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Probability and its Laws

Sample space
A sample space is a collection of all possible outcome of an experiment.it is denoted by S. For example, if E1, E2, E3 . . . . , Eare n mutually exclusive outcomes or sample events of a random experiment, then set S = {E1, E2, E3, . . . . . , En} is said to be the sample space of a random experiment.

For examples
In a tossing two fair coins at the same time
S = {HH, HT, TH, TT}
IN ROLLING A FAIR DIE, THE SAMPLE SPACE IS,
S = {1, 2, 3, 4, 5, 6}

Events or outcome
The possible results or outcomes of a random experiment are known as events or outcomes. For examples
In tossing a coin, the possible outcomes are head (H) and tail (T)
In rolling a fair die , the possible outcomes are face 1, 2, 3, 4, 5, 6.
In tossing two fair coins at the same time, the possible outcomes are HH, TT, HT, TH.
The event is said to be simple if it corresponds to a single possible outcome of a random experiment and if it corresponds to two or more than two outcomes at the same time in a single trial then it is said to be joint or composite or compound events. For example, in a rolling a die getting a face 1 is a simple event where as in a rolling a die, getting even face is a compound event.

Types of events

  • Equally likely events
    If the chance of occurrences in a random experiment is equal for two or more events, then they are said to be equally likely events. For examples,
    In a tossing of an unbiased or uniform coin, head (H) and tail (T) are equally likely events and in rolling an unbiased die, all the six faces are equally likely events.
  • Mutually exclusive events
    The events or cases are said to be mutually exclusive if the happening of one event does not lead to the happening of other events in a single trial of random experiment. For example, in rolling of fair die, all the six faces numbered are mutually exclusive events and in tossing of a fair coin, head (H) and tail (T) are mutually exclusive events.
  • Independent and dependent events
    Two or more events are said to be independent of each other if happening of one event does not affect the happening of other events or vice-versa. For 4 example, if a pack of card is playing with replacement, the occurrence of king does not affect the occurrence of other cards and vice-versa.
    Two or more events are said to be dependent of each other if happening of one event affects the happening of other events or vice-versa. For example, if a pack of card is playing without replacement, the occurrence of a king affects the occurrence of other cards and vice-versa.
  • Exhaustive events
    The total number of possible outcomes in a random experiment is called exhaustive events.it is denoted by n. for examples,
    In tossing of fair coin, n = 2.
    In tossing two fair coins, n = 22 = 4.{HH, HT, TH, TT}
    In tossing n fair coins at the same time, exhausted number of cases is equal to 2n.
  • Favourable events
    The events which favor the happening of that event are called favorable events.it is denoted by m. for example,
    Supposed a box contains 5 white and 6 black balls, the favorable number of cases for getting white balls is equal to 5.
    In tossing two coins, at the same time, the sample space S = {HH, HT, TH, TT} and the favorable number of case for getting two heads is 2 i.e. {HT, TH}.

Probability
The chance of occurrence and non-occurrence of any events in random experiment is called probability.
It is denoted by P.it lies in between 0 to 1(i.e. 0 \(\leq\) P \(\leq\) 1). If P = 0, the event is said to be impossible events and if P = 1, the event is said to be sure occurring events.to keep in mind the sum of probability in a particular situation has to be 1 (i.e. \(\sum\)P = 1).
If n be the exhaustive number of events which are equally likely and mutually exclusive and m be the number of cases to the desired event E, then the probability of occurrence of event E is given by,
P(E)= \(\frac{favourable \; number \; of \; events}{total \; numbers \; of \; events}\) = \(\frac{m}{n}\)
The probability of non-occurrence of the event E is denoted by P(E’).

Example
A box contains 50 items of which 5 items are defective .one item is selected randomly, what is the probability that it is defective?
Solution:
Total number of items (n) = 50
Favourable number of cases for defective items (m) = 5
Probability of defective item P(D) = \(\frac{m}{n}\) = \(\frac{5}{50}\) = 0.1.

