Probability: Chapter 13 Links
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Probability 13.1 NCERT Solutions focuses on the important concept of Conditional Probability — finding the probability of an event when we already have some information about the occurrence of another event.
\displaystyle P(E\mid F)=\frac{P(E\cap F)}{P(F)},\qquad P(F)\neq 0
The questions in this exercise cover the basic formula of conditional probability and its applications to different situations involving events, coins, dice, cards, family arrangements and everyday examples. Some questions also require us to find probabilities of intersections and unions before applying the conditional probability formula.
As you move through the exercise, you will see how the information given in a question changes the sample space and helps us calculate the required probability. The exercise also includes a few mixed and application-based questions along with MCQs, making it a useful practice set for strengthening the concept of Conditional Probability.
By solving these questions step by step, you will build a strong foundation for the concepts that follow in this chapter, while also improving your confidence for the CBSE Class 12 Board Exams, CUET, and other competitive examinations.
Key Concepts
Before solving Exercise 13.1, quickly revise the important concepts and formulas related to Conditional Probability. Most questions in this exercise require you to identify the given event carefully and then apply the appropriate probability formula.
1. Conditional Probability
If E and F are two events associated with the same sample space, then the probability of E given that F has already occurred is called the conditional probability of E given F.
It is denoted by \displaystyle P(E\mid F) and 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: The event after the vertical bar is the given event. It determines the new sample space for the conditional probability.
2. Conditional Probability Using Number of Outcomes
When the elementary events are equally likely, conditional probability can also be understood as
\displaystyle P(E\mid F)=\frac{n(E\cap F)}{n(F)}
In other words, once F has occurred, consider only the outcomes favourable to F. Out of these outcomes, count those which are also favourable to E.
3. Multiplication Rule of Probability
From the formula of conditional probability,
\displaystyle P(E\cap F)=P(F)\,P(E\mid F)
Similarly,
\displaystyle P(E\cap F)=P(E)\,P(F\mid E)
Therefore,
\displaystyle P(E)\,P(F\mid E)=P(F)\,P(E\mid F)
4. Conditional Probability of Union
For any two events A and B, provided \displaystyle P(F)\neq0,
\displaystyle P((A\cup B)\mid F)=P(A\mid F)+P(B\mid F)-P((A\cap B)\mid F)
If A and B are disjoint events, then \displaystyle A\cap B=\varnothing, so
\displaystyle P((A\cup B)\mid F)=P(A\mid F)+P(B\mid F)
5. Complementary Events
If E' is the complement of event E, then
\displaystyle P(E'\mid F)=1-P(E\mid F)
This is especially useful when the question asks for the probability of an event such as “at least one”, “not” or “none”.
6. Important Points
- In \displaystyle P(E\mid F), F is the given event and E is the event whose probability is required.
- When F has already occurred, consider only the outcomes belonging to F while finding \displaystyle P(E\mid F).
- The formula \displaystyle P(E\mid F)=\frac{P(E\cap F)}{P(F)} is valid only when \displaystyle P(F)\neq0.
- Do not confuse \displaystyle P(E\mid F) with \displaystyle P(F\mid E). They are generally different.
- For any event F with \displaystyle P(F)\neq0, \displaystyle P(S\mid F)=P(F\mid F)=1.
- Always identify the given information first. It tells you which event should appear after the vertical bar.
Quick Reference Table
| Concept | Formula / Result |
|---|---|
| Conditional Probability | \displaystyle P(E\mid F)=\frac{P(E\cap F)}{P(F)},\qquad P(F)\neq0 |
| Reverse Conditional Probability | \displaystyle P(F\mid E)=\frac{P(E\cap F)}{P(E)},\qquad P(E)\neq0 |
| Equally Likely Outcomes | \displaystyle P(E\mid F)=\frac{n(E\cap F)}{n(F)} |
| Multiplication Rule | \displaystyle P(E\cap F)=P(F)P(E\mid F)=P(E)P(F\mid E) |
| Conditional Probability of Union | \displaystyle P((A\cup B)\mid F)=P(A\mid F)+P(B\mid F)-P((A\cap B)\mid F) |
| Disjoint Events | \displaystyle P((A\cup B)\mid F)=P(A\mid F)+P(B\mid F) |
| Complement | \displaystyle P(E'\mid F)=1-P(E\mid F) |
| Sure Event | \displaystyle P(S\mid F)=1 |
| Same Event | \displaystyle P(F\mid F)=1,\qquad P(F)\neq0 |
| Important Condition | \displaystyle P(E\mid F) is defined only when \displaystyle P(F)\neq0 |
Before You Begin… Before solving a question, first identify the given event and the required event. Pay special attention to the phrase “given that”, as it determines the relevant sample space. Once you identify the condition clearly, choosing the correct formula and method becomes much easier.
