Class 12 Chapter 13 (Probability) Class Notes)

Probability

Introduction:

Probability: probability measures how much chance is there for my expected
result to happen.

\[{\rm{Probability  =  }}\dfrac{{{\rm{Number \,\, of \,\, Favourable \,\, Outcomes}}}}{{{\rm{Total \,\, Number \,\, of \,\, Outcomes}}}}\]

Example: Let there be a basket with 3 balls: Red, Green, Blue. If I want to pick a Red ball,
we can calculate the probability of picking a red ball.

Total number of possibilities when we pick a ball from the basket = 3 ( The ball we pick can be Red, Green or blue)

Total number of favorable possibility i.e. possibility to get a red ball = 1

Therefore, Probability of getting a red ball, \(P(red) = \dfrac{1}{3}\)

Sample space: set of all possible outcomes of a random experiment is called Sample space. It is denoted by S.

Sample point: each element of the sample space is called the sample point.
Example: for Tossing a coin, the sample space is S = {HEAD, TAIL} and and the element in sample space namely head, or tail, each of them can be called as a sample point

Event: it is the set of favorable outcome.
Example: if you throw a coin and if you are looking for head, then head is the favorable outcome.

Mutually Exclusive events: two events are said to be mutually exclusive if
there is no common element between them.

Exhaustive events: the given events are Exhaustive if when I take the
elements in those events forms the given sample space.

 

Conditional Probability:

Given a condition and you have to find the probability

Example: when I throw a dice, what is the probability that the outcome is 3 given that the outcome is odd.

The answer is 1/3 as the odd numbers are 1, 3 and 5. So we need the probability of getting the number 3 out of the numbers 1,3,5. The chance is one out of three so The probability is 1/3.

∴ if number is odd,then the probability that it is 3 is = P(3|number is odd) = 1/3

Note:But P(3) = 1/6 If there is no condition. As the probability of getting a single number named 3, out of total 6 numbers namely 1,2,3,4,5,6 is 1/6

 

Conditional Probability Definition:

Probability of event E is called the conditional probability of F given that E has already occurred, and it's defined by P(F|E).

Formula for conditional probability
P(F|E) = P(F∩E) / P(E)

 

Conditional Probability Problems:

Example: In a survey in a class it was found that the probability of a student watching ABC videos is 0.8. and the probability that a student is both topper and also watches ABC videos is 0.792. what is the probability that a student is a topper if he watches ABC videos?

Solution:
Let E denotes the event that a person watches ABC videos 
Let F denotes the event that a person is topper .

Then, P(E) = The probability that a person watches ABC videos = 0.8
P(E ∩ F ) = probability that one is both topper and also watches ABC videos = 0.792
and P(F|E) = the probability that a person is a topper if he watches ABC videos = ?
then according to Formula for conditional probability: P(F|E) = P(F∩E)/ P(E)
= 0.792/0.8
= 0.99
∴ The probability that a person is a topper if he watches ABC videos is 0.99

  

Properties of conditional probability:

• P(S|F) = 1
• P(F|F) = 1
• P((A∪B)|F) = P(A|F) + P(B|F) – P((A∩B)|F)
• P(E'|F) = 1- P(E|F)

Example: Evaluate P(A ∪ B), if 2P(A) = P(B) = 5/13 and P(A|B) =2/5
Solution: ∵ 2P(A) = P(B) = 5/13
=> P(B) = 5/13 and P(A) = 1/2 x P(B) = 1/2  × 5/13 = 5/26

Now the formula of conditional probability is P(A|B) = P( A∩B)/ P(B)
∴ 2/ 5 = P( A∩B) /P(B)  [Since it was given in question that P(A|B)=2/5 ]

=> P(A∩B) = 2/5 X P(B)
= 2/5 × 5/13
= 2/13

So, P(A∩B)= 2/13 , but we need to find P(A ∪ B)
Now is there any formula relating P(A ∪ B) and P(A∩B) ?
Yes , the formula is P(A ∪ B)=P(A)+P(B)−P(A∩B)
Now you know values of P(A)= 5/ 26 , P(B)= 5/13 and P(A∩B)= 2/13.

