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Probability defines the chances of the occurrence of an event. For example chances of getting an even number when we roll a die are 50%, then its probability is 0.5. The probability of any event lies between 0 to 1, where 0 shows the absolute impossibility of an event, and 1 shows the maximum chances of an event certainly. Class 12 Maths chapter 13 notes contain the basics of probability, conditional probability, probability distribution, and all related theorems.
The main idea behind these Class 12 Maths Chapter 13 notes is to make the learning process easier for students and make a convenient revision tool whenever they need to recall important concepts and formulas. Experienced Careers360 experts prepared these notes, following the latest CBSE syllabus. Students should go through the NCERT textbook solutions first before checking these notes. For further practice, they can use NCERT exemplar solutions.
In general terms, probability is defined as a measurement of the uncertainty of events in random experiments. Mathematically, it is the ratio of the number of outcomes to the total number of possible outcomes.
If E and F are two events associated with the same sample space of a random experiment, then the conditional probability of the event E under the condition that the event F has occurred, written as P(E∣F), is given by
P(E∣F)=P(E∩F)P(F),P(F)≠0
Property 1: P(F|F) = P(S|F) = 1
Property 2: If A and B are two events in the sample space S and F is an event of S such that P(F) ≠ 0, then
P((A ∪ B)|F) = P(A|F) + P(B|F) – P((A ∩ B)|F)
Property 3: P(E′|F) = 1 − P(E|F)
Let E and F be two events associated with an experiment sample space. 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)
Independent Events
We have introduced conditional probability, where partial information that event B provides about event A. Now we found an interesting case where the occurrence of B does not provide information on A.
i.e. P(B|A)=P(B)
No w we can deduce that
P(B∣A)=P(B∩A)P(A),P(A)≠0P(B)=P(A∩B)P(A)P(A∩B)=P(A)⋅P(B)
Example:
In a dice. If M is the event in which a number appears as a multiple of 3, and N is the event in which even numbers occur. So, find whether events M and N are independent or not?
Solution:
We know that the sample space is S = {1, 2, 3, 4, 5, 6}
M={3,6}N={2,4,6}
There are even numbers in event M also.
M∩N={6}
Therefore,
P(M)=26=13P(N)=36=12P(M∩N)=13×12=16
So, P(M∩N)=P(M)⋅P(N)
From the above relation, we can say M and N are independent events.
Partition of a Sample Space
A set of events E1,E2,…,En is said to represent a partition of a sample space S if
(a) Ei∩Ej=ϕ,i≠j;i,j=1,2,3,……,n
(b) Ei∪E2∪…∪En=S, and
(c) Each Ei≠ϕ, i. e, P(Ei)>0 for all i=1,2,…,n
Theorem of Total Probability
Let {E1,E,…,En} be a partition of the sample space S . Let A be any event associated with S, then
P(A)=∑j=1nP(Ej)P(A∣Ej)
Bayes' theorem is a theorem in probability that is used to determine the probability of the event that is related to an event that has already occurred.
The formula is
P(A∣B)=P(B∣A)P(A)P(B)
Let S be a sample space associated with a random experiment. A function R: S→R is termed a random variable.
Example:
From 52 cards well-shuffled deck two cards are drawn. Find the probability for several aces.
Solution:
Let the number of aces be A.
Therefore, 0,1 or 2 will be the values of A.
When ace doesn’t occur
P(A=0)=P( non-ace )
P(A=0)=P( non − ace )×P( non − ace )=4852×4852=144169
When ace occurs once
P(A=1)=P( ace )⋅P( no )
P(A=1)=P( ace )⋅P( non − ace )+P( non − ace )⋅P( ace )=452×4852+4852×452=24169
When ace occurs twice
P(A=2)=P( ace − ace )=452×452=1169
X | 0 | 1 | 2 |
P(X) | 144/169 | 24/169 | 1/169 |
The mean of a random variable
The mean of a random variable is used to locate the middle or the average value of the random variable.
μ=∑i=1nxiPi
The variance of a random variable
It is the expectation of the squared deviation of a random variable from its sample mean.
σ2=∑i=1n(xi−μ)2P(xi)
Bernoulli Trails
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:
If there are 7 white and 9 red balls in a container, 6 balls are drawn successively. The trials of 6 balls from the container will be Bernoulli’s trials after each draw of the ball, as replaced or not replaced in the urn.
Solution:
If he trial numbers are finite then the drawing of the ball with replace, ent will be the success for drawing a white ball is ,p = 7/16. It will be the same for all six trials. So, the balls drawing with replacements are Bernoulli trials.
If the drawing of the ball is done without replacement, then the probability of success of drawing a white ball will be 7/16 for the first trial, 6/15 for the second trial, or if a red ball is drawn for 1st time will be 7/15, and so on. From the above statement, we can say that the success will not be the same for all the trials, so the trials are not Bernoulli trials.
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)=ncrprqn−r,
where q=1−p and r=0,1,2,…,n.
Example:
Ten eggs are drawn successively with replacement from a lot containing 10% defective eggs. Find the probability that there is at least one defective egg.
Solution:
Let X denote the number of defective eggs in the 10 eggs drawn. Since the drawing is done with replacement, the trials are Bernoulli trials. X has the binomial distribution with
n=10 and 10%=10100,1 egg is defective out of 10 so 110
Therefore; q=1−p=010
Now P( at least 1 defect egg )=P(X⩾1)=1−P(X=0)
=1−10C0(910)10=1−9101010
This brings us to the end of the chapter
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