Probability:
Basic concepts of Probability:
Probability is a way to measure hoe likely something is to happen. Probability is number between 0 and 1, where probability is 0 means is not happen at all and probability is 1 means it will be definitely happen, e.g. if we tossed coin there is a 50% chance to get head and 50% chance to get tail, it can be represented in probability as 0.5 for each outcome to get head and tail. Probability is used to help us taking decision and predicting the likelihood of the event in many areas, that are science, finance and Statistics.
Now we learn the some basic concepts that used in Probability:
i) Random Experiment OR Trail: A Random Experiment is an process that get one or more possible outcomes. examples of random experiment include tossing a coin, rolling a die, drawing a card from pack of card etc. using this we specify the possible outcomes known as sample pace.
ii)Outcome: An outcome is a result of experiment. an outcome is one of the possible result that can occur when random experiment is performed. e.g. if we tossed coin the possible outcomes are either head or tail, that outcome of experiment of tossing coin (i.e. H, T). the some of probability of all possible outcome of an experiment is always equal to 1.
iii) Sample Space: Sample Space is the set of all possible outcome that occur in experiment. it is denoted as "S". e.g. if we tossed coin the sample space consist the following possible outcomes as head and tail. (i.e. H, T), in experiment of rolling die following are possible outcomes as 1,2,3,4,5,6.
iv) Event : event means we are interested in getting particular type of outcome out off set of all type of outcomes. e.g. in coin tossing experiment we get H & T but we are interested to getting H, it is getting H is event.
v)Certain Event: A certain event is an event that guaranteed to occur. or if an event A is the entire sample space then the event A is called certain event. the probability of certain event is 1.
vi)Mutually Exclusive: Mutually Exclusive events are those event that cannot occur at the same time. if two events are said to be mutually exclusive event means one event can occur then other is not occur. or if two events are said to be mutually exclusive event if both event cannot happen simultaneously , e.g. tossing a coin we get only head or tail only at one time both not happen same time.
vii)Equally Likely Event: Equally likely event are event that have same probability of occurrence. e.g. the probability of rolling a die we get 1,2,3,4,5,6 and each have same chance out occur or each have same probability.
viii) Impossible Event: the event dose not have any sample point is called impossible event. or there is no chance to occur this event. e.g. we rolling die then it is impossible to get 7. it is an impossible event. probability of impossible event is always 0.
ix)Exhaustive Event: A event is said to be exhaustive event if an event include all possible outcomes of an experiment. it covers all possible outcomes of experiment. e.g. if we toss coin two exhaustive events are Head and Tail. there is no other outcome of that experiment.
x) Union of Two Event: The event containing the sample point which are in either in event A or in event B or in both is called the union of the event A and B. And it is denoted as AUB Or A+B .
Example: if A = { 1, 2, 4 } and B = { 3, 5 }
then AUB = { 1, 2, 3, 4, 5 }
Note that Union of the event A and B means Combining the all point in the both event A and B.
xi ) Intersection of Two event: The event containing sample point of event A and event B both is called Intersection of the event A and B. and it is denotes as A⋂B.
Example: if A = { 2, 4, 8 } and B= { 2, 8, 6 , 10 }
therefore A⋂B = { 2, 8 }
Note: Intersection of Event A and B means collecting common point in the both events A and B.
Definition and Properties of Probability:
Definition: if S is a sample space and it has n sample points they are equally likely, mutually exclusive and exhaustive and an event A has m points then the ration m/n is called Probability of event A and it is denoted as P(A).
i.e.
P(A) = (m/n) = (Number of point in event A / Number of point in sample space S).
Properties: 1. Probability of certain event is 1. i.e. P(S) = 1
(Certain Event: A certain event is an event that guaranteed to occur. or if an event A is the entire sample space then the event A is called certain event. the probability of certain event is 1. )
Proof: If S has n points and A is certain event then it contain entire sample space S or sample Point, then A has n point. therefore by definition of probability,
P(A) = (Number of point in A / Number of point in S) = (n/n) = 1
2.Probability od Impossible event is zero. i.e. P(ф) = 0 (Impossible event is denoted as ф)
Proof: we known Impossible event has no point.
