Probability: Chapter 13 Links
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Probability 13.3 NCERT Solutions focuses on two important concepts — the Theorem of Total Probability and Bayes’ Theorem. The theorem of total probability helps us find the probability of an event by considering all the possible events or cases that can lead to it.
\displaystyle P(A)=\sum_{j=1}^{n}P(E_j)\,P(A\mid E_j)
Bayes’ Theorem takes us a step further. It helps us find the reverse probability — that is, when an event has already occurred, we can find the probability that it occurred due to a particular cause or case.
\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)}
The questions in this exercise cover a variety of applications of these concepts involving bags, balls, students, tests, machines, coins, cards and other real-life situations. Many questions require us to identify the different possible cases, find the corresponding conditional probabilities and then apply Bayes’ Theorem.
In the exercise of Probability 13.3, you will see how the Theorem of Total Probability helps us find the probability of an event, while Bayes’ Theorem helps us work backwards to find the probability of a particular case when the event is already known to have occurred.
By solving these questions step by step, you will strengthen your understanding of an important part of Probability and build confidence for the CBSE Class 12 Board Exams, CUET, and other competitive examinations.
Key Concepts
Before solving Exercise 13.3, quickly revise the important concepts and formulas related to the Theorem of Total Probability and Bayes’ Theorem. Most questions in this exercise require you to identify the different possible cases or causes, find the relevant conditional probabilities and then decide whether to use total probability or Bayes’ theorem.
1. Partition of a Sample Space
Events E1, E2, …, En are said to form a partition of the sample space S if they are mutually exclusive and together cover the entire sample space.
Therefore, the events satisfy
\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, exactly one of the events in the partition occurs.
For example, if a ball is drawn from one of two bags, the events “Bag I is selected” and “Bag II is selected” form a partition of the sample space.
2. Theorem of Total Probability
If E1, E2, …, En form a partition of the sample space S, and A is any event, then the probability of A is given by
\displaystyle P(A)=P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)+\cdots+P(E_n)P(A\mid E_n)
Or, in summation form,
\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.
For example, if an item can be produced by either Machine A or Machine B, the probability that a randomly selected item is defective can be found by considering the defective items produced by each machine separately.
3. Bayes’ Theorem
Bayes’ Theorem helps us find the reverse probability. Instead of finding the probability of an event A when the case Ei is known, we find the probability of the case Ei when the event A is known to have occurred.
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)}
For two events in the partition, say E1 and E2, this becomes
\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)}
This is the formula you will use frequently in this exercise.
4. Understanding Bayes’ Theorem
Bayes’ Theorem is based on the conditional probability formula
\displaystyle P(E_i\mid A)=\frac{P(A\cap E_i)}{P(A)}
Using the multiplication theorem,
\displaystyle P(A\cap E_i)=P(E_i)\,P(A\mid E_i)
and using the theorem of total probability for P(A), we get Bayes’ theorem.
So, remember the basic structure:
\displaystyle P(E_i\mid A)=\frac{\text{Probability of }E_i\text{ and }A}{\text{Total probability of }A}
5. Prior and Posterior Probability
Bayes’ theorem is often described as the formula for finding the probability of a cause when the result or event is already known.
- Prior probability: The probability of the hypothesis before the event is known. It is represented by \displaystyle P(E_i).
- Posterior probability: The conditional probability of the hypothesis after the event is known. It is represented by \displaystyle P(E_i\mid A).
- Conditional probability: \displaystyle P(A\mid E_i) tells us the probability of event A when the case Ei is already known.
For example, if a defective item is found and we need to determine whether it was produced by Machine A or Machine B, Bayes’ theorem helps us find the probability of the particular machine given that the item is defective.
6. How to Identify a Bayes’ Theorem Question
A useful way to recognise Bayes’ theorem questions is to look for a situation where the result is given first and we are asked to find the case or cause behind it.
- A ball is found to be red. Find the probability that it came from Bag I.
- An item is found to be defective. Find the probability that it was produced by Machine B.
- A test result is positive. Find the probability that the person actually has the disease.
- A student has an A grade. Find the probability that the student belongs to a particular group.
In such questions, the wording usually has the form “given that A has occurred, find the probability of Ei”, which means we need to find \displaystyle P(E_i\mid A).
7. A Simple Bayes’ Theorem Approach
For most questions in this exercise, the following steps will make the solution easier:
- Define the events: Clearly identify the possible cases or hypotheses E1, E2, …, En and the given event A.
