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Probability Miscellaneous Exercise NCERT Solutions brings together the important concepts covered throughout Chapter 13 — Probability. Unlike the individual exercises, this miscellaneous exercise contains a variety of questions that require us to identify the right concept and apply it carefully.
The questions cover conditional probability, independent events, the multiplication rule, the Theorem of Total Probability and Bayes’ Theorem. Some questions also involve repeated trials and different possible cases, making this exercise a useful revision of the concepts studied in the chapter.
\displaystyle P(A\mid B)=\frac{P(A\cap B)}{P(B)},\qquad P(B)\neq0
The Theorem of Total Probability helps us find the probability of an event by considering all the possible cases that can lead to it, while Bayes’ Theorem helps us work backwards to find the probability of a particular case when the event is already known to have occurred.
\displaystyle P(A)=\sum_{j=1}^{n}P(E_j)\,P(A\mid E_j)
\displaystyle P(E_i\mid A)=\frac{P(E_i)\,P(A\mid E_i)}{\displaystyle\sum_{j=1}^{n}P(E_j)\,P(A\mid E_j)}
One interesting question in this exercise involves A and B throwing a die alternately until one of them gets a 6. The probability of A winning can be expressed as an infinite G.P., giving us a useful connection between Probability and the concept of infinite geometric progressions studied in Class XI.
\displaystyle a+ar+ar^2+\cdots=\frac{a}{1-r},\quad -1<r<1
So, this miscellaneous exercise is not just about applying formulas. It is about recognising the situation, choosing the right concept and connecting different ideas from the chapter. By solving these questions step by step, you will revise the complete chapter and build confidence for the CBSE Class 12 Board Exams, CUET, and other competitive examinations.
Key Concepts
Before solving the Miscellaneous Exercise, quickly revise the important concepts and formulas from the complete chapter of Probability. This exercise brings together different types of questions, so the main challenge is to identify the right concept before applying the formula.
1. Conditional Probability
The conditional probability of an event E, given that another event F has already occurred, is given by
\displaystyle P(E\mid F)=\frac{P(E\cap F)}{P(F)},\qquad P(F)\neq0
Similarly,
\displaystyle P(F\mid E)=\frac{P(E\cap F)}{P(E)},\qquad P(E)\neq0
Remember that P(E|F) and P(F|E) generally have different meanings and are not equal.
Some useful results are
\displaystyle P(E'\mid F)=1-P(E\mid F)
\displaystyle P(E\cup F\mid G)=P(E\mid G)+P(F\mid G)-P(E\cap F\mid G)
2. Multiplication Theorem
From the definition of conditional probability, we get the multiplication theorem.
\displaystyle P(E\cap F)=P(E)\,P(F\mid E)
It can also be written as
\displaystyle P(E\cap F)=P(F)\,P(E\mid F)
For three events,
\displaystyle P(E\cap F\cap G)=P(E)\,P(F\mid E)\,P(G\mid E\cap F)
This theorem is particularly useful in questions involving successive selections, repeated trials and dependent events.
3. Independent Events
Two events E and F are independent if the occurrence of one does not affect the probability of the other.
For independent events,
\displaystyle P(E\cap F)=P(E)\,P(F)
Equivalently, when the relevant probabilities are defined,
\displaystyle P(E\mid F)=P(E),\qquad P(F\mid E)=P(F)
For three events E, F and G to be mutually independent, we need
\displaystyle P(E\cap F)=P(E)P(F),\quad P(F\cap G)=P(F)P(G),\quad P(G\cap E)=P(G)P(E)
and also
\displaystyle P(E\cap F\cap G)=P(E)P(F)P(G)
Important: Mutually exclusive events with non-zero probabilities cannot be independent.
4. Complementary Events
The complement of an event E is denoted by E’. Since exactly one of E and E’ occurs,
\displaystyle P(E')=1-P(E)
Similarly, for conditional probability,
\displaystyle P(E'\mid F)=1-P(E\mid F)
This is especially useful when a question asks for at least one, none, not more than or similar complementary cases.
5. Theorem of Total Probability
If E1, E2, …, En form a partition of the sample space S, then for any event A,
\displaystyle P(A)=\sum_{j=1}^{n}P(E_j)\,P(A\mid E_j)
The idea is simple: find the probability of A through each possible case and add them.
Whenever a question involves different possible sources, groups, bags, boxes or machines, first identify the different cases and then consider the probability of the required event through each case.
6. Bayes’ Theorem
Bayes’ Theorem helps us find the reverse probability. We know that an event A has occurred and want to find the probability that it occurred through a particular case Ei.
If E1, E2, …, En form a partition of the sample space and A is an event with non-zero probability, then
\displaystyle P(E_i\mid A)=\frac{P(E_i)\,P(A\mid E_i)}{\displaystyle\sum_{j=1}^{n}P(E_j)\,P(A\mid E_j)}
A useful way to recognise a Bayes’ theorem question is when the result is given and we are asked to find the case or cause behind it.
Remember the direction:
\displaystyle \text{Case}\longrightarrow\text{Event}\qquad\text{Total Probability}
\displaystyle \text{Event}\longrightarrow\text{Case}\qquad\text{Bayes' Theorem}
7. Partition of a Sample Space
Events E1, E2, …, En form a partition of the sample space S if they are mutually exclusive and together cover the entire sample space.
Therefore,
\displaystyle E_i\cap E_j=\varnothing,\qquad i\neq j
and
\displaystyle E_1\cup E_2\cup\cdots\cup E_n=S
In simple words, the partition divides the sample space into mutually exclusive cases that together include every possibility.