P(ф) = (0/n) = 0
(Impossible Event: the event dose not have any sample point is called impossible event. or there is no chance to occur this event. e.g. we rolling die then it is impossible to get 7. it is an impossible event. probability of impossible event is always 0.)
3. Probability of an event always lies between 0 and 1. i.e. 0≤ P(A) ≥ 1
Proof: if S has n point and A has m points then the number of points in event A is less than or equal to n because S is sample space it contains all point of events. and point in A is greater than 0,
0≤ m≤ n
now dividing by n, we get
0≤ (m/n) ≤ (n/n)
0≤ P(A) ≤ 1.
Hence, Probability of an event always lies between 0 and 1
Example 1. : find the probability of getting tail in coin tossing event.
Answer: In a coin tossing experiment there are only two outcomes that are head and tail, therefore the sample space is two pints head and tail only. consider event A is getting tail. therefore the event A has only one point. i.e. n(S)= n = 2 and n(A)= m=1.
P(A) = (m /n) = (1/2)= 0.5
Therefore the probability of getting tail is Equal to 0.5
Example 2. An urn contains 8 white balls and 12 black balls. find the probability of drawing a white ball.
Answer:
we know the definition of probability of an event A is P(A) = (m /n)
where s is all points in sample space, here 8 white and 12 black balls therefore the total balls are 8 + 12 = 20 i.e. total balls are 20 means sample space has 20 balls. n = 20
and m is the point in event here getting a white ball is an event A therefore the 8 white balls means n = 8
the Probability of drawing a white ball = P ( A ) = ( m / n ) = 8 / 20 = 0.4
Hence the probability of drawing a white ball is equal to 0.4
Example 3. Find the Probability that a card drawn from pack of cards is King.
Answer:
here the event A is getting king card
We know the definition of probability of an event A is P(A) = (m /n)
the total cards in pack of card is 52 i.e. n = 52
and total king card in pack of a card is 4 i.e. m = 4
the probability of getting king card = P(A) = (m /n)
P(A) = (m /n) = (4 / 52 ) = 1 / 13 = 0.0769
The Probability of drawing a card from pack of card is king is equal to 0.0769
Example 4. Find the Probability of getting 3 or 5 in a throwing a dice.
Answer: the event A is that getting 3 or 5 in throwing dice.
therefore the sample space of throwing a dice is s = { 1, 2, 3, 4, 5, 6}
the total points in sample space is 6 i.e. n= 6
and the getting 3 or 5 means m = 2
the probability od getting 3 or 5 is = P(A) = (m /n) = (2 / 6 )
P(A) = ( 1 / 3 ) = 0.3333
The Probability of getting 3 or 5 in throwing a dice is equal to 0.3333
Probability Examples:
Example 1. An urn contains 10 white and 5 black ball's. two ball are drawn from urn without replacement. find the probability that i) both ball will be of same colour. ii) both ball will be of different colours.
Answer : Given an urn contains 10 white & 5 black balls, total number of ball's is 15
i) Probability that both ball's will be of same colour.
here given that two colours of balls, therefore we consider firstly the probability of drawing two ball's is white balls without replacement.
the probability of drawing a white ball in first drawn is 10/15. (i.e. if we want to find probability of drawing white ball is calculated as number of white ball's divided by total number of ball's = 10/15). in this case we- select next ball without replacement therefore first ball is not replaced then 9 white ball's are left in urn and total number of ball's is 14. the probability of drawing second ball is white is 9/14. (i.e. if we want to find probability of drawing white ball is calculated as number of white ball's divided by total number of ball's = 9/14).
