Unit-2 Probability:- Probability is “How likely something is to happen” or “chance of occurring of an event”. Many events can't be predicted with total certainty. The best we can say is how likely they are to happen, using the idea of probability. Tossing a Coin when a coin is tossed, there are two possible outcomes: heads (H) or tails (T). We say that the probability of the coin landing H is ½. And the probability of the coin landing T is ½. Throwing Dice When a single die is thrown, there are six possible outcomes: 1, 2, 3, 4, 5, 6. The probability of any one of them is 1/6. Probability In general: Probability of an eventhappening = Number of wa yscanhappen Total numberof outcomes Example: The chances of rolling a "4" with a die. Number of ways it can happen: 1 (there is only 1 face with a "4" on it) Total number of outcomes: 6 (there are 6 faces altogether) So the probability ¿ 1 6
1. Unit-2 Probability:- Probability is How likely something is
to happen or chance of occurring of an event. Many events can't be
predicted with total certainty. The bestwe can say is how likely
they are to happen, using the idea of probability. Tossing a Coin
when a coin is tossed, there are two possible outcomes: heads (H)
or tails (T). We say that the probability of the coin landing H is
. And the probability of the coin landing T is . Throwing Dice When
a single die is thrown, there are six possible outcomes: 1, 2, 3,
4, 5, 6. The probability of any one of them is 1/6. Probability In
general: = Example:The chances of rolling a "4" with a die. Number
of ways it can happen: 1 (there is only 1 face with a "4" on it)
Total number of outcomes: 6 (there are 6 faces altogether) So the
probability = 1 6 Example:There are 5 marbles in a bag: 4 are blue,
and 1 is red. Whatis the probability that a blue marble gets
picked? Number of ways it can happen: 4 (there are 4 blues) Total
number of outcomes: 5 (there are 5 marbles in total) So the
probability = 4 5 = 0.8 Note:- Probability is always between 0 and
1
2. Probability is Just a Guide (Probability does not tell us
exactly whatwill happen, it is justa guide) Example:Toss a coin 100
times, how many Heads will come up? Probability says that heads
have a chance, so we can expect 50 Heads. But when we actually try
it we might get 48 heads, or 55 heads ... or anything really, but
in most cases it will be a number near 50. Some words havespecial
meaning in Probability: Experiment or Trial: an action where the
result is uncertain. Tossing a coin, throwing dice, seeing
whatpizza people chooseare all examples of experiments. Sample
Space: all the possibleoutcomes of an experiment Example:choosing a
card froma deck There are 52 cards in a deck (not including Jokers)
So the Sample Space is all 52 possiblecards: {Ace of Hearts, 2 of
Hearts, etc... } The Sample Space is made up of Sample Points:
Sample Point: justone of the possible outcomes Example:Deck of
Cards, the 5 of Clubs is a sample point the King of Hearts is a
sample point "King" is not a sample point. As there are 4 Kings,
that is 4 different sample points. Event:a single result of an
experiment Example Events: Getting a Tail when tossing a coin is an
event Rolling a "5" is an event. An event can include one or more
possibleoutcomes: Choosing a "King" froma deck of cards (any of the
4 Kings) is an event Rolling an "even number" (2, 4 or 6) is also
an event
3. The Sample Space is all possibleoutcomes. A Sample Point is
justone possibleoutcome. And an Event can be one or more of the
possibleoutcomes. Hey, let's usethose words, so we get used to
them: Example:Alex wants to see how many times a "double" comes up
when throwing 2 dice. Each time Alex throws the 2 dice is an
Experiment. Itis an Experiment becausethe resultis uncertain. The
Event Alex is looking for is a "double", whereboth dice havethe
same number. Itis made up of these 6 Sample Points: {1,1} {2,2}
{3,3} {4,4} {5,5} and {6,6} The Sample Space is all
possibleoutcomes (36 Sample Points): {1,1} {1,2} {1,3} {1,4} ...
{6,3} {6,4} {6,5} {6,6} These are Alex's Results: Experiment Is it
a Double? {3,4} No {5,1} No {2,2} Yes {6,3} No ... ...