8. Infinite Geometric Progression in Probability
In some repeated-trial questions, the same pattern can continue indefinitely. The probabilities of the required outcomes may then form an infinite G.P.
If \displaystyle a+ar+ar^2+ar^3+\cdots
is an infinite G.P. with -1 < r < 1, then its sum is
\displaystyle S_{\infty}=\frac{a}{1-r}
This idea is useful in repeated-trial questions where an event can occur on the first trial, third trial, fifth trial, and so on.
9. How to Approach the Miscellaneous Exercise
Since this exercise contains mixed questions, first focus on understanding the wording rather than immediately searching for a formula.
- Identify the events: Clearly define what each event represents.
- Look for the given information: Decide whether the question gives an ordinary probability, a conditional probability or a known outcome.
- Choose the concept: Decide whether to use conditional probability, multiplication theorem, independence, binomial probability, total probability or Bayes’ theorem.
- Use the complement when helpful: Expressions such as “at least one” or “at most” can sometimes be handled more easily using the complementary event.
- Check the final answer: A probability must always lie between 0 and 1.
10. Important Points to Remember
- Always distinguish between independent and mutually exclusive events.
- Do not confuse \displaystyle P(A\mid B) with \displaystyle P(B\mid A).
- Use the multiplication theorem when finding the probability of events occurring together.
- Use the Theorem of Total Probability when an event can occur through different possible cases.
- Use Bayes’ Theorem when the event is known to have occurred and the probability of its particular cause or case is required.
- For questions involving repeated attempts until a particular event occurs, check whether the resulting probabilities form an infinite G.P.
- Always define the events clearly before substituting values into a formula.
Quick Reference Table
| Concept | Formula / Result |
|---|---|
| Complementary Event | \displaystyle P(E')=1-P(E) \displaystyle P(E')+P(E)=1 |
| Conditional Probability | \displaystyle P(E\mid F)=\frac{P(E\cap F)}{P(F)},\qquad P(F)\neq0 |
| Multiplication Rule | \displaystyle P(E\cap F)=P(E)P(F\mid E) \displaystyle =P(F)P(E\mid F) |
| Independent Events | \displaystyle P(E\cap F)=P(E)P(F) \displaystyle P(E\mid F)=P(E) |
| Partition of Sample Space | \displaystyle E_i\cap E_j=\varnothing,\;i\neq j \displaystyle E_1\cup E_2\cup\cdots\cup E_n=S |
| Total Probability | \displaystyle P(A)=\sum_{j=1}^{n}P(E_j)P(A\mid E_j) |
| Total Probability — Two Cases | \displaystyle P(A)=P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2) |
| Bayes’ Theorem | P(E_i\mid A)=\frac{P(E_i)P(A\mid E_i)}{\displaystyle\sum_{j=1}^{n}P(E_j)P(A\mid E_j)} |
| Bayes’ Theorem — Two Cases | P(E_1\mid A)=\frac{P(E_1)P(A\mid E_1)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)} |
| Prior Probability | \displaystyle P(E_i) |
| Posterior Probability | \displaystyle P(E_i\mid A) |
| Reverse Probability | Given that A has occurred, find the probability of the case Ei. \displaystyle P(E_i\mid A) |
| Total Probability vs Bayes’ Theorem | \displaystyle \text{Case}\longrightarrow\text{Event} → Total Probability \displaystyle \text{Event}\longrightarrow\text{Case} → Bayes’ Theorem |
| Infinite G.P. | \displaystyle a+ar+ar^2+\cdots=\frac{a}{1-r},\qquad -1<r<1 |
Before You Begin… Before solving a question, first identify the events and the information given. Then decide which concept applies — conditional probability, multiplication rule, independence of events, the Theorem of Total Probability, Bayes’ Theorem or another result from the chapter. Reading the question carefully and identifying the right approach is the key to solving a miscellaneous exercise.
Before we begin the probability miscellaneous exercise, let us first solve an important example from the chapter that shows how repeated trials can lead to an infinite G.P.
NCERT Example: Infinite G.P. Question
NCERT Example. A and B throw a die alternately until one of them gets a 6 and wins the game. Find their respective probabilities of winning, if A starts first.
Solution:
Think First… A can win on the 1st, 3rd, 5th, … throws, while B can win on the 2nd, 4th, 6th, … throws. For A to win, all the previous throws must result in failures. This creates a repeated pattern, and the probabilities of A winning form an infinite G.P.
Let S denote the event of getting a 6 and F denote the event of not getting a 6.
For a throw of a fair die,
\displaystyle P(S)=\frac{1}{6} and \displaystyle P(F)=\frac{5}{6}
A wins on the 1st throw
A wins immediately if A gets a 6 on the first throw.
Therefore,
\displaystyle P(\text{A wins on the 1st throw})=P(S)=\frac{1}{6}
A wins on the 3rd throw
For A to win on the 3rd throw, A must fail on the 1st throw, B must fail on the 2nd throw and A must get a 6 on the 3rd throw.
Thus, the sequence is FFS.
Therefore,
\displaystyle P(\text{A wins on the 3rd throw})=P(F)P(F)P(S)
\displaystyle =\left(\frac{5}{6}\right)^2\left(\frac{1}{6}\right)
A wins on the 5th throw
For A to win on the 5th throw, the first four throws must all be failures and the 5th throw must be a success.
Thus, the sequence is FFFFS.
Therefore,
\displaystyle P(\text{A wins on the 5th throw})=P(F)^4P(S)
\displaystyle =\left(\frac{5}{6}\right)^4\left(\frac{1}{6}\right)
Similarly, A can win on the 7th, 9th, 11th, … throws.