So substitute these values in the formula, you will get value of P(A∪B).
That is P(A∪B)=P(A)+P(B)−P( A∩B)
=> P(A∪B)= 5/26 + 5/13 - 2/13
=> P(A∪B)= (5+10−4 )/26
=> P(A∪B)=11/26

∴ The value of P(A∪B) is 11/26

 

Example: Mother, father and son line up at random for a family picture
E : son on one end, F : father in middle , what is the probability that the son is on one end given that the father is in middle?

Solution:

let Mother be denoted with M, Father be denoted with F and Son be denoted with S. If the sample space for all the possible ways of arranging them is denoted by S.
Then the possible elements in S are, S = {MFS, MSF, FMS, FSM, SMF, SFM}
Now E denotes son on one end as given in question so A={MFS, FMS, SMF, SFM}
and F denoted Father in middle as given in question , so B={MFS, SFM}
Since the common elements between A and B are MFS and SFM ∴A∩B ={MFS, SFM}
Now, P(E∩F) = probability of getting two elements of A∩B I.e; MFS and SFM out of the total 6 elements MFS, MSF, FMS, FSM, SMF, SFM
= 2/6 = 1/3
and P(F)= probability of getting two elements of B i.e; MFS and SFM out of the total 6 elements MFS, MSF, FMS, FSM, SMF, SFM.
= 2/6 = 1/3
∴ P(E|F) = P(E∩F) /P(F)
= (1/3)  / (1/3)
= 1
∴The probability that the son is on one end given that the father is in middle is 1.

Example: If A and B are events such that P(A|B) = P(B|A), then
(A)A ⊂ B but A ≠ B

(B)A = B

(C)A ∩ B = Φ

(D)P(A) = P(B)

Solution:
Given in question that P(A|B) = P(B|A)
But we know P(A|B) = P( A∩B) /P(B)
and P(B|A) = P( A∩B)/ P(A)
since P(A|B) = P(B|A)
=> P( A∩B)/ P(B) = P( A∩B)/ P(A)
=> P(A) = P(B)
Thus the correct option is (D).

 

Multiplication Theorem on Probability:

Let E and F be two events associated with a sample space of an experiment.
Then 

P(E ∩ F) = P(E) P(F|E), P(E) ≠ 0
= P(F) P(E|F), P(F) ≠ 0

If E, F and G are three events associated with a sample space, then

P(E∩F∩G) = P(E) P(F|E) P(G|E∩F)

Example: In a survey in a class, the probability for a person to watch ABC videos is 0.8 and the probability for a person to be a topper, if given that he watches ABC videos is 0.99. find the probability for a person to be both topper and watches ABC videos.

Solution:
let Event E denotes the event that a person watches ABC videos
let Event F denotes the event that a person is topper
then P(E) = the probability that a person watches ABC videos =0.8
and P(F|E)= the probability that a person is a topper if he watches ABC videos=0.99
then P (E ∩ F)= the probability that a person is both topper and also watches ABC videos
then according to Multiplication Theorem on probability
P (E ∩ F) = P (E) P (F | E)
= 0.8 x 0.99
= 0.792
∴ The probability that a person is both topper and also watches ABC videos = 0.792

 

Independent Events:
Let E and F be two events associated with a sample space S. If the probability of occurrence of one of them is not affected by the occurrence of the other, then we say that the two events are independent. Thus, two events E and F will be independent, if

(a) P(F | E) = P(F), provided P (E) ≠ 0
(b) P(E | F) = P(E), provided P (F) ≠ 0

Using the multiplication theorem on probability, we have

(c) P(E ∩ F) = P(E) P(F)

Three events A, B and C are said to be mutually independent if all the
following conditions hold:

P(A ∩ B) = P(A) P(B)
P(A ∩ C) = P(A) P(C)
P(B ∩ C) = P(B) P(C)
and P(A ∩ B ∩ C) = P(A) P(B) P(C)

Example: If you throw two coins, then probability of getting head or tail in second coin is independent of probability of getting head or tail in first coin.