4. After 100 Experiments, Alex has 19 "double" Events ... is
that close to what you would expect? Dependent Events:- Event A and
B in the sample spaceS are said to be independent if occurrenceof
one of them does not influence the occurrenceof the other. i.e. B
is dependent of A if P(B/A)=P(B) Now, P(AB)=P(A).P(B|A)=P(A)P(B)
Thus events A and B are independent, if P(AB)=P(A).P(B) Conditional
Probabiliry:- Given an event B such that P(B)>0 and any other
event A, we define a conditional probability of A is given by ( | )
= ( ) () Example: Let a balanced die is tossed once. Use the
definition to find the probability of getting 1, given that an odd
number was obtained. Solution:- Probability of A given that the
event B has occurred, ( | ) = () () But ( ) = 1 2 , ( | ) = () () =
1 6 1 2 = 1 3 . Law of Total Probability Let 1, 2, be mutually
exclusive and let an event A occur only if anyoneof occurs, Then (
) = (|)() =1 Bayes Theorem Let 1, 2, be mutually exclusive event
such that ( ) > 0 for each i. Then for any event 1 such that ( )
> 0, we have ( |) = (|)() (|)() =1 = 1,2, Example:- Three urns
A, B, C have 1 white , 2 black, 3 red balls, 2 white 1 black, 1 red
balls and 4 white Example:- Let P be the probability that a man
aged y years will meet with an accident in a year. What is the
probability that a man among n men all aged y years will meet with
an accident ? Solution:-
5. Probability that a man aged y years will not meet with an
accident =1-p P (none meet with an accident) =(1 )(1 ) = (1 ) p
(atleast one man meets with an accident) = 1 (1 ) So p( atleast one
man meets with an accident, a person is chosen) = 1 [1 (1 ) ]
Example:- A box contains 4 white, 3 blue and 5 green balls. Four
balls are chosen. What is the probability that all three colors are
represented ? Solution:- The total number of balls in the box is
12. Hence the total number of ways in which 4 balls can be chosen =
12 4 = 12 11 10 9 4 3 2 1 = 495 Each color will be represented in
the following mutually exclusive ways: White Blue Green (i) 2 1 1
(ii) 1 2 1 (iii) 1 1 2 Hence the number of ways of drawing four
balls in the abovefashion = 4 2 3 1 5 1 + 4 1 3 2 5 1 + 4 1 3 1 5 2
= 90 + 60 + 120 = 270 So Required Probability = 270 495 . Random
variable:- A variable whosevalueis determined by the outcome of
randomexperiment is called randomvariable. The value of the
randomvariable will vary fromtrial to trial as the experiment is
repeated. Random variable is also called chance variableor
stochastic variable. There are two types of random variable -
discrete and continuous. A random variable has either an associated
probability distribution (discrete random variable) or probability
density function (continuous random variable). Discrete
randomvariable:- If the random variable takes on the integer values
such as 0,1,2, then it is called discrete random variable.
6. Example:- 1. The number in printing mistake in a book, the
number of telephone calls received by the phone operator on a farm
areexample of discrete random variable 2. A coin is tossed ten
times. The randomvariable X is the number of tails that are noted.
X can only take the values 0, 1, ..., 10, so X is a discrete random
variable. Continuous random variable:- If the randomvariable takes
all values, with in a certain interval, then the random is called
continuous randomvariable. Example:- 1. The amount of rainfall on a
rainy day or in a rainy season, the height and weight of
individuals are example of continuous randomvariable. 2. A light
bulb is burned until it burns out. The randomvariable Y is its
lifetime in hours. Y can take any positive real value, so Y is a
continuous random variable. Probability Distribution The
probability distribution of a discrete randomvariable is a list of
probabilities associated with each of its possiblevalues. Itis also
sometimes called the probability function or the probability mass
function. More formally, In terms of symbols if a variableX can
assumediscrete set of values 1, 2, with respective probabilities 1,
2, where 1 + 2 + + = 1, we say that a discrete probability
distribution for has been defined. The function () which has the
respectivevalues 1 , 2 , for = 1, 2, is called the probability
function or frequency function of . Note:-
7. a. 0 () 1 b. ( ) = 1 ExpectedValue:- The expected value (or
population mean) of a randomvariable indicates its averageor
central value. Itis a useful summary value (a number) of the
variable's distribution. Stating the expected value gives a general
impression of the behavior of some randomvariable without giving
full details of its probability distribution (if it is discrete) or
its probability density function (if it is continuous). Two
randomvariables with the same expected value can havevery different
distributions. There are other usefuldescriptivemeasures which
affect the shape of the distribution, for example variance. The
expected value of a randomvariable X is symbolized by E(X) or . If
is a discrete randomvariable with possible values 1, 2, with
respective probabilities 1 , 2 , where 1 + 2 + + =1the mathematical
expectation of is defined as: = ( ) = ()() Where the elements are
summed over all values of the randomvariable X. If X is a
continuous randomvariablewith probability density function f(x),
then the expected value of X is defined by: = ( ) = ( )
Example
8. Discrete case: When a die is thrown, each of the
possiblefaces 1, 2, 3, 4, 5, 6 (the 's) has a probability of 1/6
(the ()'s) of showing. Theexpected value of the face showing is
therefore: = ( ) = (1 1 6 ) + (2 1 6 ) + (3 1 6 ) + (4 1 6 ) + (5 1
6 ) + (6 1 6 ) = 3.5 Notice that, in this case, () is 3.5, which is
not a possiblevalue of . Variance of = ( ) = 2 = ( 2) [ ()]2
Standard Deviationof = Example:- Find expectation of the number of
points when a fair die is rolled. Solution:- Let be the
randomvariable showing number of points. Then = 1,2,3,4,5,6 ( = ) =
() Product 1 1 6 1 6 2 1 6 2 6 3 1 6 3 6 4 1 6 4 6 5 1 6 5 6 6 1 6
6 6 ________________ ( ) = 21 6 = 7 2 Expectation = 7 2 .