Hence, by adding all the cases, we get
\displaystyle P(\text{A wins})=\frac{1}{6}+\left(\frac{5}{6}\right)^2\left(\frac{1}{6}\right)+\left(\frac{5}{6}\right)^4\left(\frac{1}{6}\right)+\cdots
Taking \displaystyle\frac{1}{6} common,
\displaystyle P(\text{A wins})=\frac{1}{6}\left[1+\left(\frac{25}{36}\right)+\left(\frac{25}{36}\right)^2+\cdots\right]
This is an infinite G.P. with
\displaystyle a=1,\quad r=\frac{25}{36}
Since
\displaystyle -1<r<1
Therefore, using the sum of the infinite G.P. as \frac{a}{1-r}, we get
\displaystyle 1+\frac{25}{36}+\left(\frac{25}{36}\right)^2+\cdots=\frac{1}{1-\frac{25}{36}}
Therefore,
\displaystyle P(\text{A wins})=\frac{1}{6}\times\frac{1}{1-\frac{25}{36}}
\displaystyle =\frac{1}{6}\times\frac{36}{11}=\frac{6}{11}
Hence,
\displaystyle \boxed{P(\text{A wins})=\frac{6}{11}}
Since either A or B must eventually win the game,
\displaystyle P(\text{B wins})=1-P(\text{A wins})
\displaystyle =1-\frac{6}{11}=\frac{5}{11}
Therefore,
\displaystyle \boxed{P(\text{A wins})=\frac{6}{11},\quad P(\text{B wins})=\frac{5}{11}}
Maths Better Tip… In repeated-trial questions, first identify when the required event can occur. If the probabilities follow a repeated pattern, check whether they form a G.P. before adding the terms one by one.
With this example in mind, let us now solve all the NCERT questions of the Probability Miscellaneous Exercise step by step.
Question 1: Probability Miscellaneous
1. A and B are two events such that \displaystyle P(A)\neq0 Find \displaystyle P(B\mid A) , if (i) A is a subset of B, (ii) \displaystyle A\cap B=\varnothing
Solution:
We use the formula for conditional probability:
\displaystyle P(B\mid A)=\frac{P(A\cap B)}{P(A)}
(i) A is a subset of B
Since every outcome in A is also contained in B, we have
\displaystyle A\subseteq B\quad\Rightarrow\quad A\cap B=A
Therefore,
\displaystyle P(B\mid A)=\frac{P(A\cap B)}{P(A)}=\frac{P(A)}{P(A)}=1
Hence,
\displaystyle \boxed{P(B\mid A)=1}
(ii) A ∩ B = ∅
Since A and B are mutually exclusive, they cannot occur together. Therefore,
\displaystyle P(A\cap B)=0
Hence,
\displaystyle P(B\mid A)=\frac{P(A\cap B)}{P(A)}=\frac{0}{P(A)}=0
Hence,
\displaystyle \boxed{P(B\mid A)=0}
Question 2: Conditional Probability Miscellaneous
2. A couple has two children.
(i) Find the probability that both children are males, if it is known that at least one of the children is male.
(ii) Find the probability that both children are females, if it is known that the elder child is a female.
Solution:
Let M denote a male child and F denote a female child.
For two children, the possible outcomes are
\displaystyle MM,\;MF,\;FM,\;FF
Assuming that a male and a female child are equally likely, these four outcomes are equally likely.
(i) Both children are males, given that at least one child is male
Think First… Since we already know that at least one child is male, the outcome FF is excluded. So, the possible outcomes are MM, MF and FM.
Let A be the event that both children are males and B be the event that at least one child is male.
Thus, \displaystyle A=\{MM\}
and \displaystyle B=\{MM,MF,FM\}
Therefore, \displaystyle P(A)=\frac{1}{4} and \displaystyle P(B)=\frac{3}{4}
Also, \displaystyle A\cap B=A
Hence, using the formula for conditional probability,
\displaystyle P(A\mid B)=\frac{P(A\cap B)}{P(B)}
\displaystyle =\frac{P(A)}{P(B)}=\frac{\frac14}{\frac34}=\frac13
Hence,
\displaystyle \boxed{P(\text{Reqd. Probability})=\frac13}
(ii) Both children are females, given that the elder child is a female
Think First… Here the elder child is already known to be female. Therefore, only two possibilities remain for the younger child: the younger child can be male or female.
Let C be the event that both children are females and D be the event that the elder child is female.
We need to find the probability that both children are females, given that the elder child is female, i.e.,
\displaystyle P(C\mid D)
The possible outcomes for the two children are
\displaystyle FF,\;FM,\;MF,\;MM
Since the elder child is known to be female, the outcomes MF and MM are excluded. Therefore, the possible outcomes are
\displaystyle FF,\;FM
Thus,
\displaystyle P(D)=\frac{2}{4}=\frac{1}{2}
Also, both children are females only in the outcome FF. Therefore,
\displaystyle C\cap D=\{FF\}
Hence,
\displaystyle P(C\cap D)=\frac{1}{4}
Using the formula for conditional probability,
\displaystyle P(C\mid D)=\frac{P(C\cap D)}{P(D)}
\displaystyle =\frac{\frac14}{\frac12}=\frac12
Hence,
\displaystyle \boxed{P(\text{Reqd. Probability})=\frac12}
You may already be following Maths Better for the NCERT Solutions of the previous Class 12 Maths chapters, namely:
- Matrices
- Determinants
- Relations and Functions
- Inverse Trigonometric Functions
- Continuity and Differentiability
- Application of Derivatives
- Integrals
- Application of Integrals
- Differential Equations
- Vector Algebra
- Three Dimensional Geometry
- Linear Programming
Likewise, the Probability Miscellaneous Exercise brings together the important concepts covered throughout the chapter, including Conditional Probability, the Multiplication Theorem, independent events, the Theorem of Total Probability and Bayes’ Theorem. The questions also include a variety of applications involving children, bags, machines, cards, coins, repeated trials and other real-life situations. By solving these questions step by step, you will strengthen your ability to identify the right concept, set up the problem correctly and apply the appropriate probability rule. These questions provide valuable practice for the CBSE Board Examination and CUET, while helping you revise the complete Probability chapter. Now, let’s continue with the next question of the Probability Miscellaneous Exercise.
