B, then probabilities of these events are by definition
P(A) = x/n, P(B) = y/n.
Since, the events A and B are mutually exclusive (i.e., both of them cannot occur simultaneously), the x cases favourable to A are completely distinct from the y cases favourable to B. The number of cases favourable to either A or B is, therefore, (x + y).
Example 1:
What is the probability of obtaining head in a single toss of an unbiased coin?
When a coin is tossed either head or tail will come but not both. As the events are mutually exclusive and the toss was made in an unbiased coin, the outcome is expected to be equal. Therefore, probability of head = ½.
Example 2:
If two coins are tossed un-biasedly, what is the probability of occurrence of (a) 2 heads, (b) 2 tails, (c) 1 head and 1 tail, (d) at least 1 head?
Solution:
The outcome of the toss:
Probability of occurrence of either of the events will be:
Example 3:
What will be the probability of getting both heads or both tails when 2 coins are tossed?
Solution:
Out of 4 outcomes, viz., HH, HT, TH, TT, which are mutually exclusive, number of favourable cases of HH is ¼ (4 events) and that of TT is ¼.
Therefore, probability of getting both heads or both tails will be
¼ + ¼ = ½
Example 4:
What is the probability of getting an ace or a joker from a pack of 54 cards?
Solution:
Considering that the events are mutually exclusive:
probability of an ace is 4/54 = 2/27
probability of joker is 2/54 = 1/27.
Therefore, the probability of getting either an ace or a joker is
2/27 + 1/27 = 3/27 = 1/9.
Example 5:
Find out the probability of occurrence of a spade or a king from 52 cards.
Solution:
Probability of occurrence of a spade is 13/52 = 1/4.
Probability of occurrence of a king is 4/52 = 1/13.
As one of the four king is a spade and is included in 13 spade cards, then the probability of getting this card is 1/52.
Therefore, the probability of getting a king or a spade is
¼ + 1/13 – 1/52 = 13 + 4 – 1/52 = 16/52 = 4/13.
Example 6:
A bag contains 6 black balls and 4 white balls. One ball is drawn. What is the probability that the ball is black?
Solution:
Balls are numbered serially as: Black-1, 2, 3, 4, 5, 6 and White-7, 8, 9, 10. Therefore, 10 outcomes as regards the number on the selected ball, because any of the 10 balls could be drawn. As the balls are assumed to be identical except in colour, any of the balls is as likely to appear as any other ball. Of the 10 possible outcomes (mutually exclusive), only 6 cases are favourable to the event black ball. Hence
P = 6/10 = 3/5.
Example 7:
If 2 balls are drawn one after another from a bag containing 3 white and 5 black balls, what is the probability that (i) the first ball is white and the second is black; (ii) one ball is white and the other is black?
Solution:
Serial number in the balls: White-1, 2, 3; Blacks-4, 5, 6, 7, 8. First ball is selected in 8 ways because any of the 8 balls can be drawn; while the 2nd ball may be drawn in 7 ways. Hence, the 2 balls may be drawn in 8 x 7 = 56 possible ways. Since the balls are identical in all respects except in colour, 56 possible ways are mutually exclusive to one another.
(i) First ball can only be white if any of the ball numbered 1, 2, 3 is drawn, i.e., in 3 ways. Second ball will be black if any of the ball numbered 4, 5, 6, 7, 8 is drawn, i.e., 5 ways. ‘Hence the number of cases favourable to the event is 3 x 5 = 15. So,
the probability (P) = 15/56.
(ii) Number of ways of drawing a white ball and a black ball in the order (white, black) is 15. Similarly, the number of ways of drawing a white ball and a black ball in the order (black, white) is 5 x 3 = 15. Hence, the number of cases favourable to the event one ball is white and the other black irrespective of the order in which they are drawn is 15 + 15 = 30. Therefore,
P = 30/56 = 15/28.
2. Multiplication Theory:
The probability of occurrence of the event A as well as B is given by the product of (unconditional) probability of A and conditional probability of B, assuming that A has actually occurred.
Probability of (A and B) = Probability of A x Conditional probability of B, assuming A.
P(AB) = P(A). P(B/A).
This is also known as multiplication theorem.