Example: Let A and B be independent events with P(A)=0.3 and P(B) = 0.4.
Find(i) P(A ∩ B) (ii) P(A ∪ B) (iii) P(A|B) (iv) P(B|A)
Solution:
(i) we have to find the value of P(A ∩ B) and they gave P(A) =0.3 and P(B) = 0.4.
given A and B are independent events.
We know that if A and B are independent events then P(A ∩ B) = P(A) P(B)
=>P(A ∩ B) = 0.3×0.4 = 0.12

(ii)we have to find the value of P(A ∪ B)
the formula for P(A ∪ B) is P (A ∪B)=P(A)+P(B)−P( A∩B)
=> P(A ∪ B)=0.3+0.4-0.12 = 0.58

(iii) P(A|B)=?
the formula for P(A|B) is P(A|B)= P( A∩B)/ P(B)
=> P(A|B)= 0.12/ 0.4 = 0.3

(iv) P(B|A)=? 
the formula for P(B|A) is P(B|A)= P( A∩B)/ P(A)
=> P(B|A)= 0.12/0.3 = 0.4

Example:If A and B are two events such that P(A)=1/4 ,P(B)=1/2 , ,P(A∩B)=1/8
find P(not A and not B).
Solution:
we know that P(not A and not B)=P(A'∩B')
But we also know that A'∩B'=(A∪B)' => P(A'∩B')=P((A∪B))'
∴P(not A and not B)=P(A'∩B')
=P((A∪B))'
=1-P(A∪B) ( ∵ P(A)' = 1-P(A) )
=1- (P(A)+P(B)−P( A∩B)) (∵ P(A∪B)=P(A)+P(B)−P( A∩B) )
=1- ( 1/4 + 1/2 - 1/8 )
=1- 5/8
=3/8

 

Bayes' Theorem:

If E1, E2,..., En are mutually exclusive and exhaustive events associated
with a sample space, and A is any event of non zero probability, then

 

Example: In a survey in a classroom it was found that chance of a person becoming a topper is 10% . if a person is a topper, there is a 99% chance that he watches ABC videos. One who is not a topper there is 1% chance of watching ABC videos. Then, what is the Probability for one to be a Topper , given if he watches ABC videos?

Solution:
Let “A person to be a Topper” is Event A and “To watch ABC videos” is Event B

Given P(A) = 0.1 = probability for a person to be a Topper.

So P(A')= 1-P(A)=1 - 0.1= 0.9 = probability for a person to be not a Topper

Given P(B|A) = 0.99 =  probability that a person watches ABC videos, if he is a topper, 

so P(B'|A) = 1 - P(B|A) = 1 - 0.99 = 0.01 = probability that a person not watches ABC videos, if he is a topper

Given P(B|A')=0.01=probability that a person watches ABC videos , if he is not a topper 
so P(B'|A')=1-0.01=0.99=probability that a person not watches ABC videos , if he is not a topper

Therefore according to Bayes' Theorem,Probability for one to be a Topper , if he watches ABC videos is

= (0.1x0.99)/ {( (0.1x0.99)+(0.01x0.9))}
= 0.9909

∴ if one watches ABC videos , for him there is 99.09% chance to become topper.

Example: A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.

Solution:
Let E1= the event of selecting first bag.
E2=the event of selecting second bag.
A = The event of getting red ball.
Since there is equal chance of selecting first bag or selecting second bag,
P(E1)=P(E2)= 1/2

now P(A|E1)=P(Drawing a red ball from first bag )= 4/8
and P(A|E2)=P(Drawing a red ball from second bag )= 2/8 =1/4

probability that ball is drawn from the first bag given that the ball drawn is red is P(E1|A)

 

Example:Two groups are competing for the position on the Board of directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group.