Question 3: Probability Miscellaneous
3. Suppose that 5% of men and 0.25% of women have grey hair. A grey haired person is selected at random. What is the probability of this person being male? Assume that there are equal numbers of males and females.
Solution:
Think First… We know that the selected person has grey hair, but we need to find the probability that the person is male. So, this is a Bayes’ Theorem question.
Let M be the event that the selected person is male and W be the event that the selected person is female.
Let G be the event that the selected person has grey hair.
Since there are equal numbers of males and females,
\displaystyle P(M)=P(W)=\frac{1}{2}
We are given that 5% of men have grey hair. Therefore,
\displaystyle P(G\mid M)=\frac{5}{100}=\frac{1}{20}
Also, 0.25% of women have grey hair. Hence,
\displaystyle P(G\mid W)=\frac{0.25}{100}=\frac{1}{400}
We need to find the probability that the selected person is male, given that the person has grey hair, i.e.,
\displaystyle P(M\mid G)
By Bayes’ Theorem,
\displaystyle P(M\mid G)=\frac{P(M)P(G\mid M)}{P(M)P(G\mid M)+P(W)P(G\mid W)}
Substituting the values,
We get, \displaystyle P(M\mid G)=\frac{\frac12\times\frac1{20}}{\frac12\times\frac1{20}+\frac12\times\frac1{400}}
\displaystyle =\frac{\frac1{20}}{\frac1{20}+\frac1{400}}
\displaystyle =\frac{\frac{20}{400}}{\frac{20}{400}+\frac{1}{400}}=\frac{20}{21}
Hence,
\displaystyle \boxed{P(M\mid G)=\frac{20}{21}}
Maths Better Tip! Q4 is a little different from the other questions in this exercise. It involves repeated trials where each person can be either right-handed or not right-handed. This is a standard binomial probability situation.
Let X denote the number of successes in n independent trials. If the probability of a success is \displaystyle p, then the probability of a failure is \displaystyle q=1-p.
The probability of getting exactly r successes in n trials is obtained from the general term of the binomial expansion:
\displaystyle P(X=r)={}^{n}C_r\,p^r q^{\,n-r}
This comes from the general term of the binomial expansion
\displaystyle (a+b)^n=\sum_{r=0}^{n}{}^{n}C_r\,a^{\,n-r}b^r
Here, the two terms correspond to failure and success, so we take \displaystyle a=q and \displaystyle b=p. Thus, the term containing \displaystyle p^r q^{\,n-r} represents exactly r successes and n-r failures.
For Q4, being right-handed is taken as a success. Therefore, \displaystyle p=0.9 and \displaystyle q=0.1. Once this setup is clear, the required probability can be found using the binomial term and, where convenient, the complement of the required event.
Question 4: Probability Miscellaneous
4. Suppose that 90% of people are right-handed. What is the probability that at most 6 of a random sample of 10 people are right-handed?
Solution:
Think First… “At most 6” means 6 or fewer. It is easier to find the complementary probability that more than 6 people are right-handed and subtract it from 1.
Let X denote the number of right-handed people in the sample.
The probability that a person is right-handed is
\displaystyle p=0.9
Therefore, the probability that a person is not right-handed is
\displaystyle q=1-p=1-0.9=0.1
We need to find the probability that at most 6 people are right-handed, i.e.,
\displaystyle P(X\leq6)=P(X=0)+P(X=1)+\cdots+P(X=6)
Calculating all these terms directly would be lengthy. So, we use the complementary event:
\displaystyle P(X\leq6)=1-\left[P(X=7)+P(X=8)+P(X=9)+P(X=10)\right]
i.e. \displaystyle P(X\leq6)=1-P(X\geq7)
Now, for exactly r right-handed people out of 10 people, using the binomial distribution as explained above, the probability is
\displaystyle P(X=r)={}^{10}C_r(0.9)^r(0.1)^{10-r}
Therefore,
\displaystyle P(X\geq7)=\sum_{r=7}^{10}{}^{10}C_r(0.9)^r(0.1)^{10-r}
Hence, the reqd. probability is
\displaystyle \boxed{P(X\leq6)=1-\sum_{r=7}^{10}{}^{10}C_r(0.9)^r(0.1)^{10-r}}
Question 5: Probability Miscellaneous
5. If a leap year is selected at random, what is the chance that it will contain 53 Tuesdays?
Solution:
Think First… A leap year has 366 days, which means 52 complete weeks and 2 extra days. Therefore, one or two weekdays will occur 53 times, depending on the two extra days.
A leap year has
\displaystyle 366=52\times7+2
Thus, a leap year has 52 complete weeks and 2 extra days.