9. Normal Distribution In probability theory, the normal (or
Gaussian) distribution is a very commonly occurring continuous
probability distribution . Strictly, a Normalrandomvariable should
be capable of assuming any valueon the real line, though this
requirement is often waived in practice. For example, height at a
given age for a given gender in a given racial group is adequately
described by a Normal randomvariableeven though heights mustbe
positive. A continuous randomvariable X is said to follow a Normal
distribution with parameters and if it has probability density
function ( ) = (, , ) = 1 2 ( )2 22 , > 0, < < In such
situation we write ~( , 2) The distribution involves two parameters
and 2 . Properties:- a) Distribution is symmetrical. b) = , = 2
.
10. c) Mean , median and mode coincide. d) () 0 for all . e) (
) = 1 i.e. Total area under the curve = () bounded by the axix of
is 1. : a) Curvey-f(x), called normal curveis a bell shaped curve.
b) Itis symmetric about = . c) The two tails on the left and right
sides of the mean extend to infinity. d) Put = . Then is called a
standard normalvariable and its probability density function is
given by ( ) = 1 2 2 2 , . e) Mean of the standard
Normaldistributions is 0 and varianceis 1. Write ~(0,1) The
simplest case of a normaldistribution is known as the standard
normaldistribution. Areaunder the standard normal curve:- a) = 1 =
1 0.6827(Since total area under the standard normal curveis 1) . .
(1 < < 1) = 0.6827 b) = 2 = 2 0.9595 . . (2 < < 2) =
0.9595 c) = 3 = 3 0.9973 . . (3 < > 3) = 0.9973 In other
words,
11. EXAMPLE: If a randomvariable has the normaldistribution
with = 82.0 and = 4.8, Find the probabilities that it will take on
a value (a) less than 89.2 (b) greater than 78.4 (c) between 83.2
and 88.0 (d) between 73.6 and 90.4 Solution: (a) We have
12. z = 89.2 82 4.8 = 1.5 therefore the probability is 0.4332 +
0.5 = 0.9332. (b) We have = 78.4 82 4.8 = 0.75 therefore the
probability is 0.2734 + 0.5 = 0.7734 . (c) We have 1 = 83.282 4.8 =
0.25and 2 = 8882 4.8 = 1.25 therefore the probability is 0.3944
0.0987 = 0.2957. (d) We have 1 = 73.682 4.8 = 1.75and 2 = 90.482
4.8 = 1.75 hereforethe probability is 0.4599 + 0.4599 = 0.9198.
Applications of the Normal Distribution EXAMPLE: Intelligence
quotients (IQs) measured on the Stanford Revision of the
Binet-Simon Intelligence Scale are normally distributed with a mean
of 100 and a standard deviation of 16. Determine the percentage of
people who haveIQs between 115 and 140. Solution: We have 1 =
115100 16 = 0.9375 and 2 = 140100 16 = 2.5 therefore the
probability is 0.4938 0.3264 = 0.1674. Itfollows that 16.74% of
allpeople have IQs between 115 and 140. Equivalently, the
probability is 0.1674 thata randomly selected person will havean IQ
between 115 and 140. Many distribution tend to a normaldistribution
in the limit. When a variable is not normal, it can be made normal
using using suitable transformation. When the sample size is large
distrubution of simple mean, simple variance etc. approach
normality. Thus distribution forms a basis for test of
significance. Normal distribution is also called distribution of
errors.