Sample Space and Events, Probability, Laws of Probability, Conditional Probability, Bayes' Theorem

Sample Space

An experiment's sample space is a collection of all potential outcomes. S stands for it. For instance, the set S=E1, E2, E3,.........En is referred to as the sample space of a random experiment if E1, E2, E3,.........En are n mutually exclusive outcomes or sample events of a random experiment.

For examples:

• When tossing two fair coins simultaneously.

S={HH,HT,TH,TT}

• The sample space for a fair die roll is as follows:

S={1,2,3,4,5,6}

Events or Outcome

Events or outcomes are the potential outcomes or results of a random experiment. For instances:

• The two possible results of a coin flip are head (H) and tail (T) (T)
• The available outcomes of rolling a fair die are 1, 2, 3, 4, 5, and 6.
• The probable results when tossing two fair coins simultaneously are HH,TT,HT,TH.

Simple events are those that correlate to a single possible outcome of a random experiment; joint, composite, or compound events are those that correspond to two or more outcomes simultaneously in a single trial. Rolling a dice, for instance, and receiving a face 1 is a simple event, whereas getting an even face is a complex event.

Types of Events

• Equally likely events: It is argued that two or more events are equally likely if their chances of occurring in a random experiment are equal. For instance, when unbiased or uniform coins are tossed, both the heads (H) and tails (T) are equally likely outcomes, and when unbiased dice are rolled, all six faces are equally likely outcomes.

• Mutually exclusive events: If the occurrence of one event does not result in the occurrence of other events in a single trial of a random experiment, the events or cases are said to be mutually exclusive. For instance, the six numbered faces of a fair die are all mutually exclusive occurrences, as are the head (H) and tail (T) outcomes of a fair coin toss.

• Independent and dependent events: If the occurrence of one event has no bearing on the occurrence of any more occurrences, or vice versa, the two or more events are said to be independent of one another. The incidence of the king does not alter the occurrence of the other cards, for example, if a pack of cards is playing with replacements. If the occurrence of one event influences the occurrence of another event or vice versa, two or more occurrences are said to be dependent on one another. The recurrence of a king, for instance, influences the occurrence of other cards, and vice versa, if a pack of cards is being played without replacement.

• Exhaustive events: Exhaustive occurrences refers to all outcomes that could possibly occur in a random experiment. n is used to indicate it. 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: Favorable events are those which increase the likelihood of that event occurring. It is indicated by the letter m.

  • If a box contains 5 white balls and 6 black balls, then there are exactly 5 favorable scenarios in which you will obtain white balls.
  • The sample space S=HH,HT,TH,TT while throwing two coins simultaneously, and the favorable number of cases for receiving two heads is 2, or HT,TH.

Probability

Probability is the likelihood of any event occurring or not occurring in a random experiment.

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).

The probability that an event will occur is given by, if n is the exhaustive number of equally likely and mutually exclusive events and m is the number of examples for the desired event E.

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:

Five of the fifty goods in a box are faulty. What is the likelihood that a single item chosen at random is flawed?

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:

What is the likelihood that a random card from a deck of 52 will be an ace? A non-face card, a red 2 or a black 7, and iii) both.

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*4

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

Laws of Probability

• Probability law of addition (or total probability law):

Additive law of probability has two cases:

• When events are mutually exclusive:

If, in a random experiment, A and B are two events that are mutually exclusive, then 

P(A or B)= P(A U B)= P(A) + P(B)

Example:

20 balls, numbered from 1 to 20, are contained in a bag. A random ball is taken out of the bag. Calculate the likelihood that the drawn balls will be a multiple of 5 or 7.

Solution:

n= total number of balls in the bag=20

A random ball is taken out of 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

The circumstances in this case are exclusive.

Consequently, the likelihood that a multiple of 5 or 7 will be drawn is,

= P(A or B)

= P(A) + P(B)

=\{\frac{4}{20}\}+\{\frac{2}{20}\}

• When events are not mutually exclusive:

If A and B are two events that are not mutually exclusive, then

P(A or B) = P(A U B) = P(A) + P(B) – P(A ∩ B)

Example:

In banks A and B, a person has applied for the position of manager. His pick is 0.10 more likely in bank A than bank B, at 0.15. He also has a 0.09 percent chance of choosing from both banks at the same moment. What is his chance of being chosen for 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:

The likelihood of events A and B occurring together is equal to the product of their individual probabilities if A and B are independent events.

i.e. P (A and B )= P(A ∩ B)

= P(A) . P(B)

Example:

Aman and Suman attend an interview for two positions that are open at the same time. The odds of Aman and Suman being chosen are 1/7 and 1/5, respectively. How likely is it that only one will be chosen?

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:

If two events A and B are dependent, then the likelihood that they will both occur 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 are selected sequentially, one after the other, from a 52-card deck that has been properly shuffled. Find the likelihood that all the cards are kings if they are not replaced.

Solution:

Let A and B represent the outcomes of the first and second draws without replacement, respectively, in which kings are drawn.

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

When two occurrences A and B are interdependent, the likelihood that event A will occur after event B has occurred is determined 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:

60% of students who took the test passed the marketing section, 40% the economics section, and 20% the marketing and economics sections. The student chosen is chosen at random. If it is known that a student passed economics, what is the likelihood that he will pass marketing?

Solution:

The tests that a student chosen at random passes in marketing and economics are A and B, respectively. Then,

P(A)=\{\frac{60}{100}\}=0.6

P(B)=\{\frac{40}{100}\}=0.4

P(A∩B) =\{\frac{20}{100}\}=0.2

Given that he has passed economics, the necessary chance that the chosen student has passed marketing is given by

P(A/B)= P(A∩B)/P(B)

=\{\frac{0.2}{0.4}\}

=\{\frac{1}{2}\}

Bayes’ Theorem

When fresh information regarding a random experiment becomes available, we can update the probability. The Bayes theorem describes the process for updating probabilities as a result of a certain cause. In 1763, Rev. Thomas Bayes created it. It provides a probability law that connects posteriori and a priori probabilities.

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 E1 U E2 U E3 ,……., U En, such that P(A)>0, we have

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

Where,

P(A)= P(E1).P(A/E1) + P(E2).P(A/E2) + …….+ P(En).P(A/En)

Example:

Three white and four black balls as well as seven white and three black balls are contained in two identical boxes. A random box is chosen, and a ball is drawn from it. What is the likelihood that the ball is from the first box if it is white?

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/E­1)/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