The 2 extra days are always consecutive days. There are 7 possible pairs of consecutive days:
\displaystyle (\text{Monday, Tuesday}),\;(\text{Tuesday, Wednesday})\\ (\text{Wednesday, Thursday}),\;(\text{Thursday, Friday})\\ (\text{Friday, Saturday}),\;(\text{Saturday, Sunday})\\ (\text{Sunday, Monday})
All 7 pairs are equally likely.
For the leap year to contain 53 Tuesdays, Tuesday must be one of the 2 extra days.
From the 7 possible pairs above, Tuesday occurs in the following 2 pairs:
\displaystyle (\text{Monday, Tuesday}),\;(\text{Tuesday, Wednesday})
Therefore,
\displaystyle \text{Number of favourable cases}=2
and
\displaystyle \text{Total number of cases}=7
Hence, the required probability is
\displaystyle \boxed{P(\text{53 Tuesdays})=\frac{\text{Favourable cases}}{\text{Total cases}}=\frac{2}{7}}
Question 6: Miscellaneous Exercise
6. Suppose we have four boxes A, B, C and D containing coloured marbles as given below. One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A, box B or box C?
| Box | Red | White | Black |
|---|---|---|---|
| A | 1 | 6 | 3 |
| B | 6 | 2 | 2 |
| C | 8 | 1 | 1 |
| D | 0 | 6 | 4 |
Solution:
Think First… We know that the marble drawn is red, but we need to find which box it came from. So, this is a Bayes’ Theorem question.
Let E1, E2, E3 and E4 denote the events that boxes A, B, C and D, respectively, are selected.
Let R be the event that the marble drawn is red.
Since one of the four boxes is selected at random,
\displaystyle P(E_1)=P(E_2)=P(E_3)=P(E_4)=\frac{1}{4}
Each box contains 10 marbles. Therefore, the probabilities of drawing a red marble from the four boxes are
\displaystyle P(R\mid E_1)=\frac{1}{10},\; P(R\mid E_2)=\frac{6}{10},\; P(R\mid E_3)=\frac{8}{10},\; P(R\mid E_4)=\frac{0}{10}
We first find the probability of drawing a red marble using the theorem of total probability.
i.e. \displaystyle P(R)=\sum_{j=1}^{4}P(E_j)P(R\mid E_j)
\displaystyle =\frac14\left(\frac1{10}\right)+\frac14\left(\frac6{10}\right)+\frac14\left(\frac8{10}\right)+\frac14(0)
\displaystyle =\frac{1+6+8}{40}=\frac{15}{40}=\frac38
Probability that the marble was drawn from box A
By Bayes’ Theorem,
\displaystyle P(E_1\mid R)=\frac{P(E_1)P(R\mid E_1)}{P(R)}
\displaystyle =\frac{\frac14\times\frac1{10}}{\frac38}=\frac{1}{15}
Probability that the marble was drawn from box B
\displaystyle P(E_2\mid R)=\frac{P(E_2)P(R\mid E_2)}{P(R)}
\displaystyle =\frac{\frac14\times\frac6{10}}{\frac38}=\frac25
Probability that the marble was drawn from box C
\displaystyle P(E_3\mid R)=\frac{P(E_3)P(R\mid E_3)}{P(R)}
\displaystyle =\frac{\frac14\times\frac8{10}}{\frac38}=\frac8{15}
Hence,
\displaystyle \boxed{\begin{gathered}P(A\mid R)=\frac{1}{15},\quad P(B\mid R)=\frac{2}{5}\\[4pt]P(C\mid R)=\frac{8}{15}\end{gathered}}
Question 7: Bayes’ Thorem
7. Assume that the chances of a patient having a heart attack is 40%. It is also assumed that a meditation and yoga course reduce the risk of heart attack by 30% and prescription of certain drug reduces its chances by 25%. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga?
Solution:
Think First… We know that the patient suffered a heart attack, but we need to find which option the patient had chosen. So, this is a Bayes’ Theorem question.
Let E1 be the event that the patient followed a course of meditation and yoga, and E2 be the event that the patient was prescribed the drug.
Let A be the event that the patient suffers a heart attack.
Since the patient can choose either option with equal probabilities,
\displaystyle P(E_1)=P(E_2)=\frac{1}{2}
The initial probability of a heart attack is
\displaystyle P(A)=0.40
The meditation and yoga course reduces the risk by 30%. Therefore, the remaining risk is 70% of the original risk.
Hence,
\displaystyle P(A\mid E_1)=0.40(1-0.30)=0.40\times0.70=0.28
The drug reduces the risk by 25%. Therefore, the remaining risk is 75% of the original risk.
Hence,
\displaystyle P(A\mid E_2)=0.40(1-0.25)=0.40\times0.75=0.30
We need to find the probability that the patient followed a course of meditation and yoga, given that the patient suffered a heart attack, i.e.,
\displaystyle P(E_1\mid A)
By Bayes’ Theorem,
\displaystyle P(E_1\mid A)=\frac{P(E_1)P(A\mid E_1)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)}
Substituting the values,
We get, \displaystyle P(E_1\mid A)=\frac{\frac12\times0.28}{\frac12\times0.28+\frac12\times0.30}
\displaystyle =\frac{0.28}{0.28+0.30}=\frac{0.28}{0.58}=\frac{14}{29}
Hence,
\displaystyle \boxed{P(\text{Reqd. Probability})=P(E_1\mid A)=\frac{14}{29}}
Maths Better Tip… Be careful when a question says that a treatment or medicine reduces the risk by a certain percentage. If the original risk is \displaystyle 40\% and the risk is reduced by 30%, the new risk is 70% of the original risk, not 30% of it.
\displaystyle 40\%\times70\%=28\%
Similarly, a reduction of 25% means that 75% of the original risk remains.