- Find the prior probabilities: Calculate \displaystyle P(E_1),P(E_2),\ldots,P(E_n).
- Find the conditional probabilities: Calculate \displaystyle P(A\mid E_1),P(A\mid E_2),\ldots,P(A\mid E_n).
- Apply Bayes’ theorem: Substitute the values carefully in the formula.
- Simplify: Calculate the required probability and state the final answer clearly.
8. Total Probability vs Bayes’ Theorem
It is important to understand the difference between the two.
- Total Probability: We know the possible cases and find the probability of the resulting event.
- Bayes’ Theorem: We know that the resulting event has occurred and find the probability of a particular case that caused it.
\displaystyle \text{Case}\longrightarrow\text{Event}\qquad\text{Total Probability}
\displaystyle \text{Event}\longrightarrow\text{Case}\qquad\text{Bayes' Theorem}
This “forward” and “reverse” idea is one of the easiest ways to remember when Bayes’ theorem is required.
9. Important Points
- Always check whether the given events form a partition of the sample space.
- The events in a partition are mutually exclusive and their union is the entire sample space.
- Use the Theorem of Total Probability when you need to find \displaystyle P(A) from different possible cases.
- Use Bayes’ Theorem when you need to find \displaystyle P(E_i\mid A), that is, the probability of a particular case when the event A is already known.
- In Bayes’ theorem, the denominator is the total probability of the given event.
- Define events clearly before substituting values into the formula. This prevents confusion between \displaystyle P(A\mid E_i) and \displaystyle P(E_i\mid A).
- Remember: \displaystyle P(A\mid E_i) and \displaystyle P(E_i\mid A) are generally not equal.
- For questions involving tests, machines, bags, boxes or groups, first identify the possible causes/cases and then identify the given result/event.
Quick Reference Table
| Concept | Formula / Result |
|---|---|
| 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)}{\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)} |
| Conditional Probability | \displaystyle P(E_i\mid A)=\frac{P(E_i\cap A)}{P(A)} |
| Multiplication Rule | \displaystyle P(E_i\cap A)=P(E_i)P(A\mid E_i) |
| Prior Probability | \displaystyle P(E_i) |
| Posterior Probability | \displaystyle P(E_i\mid A) |
| Reverse Probability | Given the event A, 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 |
Before You Begin… Before solving a question, first identify the possible cases or causes and the given event. If you need to find the probability of the event from the different cases, use the Theorem of Total Probability. If the event has already occurred and you need to find the probability of a particular case, use Bayes’ Theorem.
Let us now solve all the NCERT questions of Probability 13.3 step by step.
Question 1: Total Probability
1. An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and it is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?
Solution:
Think First… The composition of the urn before the second draw depends on the colour of the first ball. So, consider the two possible cases — the first ball is red or the first ball is black — and use the theorem of total probability.
Let E1 be the event that the first ball drawn is red and E2 be the event that the first ball drawn is black.
Let A be the event that the second ball drawn is red.
Since there are 5 red and 5 black balls,
\displaystyle P(E_1)=\frac{5}{10}=\frac{1}{2}
Similarly,
\displaystyle P(E_2)=\frac{5}{10}=\frac{1}{2}
Case 1: The first ball is red
If the first ball is red, it is returned to the urn and 2 more red balls are added. The urn then contains 7 red and 5 black balls, i.e. 12 balls in total.
Hence,
\displaystyle P(A\mid E_1)=\frac{7}{12}
Case 2: The first ball is black
If the first ball is black, it is returned to the urn and 2 more black balls are added. The urn then contains 5 red and 7 black balls, i.e. 12 balls in total.
Hence,
\displaystyle P(A\mid E_2)=\frac{5}{12}
By the theorem of total probability,
\displaystyle P(A)=P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)
Substituting the values, we get
\displaystyle P(A)=\frac{1}{2}\times\frac{7}{12}+\frac{1}{2}\times\frac{5}{12}
\displaystyle =\frac{7}{24}+\frac{5}{24}=\frac{12}{24}=\frac{1}{2}
Hence,
\displaystyle \boxed{P(A)=\frac{1}{2}}
Question 2: Bayes’ Theorem
2. 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:
Think First… We know that the ball drawn is red, but we need to find the probability that it came from the first bag. This is a reverse probability, so we use Bayes’ Theorem.
Let E1 be the event that the first bag is selected and E2 be the event that the second bag is selected.
Let A be the event that a red ball is drawn.