Let us now solve all the NCERT questions of Probability 13.1 step by step.
Question 1: Conditional Probability
1. Given that E and F are events such that \displaystyle P(E)=0.6,\; P(F)=0.3 and \displaystyle P(E\cap F)=0.2, find \displaystyle P(E\mid F) and \displaystyle P(F\mid E)
Solution:
Think First… We need to find two conditional probabilities. In each case, the event after the vertical bar is the given event.
We are given
\displaystyle P(E)=0.6,\; P(F)=0.3,\; P(E\cap F)=0.2
Using the formula for conditional probability,
\displaystyle P(E\mid F)=\frac{P(E\cap F)}{P(F)},\quad P(F)\neq0
Substituting the given values,
\displaystyle P(E\mid F)=\frac{0.2}{0.3}=\frac{2}{3}
Now, using the same formula for \displaystyle P(F\mid E)
\displaystyle P(F\mid E)=\frac{P(E\cap F)}{P(E)},\quad P(E)\neq0
Substituting the given values, we get
\displaystyle P(F\mid E)=\frac{0.2}{0.6}=\frac{1}{3}
Therefore,
\displaystyle \boxed{P(E\mid F)=\frac{2}{3}}
and
\displaystyle \boxed{P(F\mid E)=\frac{1}{3}}
Question 2: Probability 13.1
2. Compute \displaystyle P(A\mid B), if \displaystyle P(B)=0.5 and \displaystyle P(A\cap B)=0.32
Solution:
Think First… Since we need to find \displaystyle P(A\mid B), the event B is the given event. We can directly use the formula for conditional probability.
We are given
\displaystyle P(B)=0.5,\quad P(A\cap B)=0.32
Using the formula for conditional probability,
\displaystyle P(A\mid B)=\frac{P(A\cap B)}{P(B)},\quad P(B)\neq0
Substituting the given values,
\displaystyle P(A\mid B)=\frac{0.32}{0.5}
Converting the decimals into fractions,
\displaystyle P(A\mid B)=\frac{32}{50}=\frac{16}{25}
Therefore,
\displaystyle \boxed{P(A\mid B)=\frac{16}{25}}
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 Probability 13.1 exercise introduces the important concept of Conditional Probability. Through these questions, you will learn how additional information affects the probability of an event and how to apply conditional probability to different types of experiments involving coins, dice, arrangements and other real-life situations. These questions provide valuable practice for the CBSE Board Examination and CUET, while also building the foundation for the remaining concepts of the Probability chapter. Now, let’s continue with the next question.
Question 3: Probability Exercise 13.1
3. If \displaystyle P(A)=0.8,\; P(B)=0.5 and \displaystyle P(B\mid A)=0.4, find
(i) \displaystyle P(A\cap B) (ii) \displaystyle P(A\mid B) (iii) \displaystyle P(A\cup B)
Solution:
Think First… We are given \displaystyle P(B\mid A). So, the multiplication rule can first be used to find \displaystyle P(A\cap B). Once the intersection is known, we can find the other two required probabilities.