\displaystyle 40\%\times75\%=30\%
So, whenever a risk is reduced by r%, always use \displaystyle (100-r)\% of the original risk.
Question 8: Probability Miscellaneous
8. If each element of a second order determinant is either zero or one, what is the probability that the value of the determinant is positive? Assume that the individual entries of the determinant are chosen independently, each value being assumed with probability \displaystyle \frac{1}{2}.
Solution:
Think First… Write the determinant in general form and identify when its value can be positive. Since every entry is either 0 or 1, each entry has only two possible values.
Let the second order determinant be
\displaystyle D=\begin{vmatrix}a&b\\c&d\end{vmatrix}=ad-bc
Each of the four entries can independently take the values 0 or 1, each with probability \displaystyle \frac12.
Therefore, the total number of possible determinants is
\displaystyle 2^4=16
For the determinant to be positive,
\displaystyle ad-bc>0
Since each of \displaystyle a,b,c,d is either 0 or 1, both \displaystyle ad and \displaystyle bc can only be 0 or 1.
Therefore, for \displaystyle ad-bc to be positive, we must have
\displaystyle ad=1\quad\text{and}\quad bc=0
Now, \displaystyle ad=1 is possible only when
\displaystyle a=1,\qquad d=1
So, the determinant must be of the form
\displaystyle D=\begin{vmatrix}1&b\\c&1\end{vmatrix}=1-bc
We also need \displaystyle bc=0. Since \displaystyle b and \displaystyle c can each be 0 or 1, this happens in exactly three cases:
\displaystyle (b,c)=(0,0),\qquad(0,1),\qquad(1,0)
Thus, the three favourable determinants are
\displaystyle \begin{vmatrix}1&0\\0&1\end{vmatrix},\qquad\begin{vmatrix}1&0\\1&1\end{vmatrix},\qquad\begin{vmatrix}1&1\\0&1\end{vmatrix}
Each of these has a positive determinant. Therefore,
\displaystyle \text{Number of favourable outcomes}=3
Hence, the required probability is
\displaystyle P(D>0)=\frac{3}{16}
Hence,
\displaystyle \boxed{\frac{3}{16}}
Question 9: Bayes’ Theorem and Total Probability
9. An electronic assembly consists of two subsystems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known:
\displaystyle P(A\text{ fails})=0.2,\; P(B\text{ fails alone})=0.15,\; P(A\text{ and }B\text{ fail})=0.15
Evaluate the following probabilities: (i) P(A fails|B has failed) (ii) P(A fails alone).
Solution:
Think First… Be careful with the phrase “B fails alone”. It means B fails while A does not fail. However, “B has failed” includes both cases — B fails alone and both A and B fail.
(i) P(A fails | B has failed)
We are given
\displaystyle P(B\text{ fails alone})=0.15
and
\displaystyle P(A\text{ and }B\text{ fail})=0.15
Therefore, the probability that B fails is
\displaystyle P(B\text{ fails})=P(B\text{ fails alone})+P(A\text{ and }B\text{ fail})
\displaystyle =0.15+0.15=0.30
Now,
\displaystyle P(A\text{ fails}\mid B\text{ has failed})=\frac{P(A\text{ and }B\text{ fail})}{P(B\text{ fails})}
\displaystyle =\frac{0.15}{0.30}=0.5
Hence,
\displaystyle \boxed{P(A\text{ fails}\mid B\text{ has failed})=0.5}
(ii) P(A fails alone)
We know that A fails either alone or along with B.
Therefore,
\displaystyle P(A\text{ fails})=P(A\text{ fails alone})+P(A\text{ and }B\text{ fail})
Substituting the given values,
\displaystyle 0.2=P(A\text{ fails alone})+0.15
Hence,
\displaystyle P(A\text{ fails alone})=0.2-0.15=0.05
Hence,
\displaystyle \boxed{P(A\text{ fails alone})=0.05}
Question 10: Probability
10. Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.
Solution:
Think First… The transferred ball can be either red or black. Since the ball finally drawn from Bag II is known to be red, we need to find the probability that the transferred ball was black. So, this is a Bayes’ Theorem question.
Let E1 be the event that the transferred ball is red and E2 be the event that the transferred ball is black.
Let A be the event that the ball drawn from Bag II is red.
Initially, Bag I contains 3 red and 4 black balls. Therefore,
\displaystyle P(E_1)=\frac{3}{7}
Similarly,
\displaystyle P(E_2)=\frac{4}{7}
Case 1: The transferred ball is red
In this case, Bag II will contain 5 red and 5 black balls, i.e., 10 balls in total.
Therefore,
\displaystyle P(A\mid E_1)=\frac{5}{10}=\frac{1}{2}
Case 2: The transferred ball is black
In this case, Bag II will contain 4 red and 6 black balls, i.e., 10 balls in total.
Therefore,
\displaystyle P(A\mid E_2)=\frac{4}{10}=\frac{2}{5}
We need to find
\displaystyle P(E_2\mid A)
By Bayes’ Theorem,
\displaystyle P(E_2\mid A)=\frac{P(E_2)P(A\mid E_2)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)}
Substituting the values,
We get, \displaystyle P(E_2\mid A)=\frac{\frac47\times\frac25}{\frac37\times\frac12+\frac47\times\frac25}
\displaystyle =\frac{\frac{8}{35}}{\frac{3}{14}+\frac{8}{35}}
\displaystyle =\frac{\frac{16}{70}}{\frac{15}{70}+\frac{16}{70}}=\frac{16}{31}
Hence,
\displaystyle \boxed{P(\text{Reqd. Probability})=P(E_2\mid A)=\frac{16}{31}}
By now, you have already completed the NCERT Solutions for Matrices, Determinants, Relations and Functions, Inverse Trigonometric Functions, Continuity and Differentiability, Application of Derivatives, Integrals, Application of Integrals, Differential Equations, Vector Algebra, Three Dimensional Geometry and Linear Programming. Now, we are working through Probability miscellaneous exercise with every exercise so far explained through detailed, step-by-step solutions to help you build strong concepts and prepare confidently for your CBSE Board Examinations.