hence the probability of drawing both ball's is white without replacement is = (10/15) x (9/14) =3/7.
now we consider the drawing both ball's is black ball's.
he probability of drawing a blackball in first drawn is 5/15. (i.e. if we want to find probability of drawing black ball is calculated as number of black ball's divided by total number of ball's = 5/15). in this case we select next ball without replacement therefore first ball is not replaced then 4 black ball's are left in urn and total number of ball's is 14. the probability of drawing second ball is black is 4/14. (i.e. if we want to find probability of drawing white ball is calculated as number of white ball's divided by total number of ball's = 4/14).
hence the probability of drawing both ball's is white without replacement is = (5/15) x (4/14) =1/21.
therefore The probability that drawing two ball's is same colour without replacement is sum of both ball's is white and black ball's is = (3/7) + (1/21) = 10/21.
ii) probability of drawing two ball's will be different in colours.
for finding this probability the probability of white ball drawing in first drawn is 10/15
and the probability of black ball drawing in second drawn is 5/14.
therefore probability of drawing white ball in first drawn and black ball in second drawn is = (10/15) x (5/14) = 0.23
similarly the probability of black ball drawing in first drawn is 5/15
and the probability of white ball drawing in second drawn is 10/14.
therefore probability of drawing black ball in first drawn and white ball in second drawn is = (5/15) x (10/14) = 0.23
therefore the probability of drawing both ball's have different colour without replacement is equal to sum of probability of drawing black ball in first drawn and white ball in second drawn and probability of drawing white ball in first drawn and black ball in second drawn is = 0.23+0.23 =0.46
Therefore the probability of drawing both ball's have different colour without replacement is = 0.46
Example 2. A Box contains 20 tickets numbered as 1 to 20. a ticket drawn randomly from the box. find the probability of ticket will be: i) multiple of 3, ii) a multiple of 5, iii) a multiple of 3 or 5, iv) a ticket multiple of 3 and five.
Answer: Given that a box contains 20 tickets and is numbered as 1 to 20,
i) Probability that the numbers on ticket will be a multiple of 3.
the ticket have 1 to 20 number in which the number on ticket will be multiple of 3 is 3,6,9,12,15,18.
number on tickets which are multiple of 3 is 6 numbers therefore the probability that the number on ticket will be a multiple of 3 is = 6/20
(i.e. 6 numbers are multiple of 3 and total number of tickets is 20).
Probability that the numbers on ticket will be a multiple of 3 is 6/20
ii) Probability that the number on ticket will be a multiple of 5.
the ticket have 1 to 20 number in which the numbers on ticket will be multiple of 5 is 5,10,15,20
number on tickets which are multiple of 5 is 4 numbers therefore the probability that the number on ticket will be a multiple of 5 is = 4/20
(i.e. 4 numbers are multiple of 5 and total number of tickets is 20)
Probability that the numbers on ticket will be a multiple of 5 is 4/20
iii) Probability that the number on ticket will be a multiple of 3 or 5.
the ticket have 1 to 20 number in which the numbers on ticket will be multiple of 3 or 5 is 3,6,9,12,15,18,5,10,20.
number on tickets which are multiple of 3 or 5 is 9 numbers therefore the probability that the number on ticket will be a multiple of 3 or 5 is = 9/20
(i.e. 4 numbers are multiple of 5 and 6 numbers are multiple of 3 15 is common therefore 9 numbers are multiple of 3 or 5, and total number of tickets is 20)
Probability that the numbers on ticket will be a multiple of 3 or 5 is 9/20.
iv) Probability that the number on ticket will be a multiple of 3 and 5.
the ticket have 1 to 20 number in which the numbers on ticket will be multiple of 3 and 5 is 15.
number on tickets which are multiple of 3 and 5 is 1 numbers therefore the probability that the number on ticket will be a multiple of 3 and 5 is = 1/20
(i.e. 4 numbers are multiple of 5 and 6 numbers are multiple of 3, 15 is common therefore 1 numbers are multiple of 3 and 5, and total number of tickets is 20)
Probability that the numbers on ticket will be a multiple of 3 and 5 is 1/20.