Since either of the two bags is selected at random,
\displaystyle P(E_1)=\frac{1}{2}
Similarly,
\displaystyle P(E_2)=\frac{1}{2}
Also, the probability of drawing a red ball from the first bag is
\displaystyle P(A\mid E_1)=\frac{4}{4+4}=\frac{1}{2}
The probability of drawing a red ball from the second bag is
\displaystyle P(A\mid E_2)=\frac{2}{2+6}=\frac{1}{4}
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{\frac{1}{2}\times\frac{1}{2}}{\frac{1}{2}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{4}}
\displaystyle =\frac{\frac{1}{4}}{\frac{1}{4}+\frac{1}{8}}=\frac{\frac{1}{4}}{\frac{3}{8}}=\frac{2}{3}
Hence,
\displaystyle \boxed{P(E_1\mid A)=\frac{2}{3}}
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, this exercise of Probability 13.3 builds on the concept of Conditional Probability and introduces two important ideas — the Theorem of Total Probability and Bayes’ Theorem. Through these questions, you will learn how to break a problem into different possible cases, find the probability of an event using total probability, and work backwards using Bayes’ Theorem to find the probability of a particular case or cause. These questions provide valuable practice for the CBSE Board Examination and CUET, while strengthening the concepts needed for the Probability chapter. Now, let’s continue with the next question of Probability 13.3.
Question 3: Probability 13.3
3. Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel). Previous year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is a hosteller?
Solution:
Think First… We know that the selected student has an A grade, but we need to find the probability that the student is a hosteller. This is a reverse probability, so we use Bayes’ Theorem.
Let E1 be the event that the student is a hosteller and E2 be the event that the student is a day scholar.
Let A be the event that the student attains an A grade.
We are given
\displaystyle P(E_1)=60\%=\frac{60}{100}=\frac{3}{5}
Also,
\displaystyle P(E_2)=40\%=\frac{40}{100}=\frac{2}{5}
The probability that a hosteller attains an A grade is
\displaystyle P(A\mid E_1)=30\%=\frac{3}{10}
and the probability that a day scholar attains an A grade is
\displaystyle P(A\mid E_2)=20\%=\frac{1}{5}
We need to find \displaystyle P(E_1\mid A), the probability that the student is a hosteller given that the student has an A grade.
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 given values,
We get, \displaystyle P(E_1\mid A)=\frac{\frac{3}{5}\times\frac{3}{10}}{\frac{3}{5}\times\frac{3}{10}+\frac{2}{5}\times\frac{1}{5}}
\displaystyle =\frac{\frac{9}{50}}{\frac{9}{50}+\frac{2}{25}}
\displaystyle =\frac{\frac{9}{50}}{\frac{9}{50}+\frac{4}{50}}=\frac{9}{13}
Hence,
\displaystyle \boxed{P(E_1\mid A)=\frac{9}{13}}
Question 4: Probability 13.3
4. In answering a question on a multiple choice test, a student either knows the answer or guesses. Let 3/4 be the probability that he knows the answer and 1/4 be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability 1/4, what is the probability that the student knows the answer given that he answered it correctly?
Solution:
Think First… We know that the answer is correct, but we need to find the probability that the student knew the answer. This is a reverse probability, so we use Bayes’ Theorem.
Let A be the event that the student knows the answer and B be the event that the student’s answer is correct.
We are given
\displaystyle P(A)=\frac{3}{4}
Therefore, the probability that the student guesses is
\displaystyle P(A')=\frac{1}{4}
If the student knows the answer, he will answer it correctly. Hence,
\displaystyle P(B\mid A)=1
If the student guesses, the probability of answering correctly is
\displaystyle P(B\mid A')=\frac{1}{4}
We need to find \displaystyle P(A\mid B), the probability that the student knows the answer given that he answered it correctly.
By Bayes’ Theorem,
\displaystyle P(A\mid B)=\frac{P(A)P(B\mid A)}{P(A)P(B\mid A)+P(A')P(B\mid A')}
Substituting the given values,
We get, \displaystyle P(A\mid B)=\frac{\frac{3}{4}\times1}{\frac{3}{4}\times1+\frac{1}{4}\times\frac{1}{4}}
\displaystyle =\frac{\frac{3}{4}}{\frac{3}{4}+\frac{1}{16}}=\frac{\frac{12}{16}}{\frac{13}{16}}=\frac{12}{13}
Hence,
\displaystyle \boxed{P(A\mid B)=\frac{12}{13}}
Maths Better Tip… Bayes’ Theorem is used when we know that an event has occurred and want to find the probability of the particular event or cause that led to it. First identify the possible cases, then use their prior probabilities and the corresponding conditional probabilities carefully.