Solution:
Let E1= event that the first group wins the competition
E2 = event that the second group wins the competition
A = event of introducing a new product.
Then,
P(E1) = Probability that the first group wins the competition = 0.6
P(E2) = Probability that the second group wins the competition = 0.4
P(A|E1)=Probability of introducing new product if the first group wins =0.7
P(A|E2)=Probability of introducing new product if the second group wins =0.3
The probability that the new product is introduced by the second group is given by

 

 

Random Variables and its probability distributions:

A random variable is a real valued function whose domain is the sample space of a random experiment.
The probability distribution of a random variable X is the system of numbers

where P_i > 0, i = 1 to n and P₁ + P₂ + P₃ + …..... + P_n =1

Example: Suppose that a coin is tossed twice so that the sample space is S = {HH,HT,TH,TT}. Let X represent the number of heads that can come up. So with each sample point we can associate a number for X so for HH the value of X is 2 as in HH there are 2 heads, for HT the value of X is 1 as there is only one head in HT. similarly for TH, X = 1 and for TT value of X = 0 as there are no heads in TT.

So in the sample space S = {HH,HT,TH,TT} assuming that the coin is fair,
P(HH)= ¼ , P(HT)= ¼ , P(TH)= ¼ , P(TT)= ¼
therefore P(X=0) = P(TT) = ¼
P(X=1) = P(HT ∪ TH) = P(HT) + P(TH) = ¼ + ¼ = ½
P(X=2) = P(HH) = ¼
Thus the table formed is shown below:

Example: A coin is biased so that the head is 3 times as likely to occur as tail. If the coin is tossed twice, find the probability distribution of number of tails.

Solution:
Let the probability of getting a tail in the biased coin be x.
∴ P (T) = x
⇒ P (H) = 3x
For a biased coin, P (T) + P (H) = 1
⇒ x + 3x = 1
⇒ 4x = 1
⇒ x = \(\frac{1}{4}\)
∴ P(T) =  \(\frac{1}{4}\)   and   P(H) =  \(\frac{3}{4}\)

When the coin is tossed twice, the sample space is {HH, TT, HT, TH}.
Let X be the random variable representing the number of tails.
∴ P (X = 0) = P (no tail) = P (H) × P (H)= \(\frac{3}{4}\)  × \(\frac{3}{4}\)
=  \(\frac{9}{16}\)

P (X = 1) = P (one tail) = P (HT) + P (TH) = \(\frac{3}{4} \times \frac{1}{4} + \frac{1}{4} \times \frac{3}{4}\)
= \(\frac{3}{{16}} + \frac{3}{{16}}\)
= \(\frac{3}{8}\)

P (X = 2) = P (two tails) = P (TT)= \(\frac{1}{4} \times \frac{1}{4} = \frac{1}{{16}}\)

Therefore, the required probability distribution is as follows.

 

 

Mean and Variance of a Random variable:

Let X be a random variable assuming values x₁, x₂,...., x_n with probabilities p₁, p₂, ..., p_n, respectively such that p_i ≥ 0,  

Mean of X, denoted by μ [or expected value of X denoted by E (X)] is defined as:

and variance denoted by σ² , is defined as

Standard deviation of the random variable X is defined as

Let us take the above table

Now mean is

 
= X₁P₁ + X₂P₂ + X₃P₃
= (0X¼ ) + (1 X ½)+ (2X ¼)
= 0 + ½ + ½
=1

Now Variance = σ_x² = E(X²) − (E(X))²

so to find Variance let us find E(X²) and (E(X))²

= 0 + ½ + 1
= 3/2
= 1.5

and (E(X))² = 1²=1

∴ Var(X)= E(X²) − (E(X))² = 1.5 − 1 = 0.5

now standard deviation of X = √(variance (X ))
= √0.5
≈ 0.707

Example: In a meeting, 70% of the members favour and 30% oppose a certain proposal. A member is selected at random and we take X = 0 if he opposed,
and X = 1 if he is in favour. Find E(X) and Var(X).
Solution:
It is given that P(X = 0) = 30% = 30/100 = 0.3
P(X=1)=70%= 70/100 = 0.7

Therefore, the probability distribution is as follows.

x	0	1
P(x)	0.3	0.7

Then, Mean= E(X) =

 
= 0×0.3+1×0.7
=0.7

= 02×0.3+12×0.7 =0.7

Now,
∴ Var(X)= E(X²)-(E(X))²
= 0.7−(0.7)²
= 0.7 − 0.49
= 0.21

Example:The mean of the numbers obtained on throwing a dice having written 1 on three faces, 2 on two faces and 5 on one face is (A) 1 (B) 2 (C) 5 (D) 8/3

Solution:
Let X be the random variable representing a number on the dice.

