Subjective Probability, Addition Law for M.E Events, Simple & Compound Events

Superior University Lahore

Sahiwal Campus

Department:	Information Technology
Programme:	B.S(I.T)
Course:	Statistics and Probability
Topic:	Subjective Probability, Addition Law for M.E Events, Simple & Compound Events
Submitted to:	M Zeeshan sarwar
Submitted by:	Prof. Fyaz hussain
Roll no:	028
Date of Submission:	16/11/2015

                              (shanich108@gmail.com)

Subjective Probability

DEFINITION OF 'SUBJECTIVE PROBABILITY'

A probability derived from an individual's personal judgment about whether a specific outcome is likely to occur. Subjective probabilities contain no formal calculations and only reflect the subject's opinions and past experience.

BREAKING DOWN 'SUBJECTIVE PROBABILITY'

Subjective probabilities differ from person to person. Because the probability is subjective, it contains a high degree of personal bias. An example of subjective probability could be asking New York Yankees fans, before the baseball season starts, the chances of New York winning the World Series. While there is no absolute mathematical proof behind the answer to the example, fans might still reply in actual percentage terms, such as the Yankees having a 25% chance of winning the world series.

(Read more: Subjective Probability Definition | Investopedia http://www.investopedia.com/terms/s/subjective_probability.asp#ixzz3rWsATVJu )

 

 

Addition Law for Mutually Exclusive Events

 

First go through the mutually exclusive events. These are the two or more than two events that have nothing in common. Theorem for Addition law of mutually exclusive events states that:

If there are two events say “x” and “y” and both are mutually exclusive events, then the probability that either “x” or “y” occur is the sum of the probabilities of both the events.

Suppose there are two events named as “x” and “y”, both events have nothing in common. Addition of such event is done by following the addition law for mutually exclusive events.

P(x or y) = p(x u y) = p(x) + p(y)

 

Let make it clear by solving simple examples.

EXAMPLE 1:

A single card is selected from a deck of 52 cards. Find the probability that the randomly selected card is either king or queen.

Solution:

Total number of outcomes = 52

Probability of each single card = 1/52

Now let first find the probability of queen

Probability of queen is represented by = P(Q)

There are four queens in a deck of playing cards.

Therefore, probability of queen = P(Q) = 4/52

Now, let find the probability of king

Probability of king is represented by = P(k)

In the same manner there are four kings in a deck of 52 playing cards.

Therefore, probability of queen = P(K) = 4/52

Addition law of mutually exclusive event is used, as the requirement is to find the probability of king or queen.

Therefore,

P(K or Q) = P(K u Q) = P(K) + P(Q) = 4/52 + 4/52 = 8/52

 

EXAMPLE 2:

A dice is thrown. Find the probability that the face is less than three or it is multiple of 5.

Solution:

When a dice is rolled, there are six possible outcomes

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

Let first find the probability of face that are less than 3.

A = {1, 2} = 2/6

Now second part is that the face is multiple of 5.

B = {5} = 1/6

Use addition law, as again there is nothing in common between the two events.

P (A or B) = P(A u B) = P(A) + P(B) = {1, 2} + {5} = {1, 2, 5}

P(A or B) = 2/6 + 1/6 = 1/2

Definition Of Equally Likely

The outcomes of an experiment are equally likely to occur when the probability of each outcome is equal.

Example of Equally Likely:

·         When you toss a fair coin, you are equally likely to get a head or a tail.

·         When you roll a fair die, you are equally likely to roll a 1, 2, 3, 4, 5, or 6.

Ques: Which one of the following best replaces the blank? 'A and B are playing tennis. The event of A or B winning the match is ______________.'

Choices:

A. a likely event 
B. an unlikely event
C. an equally likely event 
D. none of the above
Correct Answer: C

Solution:

Step 1: Both A and B have equal chances of winning the match.
Step 2: So, it is an equally likely event.

EVENTS

Event is the subset of the sample space or event can also be defined as the collection of either one or more than one outcomes of an experiment. It is sum of all possible outcomes of an experiment. Sample space forms after combining all the events.  Any part of the sample space can be named as an event. Event may be of one outcome or a combination of more than one outcome.

Event with a single outcome is named as simple event and an event with having two or more than two outcomes is known as compound event.

SIMPLE EVENTS:

Simple events can be defined as the single outcome of the performed experiment or it is an event which cannot be broken down any more.

COMPOUND EVENTS:

Compound events are the combination of two or more than two simple event. It can also be defined as an event that contains more than one sample points in it.

EVENT EXAMPLES:

1.     If a single face is considered when a die is rolled, then it will be simple event. For example suppose getting 5 or 6 or 3 or 2 etc… on the die when it is thrown, is called as simple event. If the event is any even number on the die, then the event is consist of points {2, 4, 6}, which is known as compound event. That compound event is consisting of three simple events i.e., {2}, {4} and {6}.

2.     Suppose two dice are rolled simultaneously, then the pair (1, 1) will be the simple event. This is so, because it is a single outcome in the sample space. If event consists of the sum of two dice is ‘’5’’ than it consists of four outcomes i.e., (1, 4), (2, 3), (3, 2), (4, 1) and this is considered to be a compound event.

3.     Suppose two coins are tossed simultaneously, then the pair (HT) will be the simple event. If condition is defined that an event should consist of at least one head then there are three outcomes. These outcomes are (HH), (HT) and (TH) and this is said to be as compound event. That compound event consists of three simple events i.e., {HH}, {HT} and {TH}.

NUMERICALS...

Q#1:

A Marble drawn At Random From a box containing 10 red, 30 white, 20 blue and 15 orange marbles. Find the probability that is:

Ø      Orange or Red

Ø      Not Red or Blue

Ø      Not Blue

Ø      White

Ø      Red, White or Blue

                

SOLUTION:

RED=10               P (RED) =10/75

WHITE=30         P (WHITE) =30/75

BLUE=20            P (BLUE) =20/75

ORANGE=15      P (ORANGE) =15/75

TOTAL=75

Ø  Orange Or Red

P (Orange Or Red) = P (Orange)+ P(Red)

                                  15/75+10/75=25/75=1/3

Ø  Not Red Or Blue

P (Not Red Or Blue) = P (White) +P (Orange)

                                      30/75+15/75=45/75=3/5

Ø  Not Blue

P (Not Blue)= P(White)+P(Red)+P(Orange)

                          30/75+10/75+15/75+15/75=11/15

Ø  White

P (White) =30/75

Ø  Red, White Or Blue

P (Red, White Or Blue) =P(Red)+P(White)+P(Blue)

                                         10/75+30/75+20/75=60/75

Q#2

Of 12 eggs in a refrigerator, 2 are bad. From these, 4 eggs are chosen at random to make cake. What are probabilities that:

Ø      Exactly one is bad

Ø      At least one is bad

SOLUTION:

Total eggs= 12

Bad eggs= 2

Good eggs= 10

n=4

P (exactly 1 bad egg) = 2C1 10C3/12C4=0.48

P (at least one is bad) = 2C1 10C3 + 2C2 10C2/12C4 =0.575

Q#3

A bag contains 14 identical balls, 4 of which are red, 5 black and 5 white. Six balls are drawn from the bag. Find the probability that

Ø      3 are red

Ø      At least two are white

SOLUTION:

N = 14

Red = 4

Black = 5

White = 5

n = 6

P (three are red) = 4C3+ 10C3/14C6= 124/3003=0.041

P (at least two are white) = 5C2. 9C4+ 5C3.9C3+ 5C4. 9C2+ 5C5.9C1/ 14C6= 2289/3003=0.7