Probability Distribution: Binomial Distribution
In this article we learn binomial distribution, but for that we see some terms:
Probability Distribution: The way in which Probabilities are distributed over the variable corresponding to values of x is called Probability Distribution.
if x is discrete random variable with probabilities Pi then the probability distribution is called discrete distribution function. if P(Xi ) is satisfying the condition that 0 ≤ P ≤ 1for all values od x and the Σ Pi = 1 then P(X) is called probability density function.
Permutation: A group of some or all given number of things are arranged with considering their order is called permutation. and it is denoted as nPr
it is formulated as nPr =n! /(n-r)!
where ! - is call factorial ,
n! = n x (n-1)x (n-2)x .........x1
e.g. 5! = 5x4x3x2x1 =120
Combination: A group of some or all given number of things are arranged without considering their order is called permutation. and it is denoted as nCr
it is formulated as nCr = (n!)/(n-r)! x r!
note that nCn =1, nC0 =1
e.g. the number of ways in which 11 students are selected from 16. is
16C11 = (16!)/(16-11)! x 11!
= (16x15x14x13x12x11!)/(5! x 11!)
= (16x15x14x13x12) / (5x4x3x2x1)
= 4368.
Binomial Distribution:
Definition: If an experiment which has only two results they can be expressed as Success and Failure & such that p+q= 1 and this experiment is repeated n times then Probability of x success is P(X = x).
P(X = x) = nCx px qn-x
here x is number of success then it takes values from 0 to n as x= 0,1,2,........n. and it is expressed as
P(X= x) = nCx px qn-x x = 0,1,2,..........n.
= 0 otherwise
the resulting probabilities corresponding to values of x is called Binomial Distribution.
Properties Of Binomial Distribution:
i) Mean of binomial distribution is np
ii) Variance of binomial distribution id npq
iii) When we use binomial distribution: the trials are repeated n times. and n is finite. each trial has only two results. sum of Probability of Success is p and Probability of Failure is q is always 1.
Examples on Binomial Distribution.
Example 1. Define Binomial Distribution, State it Mean and Variance, If a random variable X has Binomial Distribution with parameter n = 4 and p = 2/3 = 0.6666.
find i. Mean And Variance ii. P(X=2), iii. P(X<4).
Solution: The definition of Binomial Distribution is If an experiment which has only two results they can be expressed as Success and Failure & such that p+q= 1 and this experiment is repeated n times then Probability of x success is P(X = x).
P(X = x) = nCx px qn-x
here x is number of success then it takes values from 0 to n as x= 0,1,2,........n. and it is expressed as
P(X= x) = nCx px qn-x x = 0,1,2,..........n.
= 0 otherwise
the resulting probabilities corresponding to values of x is called Binomial Distribution.
I. The Mean and Variance of Binomial Distribution
i) Mean of binomial distribution is np
ii) Variance of binomial distribution id npq
now the Mean of the Binomial Distribution is = np = 4 x 0.6666 = 2.6664
for calculating variance we first find the value of q, we know that p + q = 1
therefore q = 1 - q
q = 1 - 0.6666 = 0.3334
then the Variance of binomial distribution id npq = 4 x 0.6666 x 0.3334 = 0.8889
therefore The Mean is 2.6664 and Variance is 0.8889
II. P(X=2)
for finding this value we is binomial distribution we know P(X= x) = nCx px qn-x
therefore in that distribution putting X=2 and the given values are n = 4, p = 0.6666 and q= 0.3334
P ( X = 2 ) = 4C2 0.66662 0.33344-2
P ( X = 2 ) = {4!}/{( 2! (4 - 2)! )} x 0.4444 x 0.1111
P ( X = 2 ) = 6 x 0.04937
P ( X = 2 ) = 0.2962
III. P (X < 4)
for finding this value we is binomial distribution we know P(X= x) = nCx px qn-x
P (X < 4) = P ( X ≤ 3) = P ( X = 1 ) +P ( X = 2) +P (X = 3)
P (X < 4) = 4C1 0.66661 0.33344-1 + 4C2 0.66662 0.33344-2 + 4C3 0.66663 0.33344-3
P (X < 4) = 0.09881 + 0.2962 + 0.3950
P (X < 4) = 0.790
The Mean and Variance of Binomial Distribution are 2.6664 and 0.8889 respectively,
P ( X = 2 ) = 0.2962 and P (X < 4) = 0.790
Problems on Probability and Probability Distribution.