The total number of observations is six.
∴P(X=1)= 3/6 = 1/2

P(X=2)= 2/6 = 1/3

P(X=5)= 1/6

∴ The probability distribution is as follows.

X	1	2	5
P(X)	1/2	1/3	1/6

Now, Mean=E(X)

 
=1×1/2 +2×1/3 +5×1/6
= 1/2 +2/3 +5/6
= 12/6
= 2

∴ The correct option is B.

 

Bernoulli Trials and Binomial Distribution:

Bernoulli Trials:
Trials of a random experiment are called Bernoulli trials, if they satisfy the following
conditions:
(i) There should be a finite number of trials
(ii) The trials should be independent
(iii) Each trial has exactly two outcomes: success or failure
(iv) The probability of success(or failure) remains the same in each trial.

Example: A tossed coin shows a ‘head’ or ‘tail’. If the occurrence of head is considered as success, then the occurrence of tail is a failure .so, if we consider tossing a coin two times as Bernoulli trials, because 

(i)There are finite number of trials that is 2
(ii)tossing a coin second time is completely independent of the result of first time tossing
(iii)each trial has only two outcomes namely either head or tail and
(iv)The probability of getting head or tail remains the same in each trial.

 

 

Binomial Distribution:

A random variable X taking values 0, 1, 2, ..., n is said to have a binomial distribution
with parameters n and p, if its probability distribution is given by 

P (X = r) = ⁿc_r p^r q^n–r,
where q = 1 – p and r = 0, 1, 2, ..., n.

Example: If a coin is tossed 10 times, find the probability of exactly 5 heads.

Solution: let X denote the number of heads.
So X has the binomial distribution with n = 10 an p = ½
∴ P(X = x) = ⁿC_x q^n-x p^x , x = 1,2,3....,n
∴ P(X = x) = ¹⁰c_x(1/2)^10-x(1/2)^x = ¹⁰c_x(1/2)¹⁰.
so P(X = 6) = ¹⁰c₅(1/2)¹⁰= (10 !)/ (5 !×5!) ×( 1/2¹⁰) = 63/ 256

Example:A bag consists of 10 balls each marked with one of the digits 0 to 9. If four balls are drawn successively with replacement from the bag, what is the probability that none is marked with the digit 0? 

Solution:
Let X denote the number of balls marked with the digit 0 among the 4 balls drawn.
Since the balls are drawn with replacement, the trials are Bernoulli trials.
X has a binomial distribution with n = 4 and P= 1/10

∴ Q = 1 − P =1 − 1/10 = 9/10

∴ P(X = x) = ⁿC_x q^n-x p^x , x = 1,2,3....,n
= ⁴C_x ( 9/10 )^n−x( 1/10 )^x

P (none marked with 0) = P (X = 0)
= ⁴C₀ ( 9/10 )⁴ ( 1/10 )⁰
= 1×( 9/10)⁴ 

= ( 9/10 )⁴

 

Example: Find the probability of getting 5 exactly twice in 7 throws of a dice.

Solution:
The repeated tossing of a dice are Bernoulli trials. Let X represent the number of times of getting 5 in 7 throws of the dice.
Probability of getting 5 in a single throw of the dice is P=16
∴ Q = 1−P = 1 − 1/6 = 5/6

Clearly, X has the probability distribution with n = 7 and P=16
∴P(X = x) = ⁿC_x q^n-x p^x
= ⁷C_x (5/6)^7−x (1/6)^x

P (getting 5 exactly twice) = P(X = 2)
=⁷C₂ (5/6)⁵ (1/6)²
= 21×(5/6)⁵× 1/36
= 7 /12 ×(5/6)⁵

 

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