sample space and events,probability,laws of probability,conditional probability,bayes' theorem

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…..,E_n are n mutually exclusive outcomes or sample events of a random experiment, then set S={E_1, E_2, E_3,………E_n} 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 otherevent5s or vice-versa.for4 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≤P≤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. ∑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 1:

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 2:

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 and iii) non-face card.

Solution:

Total number of cards (n) = 52

1. Total number of ace cards =4

Favourable cases for ace (m) =4

Probability (an ace) =\{\frac{m}{n}\}=\{\frac{4}{52}\}=\{\frac{1}{13}\}

1. 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*4

Probability (non-face card) =\{\frac{m}{n}\}=\{\frac{40}{52}\}=\{\frac{10}{13}\}

Laws of probability

1. 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}\}

=

• 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

1. 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 1/7 and that of Suman’s selection is 1/5. What is the probability that only one will be selected?

Solution:

Let P (A) = prob. That Aman will be selected = 1/7

P (B) = prob. That Suman will be selected = 1/5

P (\{\overline{A}\}) = prob that Aman will not be selected = 1 – 1/7= 6/7

P (\{\overline{B}\}) = prob. that Suman will not be selected = 1 – 1/5 =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 A’)

=P(A).P(\{\overline{B}\}) + P(B).P(\{\overline{A}\})

=\{\frac{1}{7}\}*\{\frac{4}{5}\}+ \{\frac{1}{5}\} *\{\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(B/A)=\{\frac{3}{51}\}

Now, the required probability is given by

P(A and B) = P(A) .P(B/A)

=\{\frac{4}{52}\} *\{\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(A/B) = P(A∩B)/P(B)

Where,

P(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(B/A)=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(A/B)= 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,E_2,……,E_n are mutually disjoint events with P (E_i) ≠0, I =1,2,3,……,n then for any event A which is a subset of E₁ U E₂ U E₃ ,……., U E_n, such that P(A)>0, we have

P(E_i/A)= \{frac{P(E_i) .P(A/E_i)}{P(A)\}

Where,

P(A)= P(E₁).P(A/E₁) + P(E₂).P(A/E₂) + …….+ P(E_n).P(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(A/E₁)=\{\frac{3}{7}\}

Probability of sdelect5ing a white ball from the second box, P(A/E₂)=\{\frac{7}{10}\}

The probability of selecting a white ball is given by,

P(A) =P(E₁).P(A/E₁) + P(E₂).P(A/E₂)

=\{\frac{1}{2}\}*\{\frac{3}{7}\} + \{\frac{1}{2}\}*\{\frac{7}{10}\}

=\{\frac{3}{14}\} +\{\frac{7}{20}\}

=\{\frac{79}{140}\}

By using Bayes theorem, the required probability is given by

P(E₁/A) = P(E₁).P(A/E_­1)/P(A)

=1/2*3/7 / 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