Example
A card is drawn from pack of 52 cards at random what is the probability that it is
i) an ace
ii) a red 2 or black 7
iii) non-face card.
Solution:
Total number of cards (n) = 52
Total number of ace cards = 4
Favourable cases for ace (m) = 4
Probability (an ace) = \(\frac{m}{n}\) = \(\frac{4}{52}\) = \(\frac{1}{13}\)
Favourable cases for red 2 or black 7 (m) = 2 + 2 = 4
Probability (a red 2 or black 7) = \(\frac{m}{n}\) = \(\frac{4}{25}\) = \(\frac{1}{13}\)
Favourable cases for non-face card (m) = 10 \(\times\) 4
Probability (non-face card) = \(\frac{m}{n}\) = \(\frac{40}{52}\) = \(\frac{10}{13}\)

Laws of probability
Additive law of probability (or total probability law)
Additive law of probability has two cases:

When events are mutually exclusive
Let A and B are two mutually exclusive events in a random experiment then,
P(A or B) = P(A U B) = P(A) + P(B)

Example
A bag contains 20 balls numbered from 1 to 20. One ball is drawn at a random from the bag. Find the probability that the drawn balls are multiple of 5 or 7.
Solution
n = total number of balls in the bag = 20
One ball is drawn at random from the bag.
Let ‘A’ denote the event of drawing a ball multiple of 5 = {5, 10, 15, 20}
m = 4
P(A) = \(\frac{m}{n}\) = \(\frac{4}{20}\) = 0.2
Let ‘B’ denote the event of drawing a ball multiple of 7 = {7,14}
m=2
P(B) = \(\frac{m}{n}\) = \(\frac{2}{20}\) = 0.1
Here the events are mutually exclusive
Hence the probability that the drawn balls is multiple of 5 or 7 is,
= P(A or B)
= P(A) + P(B)
= \(\frac{4}{20}\) + \(\frac{2}{20}\)
= \(\frac{4 + 2}{20}\)
= \(\frac{3}{10}\)

When events are not mutually exclusive
Let A and B are two non-mutually exclusive events, then
P(A or B) = P(A U B) = P(A) + P(B) – P(A ∩ B)

Example:
A person is applying for the post of manager in bank A and bank B. the probability of his selection in bank A is 0.10 and in bank B is 0.15. He has also chance of selecting in both bank at 5the same time with probability 0.09. What is the probability that he will be selected in bank A or bank B?
Solution
Let A be the event of selecting in Bank A
B be the event of selecting in bank B
P(A) = 0.10, P(B) = 0.15, P(A ∩ B) = 0.09
Probability that he will be selected in bank A or B
= P(A or B)
= P(A U B)
= P(A) + P(B) – P(A ∩ B)
= 0.10 + 0.15 – 0.09
= 0.16

Multiplicative law of probability (or compound probability law)
This also has two cases:

When events are independent
If the events A and B are independent, then the probability of their occurrences together is equal to the product of their respective probabilities.
i.e. P (A and B ) = P(A ∩ B)
= P(A) . P(B)

Example
Aman and Suman appear in an interview for two vacancies in the same post. The probability of Aman’s selection is \(\frac{1}{7}\) and that of Suman’s selection is \(\frac{1}{5}\). What is the probability that only one will be selected?
Solution:
Let P (A) = prob. That Man will be selected = \(\frac{1}{7}\)
P (B) = prob. That Suman will be selected = \(\frac{1}{7}\)
P (\(\overline{A}\)) = prob that Aman will not be selected = 1 – \(\frac{1}{7}\) = \(\frac{6}{7}\)
P (\(\overline{B}\)) = prob. that Suman will not be selected = 1 – \(\frac{1}{5}\)  = \(\frac{4}{5}\)
P (only one of them will be selected)
= P(only Aman will be selected or only Suman will be selected)
= P (A and \(\overline{B}\) and \(\overline{A’}\))
= P(A) . P(\(\overline{B}\)) + P(B) . P (\(\overline{A}\))
= \(\frac{1}{7}\) \(\times\) \(\frac{4}{5}\) + \(\frac{1}{5}\) \(\times\) \(\frac{6}{7}\)
= \(\frac{10}{35}\)

When events are dependent
Let two events A and B be dependent than the probability of getting event A and B is given by
P( A and B) = P(A) . P()
or P (A ∩ B) = P(A) . P()
Similarly,
P (A ∩ B) = P(B) . P()