Question 5: Probability 13.3
5. A laboratory blood test is 99% effective in detecting a certain disease when it is in fact present. However, the test also yields a false positive result for 0.5% of the healthy person tested (i.e. if a healthy person is tested, then with probability 0.005, the test will imply that he has the disease). If 0.1% of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?
Solution:
Think First… We know that the test result is positive, but we need to find the probability that the person actually has the disease. This is a reverse probability, so we use Bayes’ Theorem.
Let A be the event that the person has the disease and B be the event that the test result is positive.
We are given
\displaystyle P(A)=0.1\%=\frac{0.1}{100}=0.001
Therefore, the probability that the person does not have the disease is
\displaystyle P(A')=1-P(A)=1-0.001=0.999
Since the test is 99% effective in detecting the disease when it is present,
\displaystyle P(B\mid A)=99\%=\frac{99}{100}=0.99
Also, the probability of a false positive result for a healthy person is
\displaystyle P(B\mid A')=0.005
We need to find \displaystyle P(A\mid B), the probability that the person has the disease given that the test result is positive.
By Bayes’ Theorem,
\displaystyle P(A\mid B)=\frac{P(A)P(B\mid A)}{P(A)P(B\mid A)+P(A')P(B\mid A')}
Substituting the given values,
We get, \displaystyle P(A\mid B)=\frac{0.001\times0.99}{0.001\times0.99+0.999\times0.005}
\displaystyle =\frac{0.00099}{0.00099+0.004995}
\displaystyle =\frac{0.00099}{0.005985}=\frac{99}{598.5}=\frac{22}{133}
Hence,
\displaystyle \boxed{P(A\mid B)=\frac{22}{133}}
Question 6: Exercise 13.3
6. There are three coins. One is a two headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin?
Solution:
Think First… We know that the selected coin shows heads, but we need to find the probability that it was the two-headed coin. This is a reverse probability, so we use Bayes’ Theorem.
Let A be the event that the two-headed coin is selected, B be the event that the biased coin is selected and C be the event that the unbiased coin is selected.
Let H be the event that the coin shows heads.
Since one of the three coins is chosen at random,
\displaystyle \therefore P(A)=P(B)=P(C)=\frac{1}{3}
Also, the probability of getting heads with the two-headed coin is
\displaystyle P(H\mid A)=1
For the biased coin,
\displaystyle P(H\mid B)=75\%=\frac{3}{4}
For the unbiased coin,
\displaystyle P(H\mid C)=\frac{1}{2}
We need to find \displaystyle P(A\mid H), the probability that the two-headed coin was selected given that the result is heads.
By Bayes’ Theorem,
\displaystyle P(A\mid H)=\frac{P(A)P(H\mid A)}{P(A)P(H\mid A)+P(B)P(H\mid B)+P(C)P(H\mid C)}
Substituting the given values, we get
\displaystyle P(A\mid H)=\frac{\frac{1}{3}\times1}{\frac{1}{3}\times1+\frac{1}{3}\times\frac{3}{4}+\frac{1}{3}\times\frac{1}{2}}
\displaystyle =\frac{1}{1+\frac{3}{4}+\frac{1}{2}}=\frac{1}{\frac{9}{4}}=\frac{4}{9}
Hence,
\displaystyle \boxed{P(A\mid H)=\frac{4}{9}}
Question 7: Probability 13.3
7. An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accident are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?
Solution:
Think First… We know that an insured person has met with an accident, but we need to find the probability that he is a scooter driver. This is a reverse probability, so we use Bayes’ Theorem.
Let A, B and C be the events that the insured person is a scooter driver, car driver and truck driver respectively.
Let D be the event that the insured person meets with an accident.
The total number of insured persons is
\displaystyle 2000+4000+6000=12000
Therefore,
\displaystyle P(A)=\frac{2000}{12000}=\frac{1}{6}
\displaystyle P(B)=\frac{4000}{12000}=\frac{1}{3}
and
\displaystyle P(C)=\frac{6000}{12000}=\frac{1}{2}
We are also given that
\displaystyle P(D\mid A)=0.01,\quad P(D\mid B)=0.03,\quad P(D\mid C)=0.15
We need to find \displaystyle P(A\mid D), the probability that the person is a scooter driver given that he has met with an accident.