We are given
\displaystyle P(A)=0.8,\; P(B)=0.5,\; P(B\mid A)=0.4
(i) Finding \displaystyle P(A\cap B)
Using the multiplication rule of probability,
\displaystyle P(A\cap B)=P(A)\,P(B\mid A)
Substituting the given values,
\displaystyle P(A\cap B)=0.8\times0.4=0.32
Therefore,
\displaystyle \boxed{P(A\cap B)=0.32}
(ii) Finding \displaystyle P(A\mid B)
Using the formula for conditional probability,
\displaystyle P(A\mid B)=\frac{P(A\cap B)}{P(B)}
We already have
\displaystyle P(A\cap B)=0.32
Therefore,
\displaystyle P(A\mid B)=\frac{0.32}{0.5}=0.64
Therefore,
\displaystyle \boxed{P(A\mid B)=0.64}
(iii) Finding \displaystyle P(A\cup B)
Using the addition rule of probability,
\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)
Substituting the known values, we get
\displaystyle P(A\cup B)=0.8+0.5-0.32
\displaystyle P(A\cup B)=0.98
Hence,
\displaystyle \boxed{P(A\cup B)=0.98}
Question 4: Probability 13.1
4. Evaluate \displaystyle P(A\cup B), if \displaystyle 2P(A)=P(B)=\frac{5}{13} and \displaystyle P(A\mid B)=\frac25
Solution:
Think First… First, use the given relation to find \displaystyle P(A) and \displaystyle P(B). Then use the conditional probability formula to find \displaystyle P(A\cap B). Finally, apply the addition rule of probability.
We are given
\displaystyle 2P(A)=P(B)=\frac{5}{13},\quad P(A\mid B)=\frac25
From \displaystyle 2P(A)=\frac{5}{13}
We get, \displaystyle P(A)=\frac{5}{26}
Also, \displaystyle P(B)=\frac{5}{13}
Using the formula for conditional probability,
\displaystyle P(A\mid B)=\frac{P(A\cap B)}{P(B)}
Therefore,
\displaystyle P(A\cap B)=P(B)\,P(A\mid B)
Substituting the given values,
\displaystyle P(A\cap B)=\frac{5}{13}\times\frac25=\frac{2}{13}
Now, using the addition rule of probability,
\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)
Substituting the values,
\displaystyle P(A\cup B)=\frac{5}{26}+\frac{5}{13}-\frac{2}{13}
Taking the LCM as 26, we get
\displaystyle P(A\cup B)=\frac{5}{26}+\frac{10}{26}-\frac{4}{26}=\frac{11}{26}
Hence,
\displaystyle \boxed{P(A\cup B)=\frac{11}{26}}
Maths Better Tip… In conditional probability, the words “given that” are the key. First identify the given event and consider only the outcomes satisfying that condition. Then find the favourable outcomes and calculate the required probability. In short, the given condition changes the sample space you need to consider.
Question 5: Probability 13.1
5. If \displaystyle P(A)=\frac{6}{11},\; P(B)=\frac{5}{11} and \displaystyle P(A\cup B)=\frac{7}{11}, find
(i) \displaystyle P(A\cap B) (ii) \displaystyle P(A\mid B) (iii) \displaystyle P(B\mid A)
Solution:
Think First… We are given the probabilities of A, B and their union. So, we can first use the addition rule to find \displaystyle P(A\cap B). We can then use this value to find the two required conditional probabilities.
We are given
\displaystyle P(A)=\frac{6}{11},\; P(B)=\frac{5}{11},\; P(A\cup B)=\frac{7}{11}
(i) Finding \displaystyle P(A\cap B)
Using the addition rule of probability,
\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)
Rearranging, we have
\displaystyle P(A\cap B)=P(A)+P(B)-P(A\cup B)
Substituting the given values, we get
\displaystyle P(A\cap B)=\frac{6}{11}+\frac{5}{11}-\frac{7}{11}=\frac{4}{11}
Therefore,
\displaystyle \boxed{P(A\cap B)=\frac{4}{11}}
(ii) Finding \displaystyle P(A\mid B)
Using the formula for conditional probability,
\displaystyle P(A\mid B)=\frac{P(A\cap B)}{P(B)}
Substituting the values,
\displaystyle P(A\mid B)=\frac{\frac{4}{11}}{\frac{5}{11}}=\frac{4}{5}
Hence,
\displaystyle \boxed{P(A\mid B)=\frac{4}{5}}
(iii) Finding \displaystyle P(B\mid A)
Using the formula for conditional probability,
\displaystyle P(B\mid A)=\frac{P(A\cap B)}{P(A)}
Substituting the values, we get
\displaystyle P(B\mid A)=\frac{\frac{4}{11}}{\frac{6}{11}}=\frac{2}{3}
Hence,
\displaystyle \boxed{P(B\mid A)=\frac{2}{3}}
Question 6: Exercise 13.1
6. A coin is tossed three times, where
- (i) \displaystyle E: head on third toss, \displaystyle F: heads on first two tosses
- (ii) \displaystyle E: at least two heads, \displaystyle F: at most two heads
- (iii) \displaystyle E: at most two tails, \displaystyle F: at least one tail
Determine \displaystyle P(E\mid F)
Solution:
Think First… In each case, the event after the vertical bar is the given event. Since the coin is fair and is tossed three times, there are \displaystyle 2^3=8 equally likely outcomes.