Example
Two cards drawn successively one after the other from a well-shuffled pack of 52 card. If the cards are not replaced, find the probability that all of them are kings.
Solution:
Let A and B be the events of drawing kings in the first and second draw without replacement respectively.
Therefore, probability of drawing a king in the first draw, P(A) = \(\frac{4}{52}\)
If the card drawn is not replaced, the second card is drawn from the remaining 51 cards. Similarly one king has drawn, now, there are three remaining kings.
Probability of drawing a king in the second draw, P(\(\frac{B}{A}\)) = \(\frac{3}{51}\)
Now, the required probability is given by
P (A and B) = P(A) . P(\(\frac{B}{A}\))
= \(\frac{4}{52}\) \(\times\) \(\frac{3}{51}\)
= \(\frac{12}{2652}\)

Conditional probability
Two events A and B are dependent, then probability of getting event A when event B has already happened is given by
P(\(\frac{A}{B}\)) = \(\frac{P(A∩B)}{P(B)}\)
Where,
P(\(\frac{A}{B}\)) = Conditional probability of A when B has already occurred
P(A∩B) = Joint probability between A and B
P(B) = Marginal probability of B.
Similarly,
P(\(\frac{B}{A}\)) = \(\frac{P(A∩B)}{P(A)}\)

Example
In an examination, 60% of the students have passed in marketing, 40% of the students have passed in economics and 20% have passed in both marketing and economics. A students is selected at random. What is the probability that the students has passed in marketing if it is known that he has passed in economics.
Solution:
A and B be the events that a student selected at random pass in the marketing and pass in economics respectively. Then,
P(A) = \(\frac{60}{100}\) = 0.6
P(B) = \(\frac{40}{100}\) = 0.4
P(A∩B) = \(\frac{20}{100}\) = 0.2
The required probability that the student selected has passed in marketing given that he has passed in economics is given by
P(\(\frac{A}{B}\)) = \(\frac{P(A∩B)}{P(B)}\)
= \(\frac{0.2}{0.4}\)
= \(\frac{1}{2}\)

Bayes’ theorem
We can revise probabilities when new information concerning a random experiment is available. The procedure for revising probabilities due to a specific cause is known as Bayes’ theorem. It was developed by Rev Thomas Bayes in 1763. It gives a probability law relating posteriori probability to a priori probability.

Definition
If E1,E2,……, En are mutually disjoint events with P (Ei) ≠ 0, I = 1,2,3,……, n then for any event A which is a subset of EU EU E3 ,……., U En, such that P(A) > 0, we have
P(\(\frac{E_i}{A}\)) = \(\frac{P(E_i) . P(\frac{A}{E_i})}{P(A)}\)
Where,
P(A) = P(E1) . P(\(\frac{A}{E_1}\) + P(E2)) . P(\(\frac{A}{E_2}\) + …….+ P(En)) . P(\(\frac{A}{E_n}\))

Example
There are two identical boxes containing 3 white and 4 black balls and 7 white and 3 black balls. A box is selected at random and ball is drawn from it. If the ball is white ,what is the probability that it is from the first box?
Solution
Let E1 and E2 be the events of selecting first and second boxes respectively.
Probability of selecting the first box, P(E1) = \(\frac{1}{2}\)
Probability of selecting the second box, P(E2) = \(\frac{1}{2}\)
Let A be the events that the ball drawn is white. Then,
Probability of sdelect5ing a white ball from the first box, P(\(\frac{A}{E_1}\)) = \(\frac{3}{7}\)
Probability of sdelect5ing a white ball from the second box, P(\(\frac{A}{E_2}\)) = \(\frac{7}{10}\)
The probability of selecting a white ball is given by,
P(A) = P(E1) . P(\(\frac{A}{E_1}\)) + P(E2) . P(\(\frac{A}{E_2}\))
= \(\frac{1}{2}\) \(\times\) \(\frac{3}{7}\) + \(\frac{1}{2}\) \(\times\) \(\frac{7}{10}\)
= \(\frac{3}{14}\) + \(\frac{7}{20}\)
= \(\frac{79}{140}\)
By using Bayes theorem, the required probability is given by
P(\(\frac{E_1}{A}\)) = P(E1) . P(\(\frac{(\frac{A}{E_1})}{P(A)}\))
= \(\frac{1}{2}\) \(\times\) \(\frac{\frac{3}{7}}{\frac{79}{140}}\)
= 0.379


Reference

Chaudary, A.K. (2061).Business statistics. kathmandu:Bhundipuran Prakshan
Dhakal Bashanta (2014).Business Statistics,Buddha academic publisher
Sthapit, Azaya Bikram(2006),Business Statistics,Asmita publication


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