Proof:
In a random experiment n mutually exclusive events are there among which x cases are favourable to an event A. So, the unconditional probability of A is
P(A) = x/n
of the x cases, let y cases be favourable to another event B also; i.e., the number of cases favourable to A as well as B is y. Hence by definition
P(AB) = y/r
Once A has occurred, the occurrence of B is limited to only y cases out of x (in which A occurs). So, the conditional probability of B, assuming that A has already occurred is
When the events are independent (A and B), the probability of join occurrence is given by the product of their separate probabilities.
Thus,
P(A and B) = p(A) x p(B).
Example 8:
What will be the probability of getting 2 tails when 2 coins are tossed independently?
Solution:
Probability of getting tail in 1st toss (event A) of the coin = ½
Probability of getting tail in 2nd toss (event B) of the coin = ½
Therefore, probability of getting tails with both coins is
P(A and B) = P(A) x P(B) = ½ x ½ = ¼ .
Example 9:
Two cards are drawn from a full pack of 52 cards.
Find out the probability that:
(i) Both are black cards;
(ii) One is a spade and the other is a club.
Solution:
The first card may be drawn in 52 ways and corresponding to each way of drawing the 1st card, the 2nd card may be drawn in 51 ways. Hence, the total number of cases, considering the order, is 52 x 51 = 2652.
(i) Since the 1st black card can be drawn in 26 ways (13 spades + 13 clubs), the number of cases favourable to two black cards is 26 x 25 = 650 (independent events)
P = 650/2652 = 25/102.
(ii) Number of cases favourable to the order spade-club is 13 x 13 = 169; similarly, the number of cases favourable to the order club-spade is also 169. Therefore, the number of cases favourable to one club and one spade is (169 + 169) =338.
P = 338/2652 = 13/102
Example 10:
What is the probability that all 4 children in a family have different birthdays?
Solution:
The 1st child may be born on any of 365 days of the year, 2nd also on any of the 365 days and similarly the 3rd and 4th child. Hence, the total number of possible ways in which the 4 children have birthdays is 365 x 365 x 365 x 365. These cases are mutually exclusive and as regards to the number of favourable cases out of these, it can happen that the 1st child may have any of the 365 days of the year as its birthday.
In order that the 2nd child has a birthday different from that of the 1st, it should have been born on any of the 364 remaining days of the year; similarly 3rd on any 363 and 4th on any 362 remaining days. Hence, the number of cases favourable to the event different birthdays is 365 x 364 x 363 x 362
P = 365 x 364 x 363 x 362/ 365 x 365 x 365 x 365 = 0.984.
Example 11:
Four persons A, B, C, D occupy sets in a row at random. What is the probability that A and B sit next to each other?
Solution:
Four persons arrange themselves in a row, without restriction in 4! = 1 x 2 x 3 x 4 = 24 ways. Considering A and B together, they can arrange themselves in3′ x 2 = (3 x 2 x 1) x 2 = 12 ways; because A may be to the left or to the right of B.
p = 12/24 = ½.
Example 12:
Four cards are in 4 drawals consecutively from a pack of 52 cards without replacing cards after drawal. What is the probability of drawing a queen in each drawal?
Solution:
Probability of a queen in 1st drawal =4/52
Probability of a queen in 2nd drawal = 3/51
Probability of a queen in 3rd drawal = 2/51
Probability of a queen in 4th drawal = 1/50
Therefore, the probability of getting a queen in each drawal is
Probability of a combination of two independent events can be determined from the formula:
where, p = probability of a combination of independent events,
n = total number of independent events,
s = number of one of the two mutually exclusive events,
t = number of the other mutually exclusive event in the combination,
p = probability of the first of the two mutually exclusive events,
q = probability of the other mutually exclusive event, and
! = factorial.
Example 13:
If a coin is tossed 5 times what will be the probability of combination of 3H2T.
Solution:
In this case n = 5 (number of events), s = 3 (number of H, one of the mutually exclusive event), t = 2 (the number of t, the other exclusive event), p = ½ (probability of H) and q = ½ probability of T.
The problem can be also be solved by using the binomial expansion.
The expansion of this binomial yields:
(a + b)5 = a5 + a4b + a3b2 + a2b3 + ab4 + b5.
To complete the binomial, coefficient are written either from Pascal’s triangle or from the formula n!/s!t!, where n = index of binomial, s = index of a term, t = index of b term and l = factorial.