Using the formula for conditional probability,
\displaystyle P(E\mid F)=\frac{P(E\cap F)}{P(F)}
(i) E: head on third toss, F: heads on first two tosses
The sample space is
\displaystyle \begin{aligned} S=\{&HHH,HHT,HTH,HTT,\\ &THH,THT,TTH,TTT\} \end{aligned}
The event F means that the first two tosses are heads. Therefore,
\displaystyle F=\{HHH,HHT\}
The event E means that the third toss is a head. Hence,
\displaystyle E=\{HHH,HTH,THH,TTH\}
Therefore, the common outcomes are
\displaystyle E\cap F=\{HHH\}
Thus, \displaystyle P(F)=\frac{2}{8}=\frac14
and \displaystyle P(E\cap F)=\frac18
\displaystyle \therefore P(E\mid F)=\frac{\frac18}{\frac14}=\frac12
Hence,
\displaystyle \boxed{P(E\mid F)=\frac12}
(ii) E: at least two heads, F: at most two heads
The event E means two or three heads. Therefore,
\displaystyle E=\{HHT,HTH,THH,HHH\}
The event F means zero, one or two heads. Thus,
\displaystyle F=\{HTT,THT,TTH,HTH,THH,HHT,TTT\}
The outcomes common to both events are those with exactly two heads:
\displaystyle E\cap F=\{HHT,HTH,THH\}
Therefore,
\displaystyle P(F)=\frac78,\quad P(E\cap F)=\frac38
So, \displaystyle P(E\mid F)=\frac{\frac38}{\frac78}=\frac37
Hence,
\displaystyle \boxed{P(E\mid F)=\frac37}
(iii) E: at most two tails, F: at least one tail
The event E means zero, one or two tails. Thus, it contains all outcomes except \displaystyle TTT
Therefore,
\displaystyle E=\{HHH,HHT,HTH,THH,HTT,THT,TTH\}
The event F means at least one tail. Hence,
\displaystyle F=\{HHT,HTH,THH,HTT,THT,TTH,TTT\}
The outcomes common to E and F are all the outcomes having one or two tails:
\displaystyle E\cap F=\{HHT,HTH,THH,HTT,THT,TTH\}
Therefore,
\displaystyle P(F)=\frac78,\quad P(E\cap F)=\frac68=\frac34
Thus, \displaystyle P(E\mid F)=\frac{\frac34}{\frac78}=\frac67
Hence,
\displaystyle \boxed{P(E\mid F)=\frac67}
Question 7: Probability 13.1
7. Two coins are tossed once, where
- (i) \displaystyle E: tail appears on one coin, \displaystyle F: one coin shows head
- (ii) \displaystyle E: no tail appears, \displaystyle F: no head appears
Determine \displaystyle P(E\mid F)
Solution:
Think First… For two coins, the sample space consists of four equally likely outcomes. We first identify the events E and F in each case and then use the formula for conditional probability.