Many of these questions are also available in video format on my YouTube Channel, @MathsBetter, where each solution is explained in a simple and student-friendly manner. Now, let’s proceed to the next question of Probability miscellaneous exercise.
Question 11: Probability Miscellaneous – MCQ
11. If A and B are two events such that \displaystyle P(A)\neq0 and \displaystyle P(B\mid A)=1, then which of the following is correct?
- (A) \displaystyle A\subset B
- (B) \displaystyle B\subset A
- (C) \displaystyle B=\varnothing
- (D) \displaystyle A=\varnothing
Solution
Think First… Since P(B|A)=1, whenever A occurs, B must also occur.
Using the formula for conditional probability,
\displaystyle P(B\mid A)=\frac{P(A\cap B)}{P(A)}=1
Since \displaystyle P(A)\neq0, we get
\displaystyle P(A\cap B)=P(A)
This means that the occurrence of A always lies within the occurrence of B. Therefore,
\displaystyle A\subset B
✅️ Hence, the correct answer is (A).
Question 12: Probability Miscellaneous – MCQ
12. If \displaystyle P(A\mid B)>P(A), then which of the following is correct?
- (A) \displaystyle P(B\mid A)\lt P(B)
- (B) \displaystyle P(A\cap B)\lt P(A)\cdot P(B)
- (C) \displaystyle P(B\mid A)\gt P(B)
- (D) \displaystyle P(B\mid A)=P(B)
Solution
Think First… The given condition tells us that B makes A more likely. We need to see what this implies about \displaystyle P(B\mid A) and choose the correct option.
We are given
\displaystyle P(A\mid B)\gt P(A)
Using the formula for conditional probability,
\displaystyle P(A\mid B)=\frac{P(A\cap B)}{P(B)}
Therefore,
\displaystyle \frac{P(A\cap B)}{P(B)}\gt P(A)
Multiplying both sides by \displaystyle P(B), we get
\displaystyle P(A\cap B)\gt P(A)P(B)
Now, using the conditional probability formula in the other direction,
\displaystyle P(B\mid A)=\frac{P(A\cap B)}{P(A)}
Dividing both sides of the inequality by \displaystyle P(A), we get
\displaystyle \frac{P(A\cap B)}{P(A)}\gt P(B)
Hence,
\displaystyle P(B\mid A)\gt P(B)
Therefore, option (C) is correct.
✅️ Hence, the correct answer is (C).
Question 13: Probability Miscellaneous – MCQ
13. If A and B are any two events such that \displaystyle P(A)+P(B)-P(\text{A and B})=P(A), then which of the following is correct?
- (A) \displaystyle P(B\mid A)=1
- (B) \displaystyle P(A\mid B)=1
- (C) \displaystyle P(B\mid A)=0
- (D) \displaystyle P(A\mid B)=0
Solution
Think First… Start with the given equation and simplify it. The result will directly give us the conditional probability required in the options.
We are given
\displaystyle P(A)+P(B)-P(\text{A and B})=P(A)
Writing “A and B” as \displaystyle A\cap B,
\displaystyle P(A)+P(B)-P(A\cap B)=P(A)
Cancelling \displaystyle P(A) from both sides, we get
\displaystyle P(B)=P(A\cap B)
Dividing both sides by \displaystyle P(B),
\displaystyle \frac{P(A\cap B)}{P(B)}=1
Using the formula for conditional probability,
\displaystyle P(A\mid B)=\frac{P(A\cap B)}{P(B)}
Therefore,
\displaystyle P(A\mid B)=1
✅️ Hence, the correct answer is (B).
Common Mistakes to Avoid
- Confusing conditional probabilities: Do not confuse \displaystyle P(A\mid B) with \displaystyle P(B\mid A). They are generally not equal. Always identify which event is given and which event is to be found.
- Ignoring the condition given in the question: When a question says “given that” an event has occurred, the sample space has effectively been restricted to that event. Use the appropriate conditional probability formula rather than the ordinary probability.
- Confusing “alone” with “and”: “A fails alone” means A occurs but B does not. On the other hand, “A and B fail” means both occur. For example, \displaystyle P(A\text{ fails})=P(A\text{ fails alone})+P(A\text{ and }B\text{ fail}).
- Using Bayes’ Theorem in the wrong direction: If the result is already known and the question asks for the probability of a particular case or cause, the required probability is of the form \displaystyle P(E_i\mid A), not \displaystyle P(A\mid E_i).
- Forgetting all possible cases: When an event can occur through different cases, include every relevant case. For example, the total probability of A is \displaystyle P(A)=\sum_{j=1}^{n}P(E_j)P(A\mid E_j).
- Confusing independence with conditional probability: If A and B are independent, then \displaystyle P(A\mid B)=P(A) and \displaystyle P(B\mid A)=P(B), whenever the conditional probabilities are defined. Do not assume independence merely because two events are unrelated in the wording.
- Misreading “at least” and “at most”: In counting or repeated-trial questions, translate the wording carefully. “At most 6” means 0, 1, 2, …, 6, while “at least 6” means 6, 7, …, up to the maximum possible value.