Example 1. if the mean and standard deviation of binomial random variable are 16 and 2 respectively. find the values of n and p.
Solution: Here given that the values of mean and standard deviation of binomial variable as
mean = 16 and S.D. = 2
we know the mean and S.D. of binomial distribution is mean = np and S.D. = √npq
therefore mean = np =16
and S.D. =√npq = 2
for finding values of n and p we firstly find variance = (S.D.)^2 = npq= (2)^2 = 4
now we take ratio of mean and variance as variance / mean = npq/np =4/16
we get q = 4/16 = 1/4 = 0.25
and we know that q = 1-p or p = 1-q = 1-0.25= 0.75
q= 0.25 and p = 0.75 putting p = 0.75 in np = 16
n x 0.75 = 16
n = 16/0.75
n = 21.3333
n = 21 because the number of observation is not expressed in fractions it is approximately 21
therefore n = 21 and p = 0.75
Example 2. P(Ā)=0.4 and P(Bc) =0.7 find P (A U B ) if i) A & B are exclusive event and ii) A & B are independent even.
Solution: here given that P(Ā)=0.4 and P(Bc) =0.7 form this we finding P(A) and P(B)
P(A) = 1- P(Ā) = 1-0.4 =0.6
and P(B) = 1- P(Bc)= 1-0.7 = 0.3
i) A & B are exclusive event then P (A ∩ B ) = 0
P (A U B ) = P(A) +P(B) = 0.6 +0.3= 0.9
and ii) A & B are independent even. then P (A ∩ B ) = P(A) X P(B)
Therefore P (A U B ) = P(A)+P(B) - P (A ∩ B )
P (A U B ) = P(A) + P(B) - {P(A) X P(B) }
= 0.6+0.3-{0.6 X 0.3)
= 0.9 - 0.18
P (A U B ) = 0.72
where P(Bc) is complement event of event B and P(Ā) is complement event of event A
if A & B are exclusive event then P(A U B) = 0.9 and A & B are independent even then
P(A U B) =0.72
Example 3. P(Ā)=0.5 and P(Bc) =0.6 find P (A U B ) if i) A & B are exclusive event and
ii) A & B are independent even.
Solution: here given that P(Ā)=0.4 and P(Bc) =0.7 form this we finding P(A) and P(B)
P(A) = 1- P(Ā) = 1-0.5 =0.5
and P(B) = 1- P(Bc)= 1-0.6 = 0.4
i) A & B are exclusive event then P (A ∩ B ) = 0
P (A U B ) = P(A) +P(B) = 0.5 + 0.4 = 0.9
and ii) A & B are independent even. then P (A ∩ B ) = P(A) X P(B)
Therefore P (A U B ) = P(A)+P(B) - P (A ∩ B )
P (A U B ) = P(A) + P(B) - {P(A) X P(B) }
= 0.5 + 0.4 -{ 0.5 X 0.4)
= 0.9 - 0.20
P (A U B ) = 0.7
where P(Bc) is complement event of event B and P(Ā) is complement event of event A
if A & B are exclusive event then P(A U B) = 0.9 and A & B are independent even then
P(A U B) =0.7
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