The sample space is
\displaystyle S=\{HH,HT,TH,TT\}
(i) E: tail appears on one coin, F: one coin shows head
The event E means that exactly one tail appears. Therefore,
\displaystyle E=\{HT,TH\}
The event F means that exactly one head appears. Hence,
\displaystyle F=\{HT,TH\}
Thus, \displaystyle E=F
Therefore,
\displaystyle \therefore E\cap F=\{HT,TH\}
Since all four outcomes are equally likely,
\displaystyle P(F)=\frac{2}{4}=\frac12
and
\displaystyle P(E\cap F)=\frac{2}{4}=\frac12
Using the formula for conditional probability,
\displaystyle P(E\mid F)=\frac{P(E\cap F)}{P(F)}
Substituting the values, we get
\displaystyle P(E\mid F)=\frac{\frac12}{\frac12}=1
Hence,
\displaystyle \boxed{P(E\mid F)=1}
(i) E: no tail appears, F: no head appears
The event E means that both coins show heads. Therefore,
\displaystyle E=\{HH\}
The event F means that both coins show tails. Hence,
\displaystyle F=\{TT\}
There is no common outcome in the two events. Thus,
\displaystyle E\cap F=\varnothing
Therefore,
\displaystyle P(E\cap F)=0
Also, \displaystyle P(E)=\frac14,\quad P(F)=\frac14
Using the formula for conditional probability,
\displaystyle P(E\mid F)=\frac{P(E\cap F)}{P(F)}=\frac{0}{\frac14}=0
Hence,
\displaystyle \boxed{P(E\mid F)=0}
Maths Better Tip… Always check the denominator before using the conditional probability formula. In \displaystyle P(A\mid B)=\frac{P(A\cap B)}{P(B)}, the given event is B, so \displaystyle P(B) must be non-zero. If the given event has probability zero, the conditional probability is not defined.
Question 8: Probability Ex. 13.1
8. A die is thrown three times,
\displaystyle E: 4 appears on the third toss, \displaystyle F: 6 and 5 appear respectively on the first two tosses.
Determine \displaystyle P(E\mid F)
Solution:
Think First… The condition F tells us that the first toss is 6 and the second toss is 5. Therefore, once F has occurred, only the result of the third toss is still unknown. We need to find the probability that this third toss is 4.
We know, the sample space for three throws of a die has
\displaystyle 6^3=216
equally likely outcomes.
The event F means that the first two tosses are fixed as 6 and 5. The third toss can be any of the six possible numbers.
Therefore,
\displaystyle \begin{aligned} F=\{&(6,5,1),(6,5,2),(6,5,3),\\ &(6,5,4),(6,5,5),(6,5,6)\} \end{aligned}
Thus,
\displaystyle P(F)=\frac{6}{216}=\frac{1}{36}
The event E means that 4 appears on the third toss. For both E and F to occur, the outcome must be
\displaystyle E\cap F=\{(6,5,4)\}
Hence,
\displaystyle P(E\cap F)=\frac{1}{216}
Using the formula for conditional probability,
\displaystyle P(E\mid F)=\frac{P(E\cap F)}{P(F)}
Substituting the values, we get
\displaystyle P(E\mid F)=\frac{\frac{1}{216}}{\frac{6}{216}}=\frac16
Hence,
\displaystyle \boxed{P(E\mid F)=\frac16}
Question 9: Probability 13.1
9. Mother, father and son line up at random for a family picture, where
\displaystyle E: son on one end, \displaystyle F: father in middle.
Determine \displaystyle P(E\mid F)
Solution:
Think First… If the father is in the middle, the mother and son must occupy the two ends. Therefore, whenever F occurs, E also occurs. We can verify this using the formula for conditional probability.
There are three people: mother (M), father (F) and son (S).
The total number of possible arrangements is
\displaystyle 3!=6
So, the sample space consists of the six arrangements
\displaystyle S=\{MFS,MSF,FMS,FSM,SMF,SFM\}
The event F means that the father is in the middle. Therefore,
\displaystyle F=\{MFS,SFM\}
In both these arrangements, the son is at one end. Hence,
\displaystyle E\cap F=\{MFS,SFM\}
Thus, \displaystyle P(F)=\frac{2}{6}=\frac13
and \displaystyle P(E\cap F)=\frac{2}{6}=\frac13
Using the formula for conditional probability,
\displaystyle P(E\mid F)=\frac{P(E\cap F)}{P(F)}
Substituting the values, we get
\displaystyle P(E\mid F)=\frac{\frac13}{\frac13}=1
Hence,
\displaystyle \boxed{P(E\mid F)=1}
Question 10: Probability 13.1
10. A black and a red dice are rolled.
(a) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.