By Bayes’ Theorem,
\displaystyle P(A\mid D)=\frac{P(A)P(D\mid A)}{P(A)P(D\mid A)+P(B)P(D\mid B)+P(C)P(D\mid C)}
Substituting the given values, we get
We get, \displaystyle P(A\mid D)=\frac{\frac{1}{6}\times0.01}{\frac{1}{6}\times0.01+\frac{1}{3}\times0.03+\frac{1}{2}\times0.15}
\displaystyle =\frac{\frac{1}{6}\times\frac{1}{100}}{\frac{1}{6}\times\frac{1}{100}+\frac{1}{3}\times\frac{3}{100}+\frac{1}{2}\times\frac{15}{100}}
\displaystyle =\frac{1}{1+6+45}=\frac{1}{52}
Hence,
\displaystyle \boxed{P(A\mid D)=\frac{1}{52}}
Maths Better Tip… Do not confuse the Theorem of Total Probability with Bayes’ Theorem. Total probability is used to find the probability of an event through different possible cases, while Bayes’ Theorem is used when the event has already occurred and we need to find the probability of a particular case or cause. In short, Total Probability works forward, while Bayes’ Theorem helps us work backward.
Question 8: Probability 13.3
8. A factory has two machines A and B. Past record shows that machine A produced 60% of the items of output and machine B produced 40% of the items. Further, 2% of the items produced by machine A and 1% produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine B?
Solution:
Think First… We know that the selected item is defective, but we need to find the probability that it was produced by machine B. This is a reverse probability, so we use Bayes’ Theorem.
Let E1 and E2 be the events that an item is produced by machine A and machine B respectively.
Let A be the event that the item is defective.
We are given
\displaystyle P(E_1)=60\%=\frac{3}{5}
and
\displaystyle P(E_2)=40\%=\frac{2}{5}
Also, the probability that an item produced by machine A is defective is
\displaystyle P(A\mid E_1)=2\%=\frac{1}{50}
and the probability that an item produced by machine B is defective is
\displaystyle P(A\mid E_2)=1\%=\frac{1}{100}
We need to find \displaystyle P(E_2\mid A), the probability that the defective item was produced by machine B.
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 given values,
We get, \displaystyle P(E_2\mid A)=\frac{\frac{2}{5}\times\frac{1}{100}}{\frac{3}{5}\times\frac{1}{50}+\frac{2}{5}\times\frac{1}{100}}
\displaystyle =\frac{\frac{2}{500}}{\frac{3}{250}+\frac{2}{500}}
\displaystyle =\frac{2}{6+2}=\frac{1}{4}
Hence,
\displaystyle \boxed{P(E_2\mid A)=\frac{1}{4}}
Question 9: Bayes’ Theorem
9. 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:
Think First… We know that the new product has been introduced, but we need to find the probability that it was introduced by the second group. This is a reverse probability, so we use Bayes’ Theorem.
Let A and B be the events that the first group and second group win respectively.
Let C be the event that a new product is introduced.
We are given
\displaystyle P(A)=0.6,\quad P(B)=0.4
Also,
\displaystyle P(C\mid A)=0.7,\quad P(C\mid B)=0.3
We need to find \displaystyle P(B\mid C), the probability that the second group introduced the new product given that a new product was introduced.
By Bayes’ Theorem,
\displaystyle P(B\mid C)=\frac{P(B)P(C\mid B)}{P(A)P(C\mid A)+P(B)P(C\mid B)}
Substituting the given values, we get
\displaystyle P(B\mid C)=\frac{0.4\times0.3}{0.6\times0.7+0.4\times0.3}
\displaystyle =\frac{0.12}{0.42+0.12}=\frac{0.12}{0.54}=\frac{2}{9}
Hence,
\displaystyle \boxed{P(B\mid C)=\frac{2}{9}}
Question 10: Probability 13.3
10. Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. And If she gets 1, 2, 3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die?
Solution:
Think First… We know that the girl obtained exactly one head, but we need to find the probability that she had thrown 1, 2, 3 or 4. This is a reverse probability, so we use Bayes’ Theorem.
Let A be the event that the outcome on the die is 5 or 6 and B be the event that the outcome on the die is 1, 2, 3 or 4.
Let C be the event that exactly one head is obtained.