- Forgetting the complement in repeated-trial questions: If the required probability is easier to find through the complement, use \displaystyle P(E)=1-P(E'). For example, “at most 6” out of 10 can be found as 1 minus the probability of 7, 8, 9 or 10.
- Counting outcomes incorrectly: When each entry or object can take only a fixed number of values, first find the total number of possible outcomes. For example, if 4 independent entries can each be 0 or 1, the total number of possible arrangements is \displaystyle 2^4=16.
- Making an error in determinant conditions: For a second order determinant \displaystyle \begin{vmatrix}a&b\\c&d\end{vmatrix}=ad-bc, first write its value in this form and then carefully determine when it is positive, negative or zero.
- Using the wrong probability after a transfer: In bag-and-ball questions, update the composition of the bag after the transfer before finding the probability of the next draw. Do not use the original numbers after a ball has been transferred.
- Forgetting that the transferred ball can have different colours: In transfer questions, consider each possible colour of the transferred ball as a separate case. If the final colour is known, Bayes’ Theorem may be required to work backwards and find the probability of the transferred colour.
- Converting percentages incorrectly: Convert percentages carefully before using them in probability calculations. For example, \displaystyle 5\%=0.05 and \displaystyle 0.25\%=0.0025.
- Ignoring the complement of an event: Remember that \displaystyle P(E')=1-P(E). This is especially useful when the question asks for the probability of an event not occurring or when the complement is easier to calculate.
- Using the infinite GP formula without checking the condition: For an infinite geometric progression \displaystyle a+ar+ar^2+\cdots, the sum \displaystyle \frac{a}{1-r} is valid only when \displaystyle \left|r\right|\lt1.
- Stopping the solution too early: In application-based questions, define the events or cases first, write the relevant probability formula, substitute the values and simplify. A clear setup helps avoid mistakes even when the final calculation is simple.
Continue Learning
With the completion of the Probability Miscellaneous Exercise, you have now revised and applied the important concepts covered throughout Chapter 13 — Probability. The questions in this exercise bring together conditional probability, the multiplication theorem, independence, the theorem of total probability and Bayes’ theorem, along with several applications of probability.
Before moving on to the next chapter, make sure you are comfortable with the following key ideas:
- Using the formula for conditional probability: \displaystyle P(A\mid B)=\frac{P(A\cap B)}{P(B)}, where \displaystyle P(B)\neq0.
- Using the multiplication theorem: \displaystyle P(A\cap B)=P(A)P(B\mid A)=P(B)P(A\mid B).
- Understanding independent events and the result \displaystyle P(A\cap B)=P(A)P(B).
- Using the theorem of total probability when an event can occur through different possible cases: \displaystyle P(A)=\sum_{j=1}^{n}P(E_j)P(A\mid E_j).
- Recognising when a question requires Bayes’ theorem — especially when an event is already known to have occurred and we need to find the probability of a particular case or cause.
- Using Bayes’ theorem correctly: \displaystyle P(E_i\mid A)=\frac{P(E_i)P(A\mid E_i)}{\displaystyle\sum_{j=1}^{n}P(E_j)P(A\mid E_j)}.
- Distinguishing carefully between \displaystyle P(A\mid E_i) and \displaystyle P(E_i\mid A).
- Using the complement of an event when it makes the calculation easier: \displaystyle P(E')=1-P(E).
- Reading probability questions carefully, especially words such as “given that”, “at least”, “at most”, “alone” and “both”.
- Defining events clearly and identifying the possible cases or causes before starting the calculation in application-based questions.
- Checking conditions carefully before applying a formula, such as \displaystyle P(B)\neq0 for conditional probability and \displaystyle \left|r\right|\lt1 when finding the sum of an infinite GP.
The Miscellaneous Exercise is designed to bring together the concepts from the entire chapter. By solving these questions, you have not only revised the important formulas but also practised deciding which concept to use and when to use it — an essential skill for the CBSE Class 12 Board Exams, CUET, and other competitive examinations.
Explore More
A strong understanding of conditional probability, the multiplication theorem, independence, the theorem of total probability and Bayes’ theorem forms the foundation for solving a wide variety of probability problems. While solving questions, first understand the given information and identify the events involved. Then choose the appropriate probability rule instead of trying to apply a formula directly.
While solving questions, carefully analyse the wording, especially phrases such as “given that”, “at least”, “at most”, “alone” and “both”. In application-based problems involving bags, machines, tests, cards, coins and other real-life situations, define the events clearly and identify all possible cases before starting the calculation. Always pay attention to the direction of conditional probability and do not confuse \displaystyle P(A\mid B) with \displaystyle P(B\mid A). With regular practice, you will become more confident in recognising the appropriate probability concept, setting up the problem correctly and solving it accurately.
With this, Probability Miscellaneous and the whole Chapter 13 is complete. Keep practising different types of questions, revise the important formulas regularly and focus on understanding why a particular probability rule is used. This will help you approach CBSE Class 12 Board Exams, CUET and other competitive examinations with greater confidence.
And also with this, the entire Class 12 NCERT Mathematics syllabus is complete! 🎉
From Relations and Functions to Probability, we have worked through all the NCERT chapters step by step. This completes the full set of Class 12 NCERT Mathematics Solutions on Maths Better — a huge milestone! 🙌
But this is not the end of the journey. It is the foundation for what comes next — revision, practice, PYQs, MCQs, NCERT Exemplar questions and much more. Keep practising, keep revising and, most importantly, keep making your Maths Better! 💙