(b) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.
Solution:
Think First… In each part, the information given after “given that” restricts the possible outcomes. We should first consider only those outcomes satisfying the given condition and then find the favourable outcomes.
(a) Black die resulted in a 5
Let
\displaystyle E: the sum of the numbers on the two dice is greater than 9
and
\displaystyle F: the black die resulted in 5.
Since the black die has resulted in 5, the red die can show any of the six numbers. Thus, the possible outcomes under the condition F are
\displaystyle F=\{(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\}
For the sum to be greater than 9, with the black die showing 5, the red die must show a number greater than 4.
Therefore,
\displaystyle E\cap F=\{(5,5),(5,6)\}
Hence,
\displaystyle P(F)=\frac{6}{36}=\frac16
and
\displaystyle P(E\cap F)=\frac{2}{36}=\frac1{18}
Using the formula for conditional probability,
\displaystyle P(E\mid F)=\frac{P(E\cap F)}{P(F)}
Substituting the values,
\displaystyle P(E\mid F)=\frac{\frac1{18}}{\frac16}=\frac13
Hence,
\displaystyle \boxed{P(E\mid F)=\frac13}
(i) Red die resulted in a number less than 4
Let
\displaystyle E: the sum of the numbers on the two dice is 8
and
\displaystyle F: the red die resulted in a number less than 4.
The red die can show 1, 2 or 3. Therefore, the possible outcomes under the condition F are
\displaystyle \begin{aligned} F=\{&(1,1),(2,1),(3,1),(4,1),(5,1),(6,1),\\ &(1,2),(2,2),(3,2),(4,2),(5,2),(6,2),\\ &(1,3),(2,3),(3,3),(4,3),(5,3),(6,3)\} \end{aligned}
Thus, there are 18 outcomes satisfying the given condition.
For the sum to be 8, the corresponding black die values must be 7, 6 and 5 when the red die shows 1, 2 and 3 respectively. Since a die cannot show 7, only the following two outcomes are possible:
\displaystyle E\cap F=\{(6,2),(5,3)\}
Hence,
\displaystyle P(F)=\frac{18}{36}=\frac12
and
\displaystyle P(E\cap F)=\frac{2}{36}=\frac1{18}
Using the formula for conditional probability,
\displaystyle P(E\mid F)=\frac{P(E\cap F)}{P(F)}
Substituting the values,
\displaystyle P(E\mid F)=\frac{\frac1{18}}{\frac12}=\frac19
Hence,
\displaystyle \boxed{P(E\mid F)=\frac19}
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 13.1.
Question 11: Probability 13.1
11. A fair die is rolled. Consider events \displaystyle E=\{1,3,5\},\; F=\{2,3\} and \displaystyle G=\{2,3,4,5\}. Find
(i) \displaystyle P(E\mid F) and \displaystyle P(F\mid E)
(ii) \displaystyle P(E\mid G) and \displaystyle P(G\mid E)
(iii) \displaystyle P((E\cup F)\mid G) and \displaystyle P((E\cap F)\mid G)
Solution:
Think First… Since the die is fair, each of the six outcomes has probability \displaystyle \frac16. For each conditional probability, first find the required intersection and then use \displaystyle P(A\mid B)=\frac{P(A\cap B)}{P(B)}.