Since the die is fair,
\displaystyle P(A)=\frac{2}{6}=\frac{1}{3}
and
\displaystyle P(B)=\frac{4}{6}=\frac{2}{3}
If the outcome is 5 or 6, the coin is tossed three times. The possible outcomes are
\displaystyle \{HHH,HHT,HTH,THH,TTH,THT,HTT,\\ TTT\}
Exactly one head occurs in 3 of these 8 outcomes. Therefore,
\displaystyle P(C\mid A)=\frac{3}{8}
If the outcome is 1, 2, 3 or 4, the coin is tossed once. Therefore,
\displaystyle P(C\mid B)=\frac{1}{2}
We need to find \displaystyle P(B\mid C), the probability that the girl threw 1, 2, 3 or 4 given that exactly one head was obtained.
By Bayes’ Theorem,
\displaystyle P(B\mid C)=\frac{P(B)P(C\mid B)}{P(A)P(C\mid A)+P(B)P(C\mid B)}
Substituting the given values, we get
\displaystyle P(B\mid C)=\frac{\frac{2}{3}\times\frac{1}{2}}{\frac{1}{3}\times\frac{3}{8}+\frac{2}{3}\times\frac{1}{2}}
\displaystyle =\frac{\frac{1}{3}}{\frac{1}{8}+\frac{1}{3}}=\frac{\frac{1}{3}}{\frac{11}{24}}=\frac{8}{11}
Hence,
\displaystyle \boxed{P(B\mid C)=\frac{8}{11}}
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, with every exercise 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 13.3.
Question 11: Probability 13.3
11. A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, whereas the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B is on the job for 30% of the time and C is on the job for 20% of the time. A defective item is produced, what is the probability that it was produced by A?
Solution:
Think First… We know that the item is defective, but we need to find the probability that it was produced by operator A. This is a reverse probability, so we use Bayes’ Theorem.
Let A, B and C be the events that the item is produced by operators A, B and C respectively.
Let D be the event that the item is defective.
We are given
\displaystyle P(A)=50\%=\frac{1}{2},\quad P(B)=30\%=\frac{3}{10},\quad P(C)=20\%=\frac{1}{5}
Also, the probabilities of producing a defective item by three machines respectively are
\displaystyle P(D\mid A)=1\%=\frac{1}{100}
\displaystyle P(D\mid B)=5\%=\frac{1}{20}
and
\displaystyle P(D\mid C)=7\%=\frac{7}{100}
We need to find \displaystyle P(A\mid D), the probability that the defective item was produced by operator A.
By Bayes’ Theorem,
\displaystyle P(A\mid D)=\frac{P(A)P(D\mid A)}{P(A)P(D\mid A)+P(B)P(D\mid B)+P(C)P(D\mid C)}
Substituting the given values,
We get, \displaystyle P(A\mid D)=\frac{\frac{1}{2}\times\frac{1}{100}}{\frac{1}{2}\times\frac{1}{100}+\frac{3}{10}\times\frac{1}{20}+\frac{1}{5}\times\frac{7}{100}}
\displaystyle =\frac{\frac{1}{200}}{\frac{1}{200}+\frac{3}{200}+\frac{7}{500}}
\displaystyle =\frac{5}{5+15+14}=\frac{5}{34}
Hence,
\displaystyle \boxed{P(A\mid D)=\frac{5}{34}}
Question 12: Probability
12. A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.
Solution:
Think First… We know that the two cards drawn are diamonds, but we need to find the probability that the lost card was a diamond. This is a reverse probability, so we use Bayes’ Theorem.
Let A be the event that the lost card is a diamond and B be the event that the two cards drawn are both diamonds.
There are 13 diamonds in a pack of 52 cards. Therefore,
\displaystyle P(A)=\frac{13}{52}=\frac{1}{4}
Hence, the probability that the lost card is not a diamond is
\displaystyle P(A')=1-P(A)=1-\frac{1}{4}=\frac{3}{4}
Case 1: The lost card is a diamond
If the lost card is a diamond, there are 12 diamonds among the remaining 51 cards. Therefore,
\displaystyle P(B\mid A)=\frac{12}{51}\times\frac{11}{50}=\frac{22}{425}
Case 2: The lost card is not a diamond
If the lost card is not a diamond, all 13 diamonds are among the remaining 51 cards. Therefore,
\displaystyle P(B\mid A')=\frac{13}{51}\times\frac{12}{50}=\frac{26}{425}
We need to find \displaystyle P(A\mid B), the probability that the lost card was a diamond given that the two cards drawn are both diamonds.
By Bayes’ Theorem,
\displaystyle P(A\mid B)=\frac{P(A)P(B\mid A)}{P(A)P(B\mid A)+P(A')P(B\mid A')}
Substituting the given values, we get
\displaystyle P(A\mid B)=\frac{\frac{1}{4}\times\frac{22}{425}}{\frac{1}{4}\times\frac{22}{425}+\frac{3}{4}\times\frac{26}{425}}
\displaystyle =\frac{22}{22+78}=\frac{22}{100}=\frac{11}{50}
Hence,
\displaystyle \boxed{P(A\mid B)=\frac{11}{50}}
Maths Better Tip! When cards or objects are drawn one after another, always check whether there is replacement. If the question specifically says that the card or object is replaced before the next draw, the probabilities may remain unchanged. When cards are simply drawn or withdrawn one after another without mentioning replacement, it is generally treated as without replacement. In such questions, remember that the total number of cards and the number of favourable cards change after each draw.
There can also be more than one way to calculate the probability. For example, in Q12, after one card is lost, there are \displaystyle 51 cards remaining. The probability that the two drawn cards are both diamonds, when the lost card is a diamond, can be found using combinations as
\displaystyle \frac{{}^{12}C_2}{{}^{51}C_2}
Similarly, when the lost card is not a diamond, the corresponding probability is
\displaystyle \frac{{}^{13}C_2}{{}^{51}C_2}
So, for drawing objects without replacement, you can often choose between the successive probability method and the combination method, depending on which makes the calculation clearer.
Question 13: Probability 13.3 – MCQ
13. Probability that A speaks truth is \displaystyle \frac{4}{5}. A coin is tossed. A reports that a head appears. The probability that actually there was head is
- (A) \displaystyle \frac{4}{5}
- (B) \displaystyle \frac{1}{2}
- (C) \displaystyle \frac{1}{5}
- (D) \displaystyle \frac{2}{5}
Solution:
Think First… We know that A has reported that a head appears, but we need to find the probability that the coin actually showed a head. This is a reverse probability, so we use Bayes’ Theorem.
Let E1 be the event that the coin shows a head and E2 be the event that the coin shows a tail.
Let B be the event that A reports that a head appears.
Since the coin is fair,
\displaystyle P(E_1)=P(E_2)=\frac{1}{2}
We are given that A speaks the truth with probability \displaystyle \frac{4}{5}. Therefore, if the coin actually shows a head, A reports a head with probability
\displaystyle P(B\mid E_1)=\frac{4}{5}
If the coin actually shows a tail, A can report a head only when A does not speak the truth. Hence,
\displaystyle P(B\mid E_2)=1-\frac{4}{5}=\frac{1}{5}
We need to find \displaystyle P(E_1\mid B), the probability that the coin actually showed a head given that A reported a head.
By Bayes’ Theorem,
\displaystyle P(E_1\mid B)=\frac{P(E_1)P(B\mid E_1)}{P(E_1)P(B\mid E_1)+P(E_2)P(B\mid E_2)}
Substituting the given values,
\displaystyle P(E_1\mid B)=\frac{\frac{1}{2}\times\frac{4}{5}}{\frac{1}{2}\times\frac{4}{5}+\frac{1}{2}\times\frac{1}{5}}
\displaystyle =\frac{\frac{4}{10}}{\frac{4}{10}+\frac{1}{10}}=\frac{4}{5}
✅️ Hence, the correct answer is (A).
Question 14: Probability 13.3 – MCQ
14. If A and B are two events such that \displaystyle A\subset B and \displaystyle P(B)\neq0, then which of the following is correct?
- (A) \displaystyle P(A\mid B)=\frac{P(B)}{P(A)}
- (B) \displaystyle P(A\mid B)\lt P(A)
- (C) \displaystyle P(A\mid B)\geq P(A)
- (D) None of these
Solution
Think First… Since A is a subset of B, whenever A occurs, B must also occur.
Since \displaystyle A\subset B, we have
\displaystyle A\cap B=A
Using the formula for conditional probability,
\displaystyle P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{P(A)}{P(B)}\qquad ...(1)
Since \displaystyle P(B)\leq1 and \displaystyle P(B)\neq0,
\displaystyle \frac{1}{P(B)}\geq1
Since \displaystyle P(A)\geq0, multiplying both sides by \displaystyle P(A) gives
\displaystyle \frac{P(A)}{P(B)}\geq P(A)\qquad ...(2)
Therefore, from (1) and (2), we get
\displaystyle P(A\mid B)=\frac{P(A)}{P(B)}\geq P(A)
✅️ Hence, the correct answer is (C).
Common Mistakes to Avoid
- Confusing Total Probability with Bayes’ Theorem: Use the Theorem of Total Probability when finding the probability of an event from different possible cases. Use Bayes’ Theorem when the event has already occurred and you need to find the probability of a particular case.
- Confusing conditional probabilities: Do not confuse \displaystyle P(A\mid B) with \displaystyle P(B\mid A). They are generally not equal. Always read the question carefully to identify which event is given and which event is to be found.
- Not identifying the cases correctly: Before applying Bayes’ Theorem, clearly identify the possible cases or causes \displaystyle E_1,E_2,\ldots,E_n and the given event A. This makes it much easier to choose the correct probabilities.
- Forgetting the prior probabilities: In Bayes’ Theorem, do not use only the conditional probabilities. The formula also includes the probabilities of the different cases, such as \displaystyle P(E_i).
- Using the wrong denominator in Bayes’ Theorem: The denominator represents the total probability of the given event. For a partition \displaystyle E_1,E_2,\ldots,E_n, use \displaystyle P(A)=\sum_{j=1}^{n}P(E_j)P(A\mid E_j).
- Forgetting that the cases form a partition: In questions based on Bayes’ Theorem, make sure the possible cases are mutually exclusive and together cover the entire sample space. Thus, \displaystyle E_i\cap E_j=\varnothing for \displaystyle i\neq j and \displaystyle E_1\cup E_2\cup\cdots\cup E_n=S.
- Using the wrong direction in the formula: If the question says “given that A has occurred, find the probability that it came from Ei”, the required probability is \displaystyle P(E_i\mid A), not \displaystyle P(A\mid E_i).
- Ignoring the different cases: When an event can occur through several possible cases, do not calculate its probability using only one case. For example, \displaystyle P(A)=P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)+\cdots.
- Mixing up “cause” and “result”: In a typical Bayes’ Theorem question, the cause or case is known before the experiment, while the result or event is observed afterwards. Bayes’ Theorem works backwards from the observed result to the likely cause.
- Converting percentages incorrectly: When probabilities are given as percentages, convert them carefully before substituting. For example, \displaystyle 0.5\%=0.005, while \displaystyle 5\%=0.05.
- Skipping event definitions: In application-based questions, do not jump directly into the formula. Clearly define the events first. For example, let A be the event that the item is defective and E1, E2 be the events that it came from two different machines.
Continue Learning
With the completion of Probability 13.3, you have now covered the important ideas of the Theorem of Total Probability and Bayes’ Theorem. In this exercise, you learned how to divide a problem into different possible cases, find the probability of an event through these cases and use Bayes’ Theorem to work backwards from a known event to the most likely case or cause.
Before moving on to the Miscellaneous Exercise, make sure you are comfortable with the following concepts, as they will be used frequently in the questions that follow:
- Understanding how events E1, E2, …, En can form a partition of the sample space.
- Using the Theorem of Total Probability: \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 has already 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).
- Identifying the prior probability, the conditional probability and the posterior probability in a Bayes’ Theorem problem.
- Understanding the difference between the forward direction of total probability and the reverse direction of Bayes’ Theorem.
- Converting percentages and real-life information into probabilities carefully before applying the formulas.
- Defining events clearly before starting the calculation, especially in application-based questions involving bags, machines, tests, cards and other real-life situations.
These ideas complete the main concepts of Probability covered in this chapter. The upcoming Miscellaneous Exercise brings together conditional probability, multiplication theorem, independence, total probability and Bayes’ Theorem, giving you an opportunity to apply everything you have learned so far.
Explore More
A strong understanding of the Theorem of Total Probability and Bayes’ Theorem forms the foundation for solving many probability problems. While solving questions, first identify the possible cases and the given event. If you need to find the probability of an event through different cases, use \displaystyle P(A)=\sum_{j=1}^{n}P(E_j)P(A\mid E_j). If the event has already occurred and you need to find the probability of a particular case, use Bayes’ Theorem.
While solving questions, carefully analyse the given information, especially in application-based problems involving bags, machines, tests, cards, coins and other real-life situations. Pay particular attention to the direction of the conditional probability and do not confuse \displaystyle P(A\mid E_i) with \displaystyle P(E_i\mid A). Always define the events clearly before applying the formula. With regular practice, you will be able to recognise Total Probability and Bayes’ Theorem questions quickly and apply the appropriate probability rule accurately and confidently as you move into the Miscellaneous Exercise.
All the best and keep learning 👍