The sample space is
\displaystyle S=\{1,2,3,4,5,6\}
We are given
\displaystyle E=\{1,3,5\},\; F=\{2,3\},\; G=\{2,3,4,5\}
Using the formula for conditional probability,
\displaystyle P(A\mid B)=\frac{P(A\cap B)}{P(B)}
(i) Finding \displaystyle P(E\mid F) and \displaystyle P(F\mid E)
First,
\displaystyle E\cap F=\{3\}
Therefore,
\displaystyle P(E\cap F)=\frac16,\; P(F)=\frac26=\frac13,\quad P(E)=\frac36=\frac12
Hence,
\displaystyle P(E\mid F)=\frac{P(E\cap F)}{P(F)}=\frac{\frac16}{\frac26}=\frac12
Similarly,
\displaystyle P(F\mid E)=\frac{P(E\cap F)}{P(E)}=\frac{\frac16}{\frac36}=\frac13
Therefore,
\displaystyle \boxed{P(E\mid F)=\frac12;\; P(F\mid E)=\frac13}
(ii) Finding \displaystyle P(E\mid G) and \displaystyle P(G\mid E)
Now,
\displaystyle E\cap G=\{3,5\}
Therefore,
\displaystyle P(E\cap G)=\frac26=\frac13,\; P(G)=\frac46=\frac23,\quad P(E)=\frac36=\frac12
Using the formula for conditional probability,
\displaystyle P(E\mid G)=\frac{P(E\cap G)}{P(G)}=\frac{\frac13}{\frac23}=\frac12
Similarly,
\displaystyle P(G\mid E)=\frac{P(E\cap G)}{P(E)}=\frac{\frac13}{\frac12}=\frac23
Therefore,
\displaystyle \boxed{P(E\mid G)=\frac12:\; P(G\mid E)=\frac23}
(iii) Finding \displaystyle P((E\cup F)\mid G) and \displaystyle P((E\cap F)\mid G)
First, find the union of E and F:
\displaystyle E\cup F=\{1,2,3,5\}
Now,
\displaystyle (E\cup F)\cap G=\{2,3,5\}
Therefore,
\displaystyle P((E\cup F)\cap G)=\frac36=\frac12
Using the formula for conditional probability,
\displaystyle P((E\cup F)\mid G)=\frac{P((E\cup F)\cap G)}{P(G)}
Substituting the values,
\displaystyle P((E\cup F)\mid G)=\frac{\frac12}{\frac23}=\frac34
Next,
\displaystyle E\cap F=\{3\}
Hence,
\displaystyle (E\cap F)\cap G=\{3\}
Therefore,
\displaystyle P((E\cap F)\cap G)=\frac16
Using the formula for conditional probability,
\displaystyle P((E\cap F)\mid G)=\frac{P((E\cap F)\cap G)}{P(G)}
Thus,
\displaystyle P((E\cap F)\mid G)=\frac{\frac16}{\frac23}=\frac14
Hence,
\displaystyle \boxed{P((E\cup F)\mid G)=\frac34;\; P((E\cap F)\mid G)=\frac14}
Question 12: Probability
12. Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that
- (i) the youngest is a girl,
- (ii) at least one is a girl?
Solution:
Think First… The order of the children matters here because the question refers to the youngest child. So, we must distinguish between the older and younger child when writing the sample space.
Let b denote a boy and g denote a girl.
Since each child is equally likely to be a boy or a girl, the sample space for two children is
\displaystyle S=\{(b,b),(b,g),(g,b),(g,g)\}
Each of these four outcomes is equally likely.
(i) Given that the youngest is a girl
Let E be the event that both children are girls.
Then,
\displaystyle E=\{(g,g)\}
Let F be the event that the youngest child is a girl.
Since the second entry represents the youngest child,
\displaystyle F=\{(b,g),(g,g)\}
Therefore,
\displaystyle E\cap F=\{(g,g)\}
Thus, \displaystyle P(F)=\frac24=\frac12
and \displaystyle P(E\cap F)=\frac14
Using the formula for conditional probability,
\displaystyle P(E\mid F)=\frac{P(E\cap F)}{P(F)}
Substituting the values,
\displaystyle P(E\mid F)=\frac{\frac14}{\frac12}=\frac12
Hence,
\displaystyle \boxed{P(E\mid F)=\frac12}
(i) Given that at least one child is a girl
Again, let E be the event that both children are girls.
Thus, \displaystyle E=\{(g,g)\}
Let F be the event that at least one child is a girl.
The only outcome in which there is no girl is \displaystyle (b,b). Therefore,
\displaystyle F=\{(b,g),(g,b),(g,g)\}
Therefore,
\displaystyle E\cap F=\{(g,g)\}
Thus, \displaystyle P(F)=\frac34
and \displaystyle P(E\cap F)=\frac14
Using the formula for conditional probability,
\displaystyle P(E\mid F)=\frac{P(E\cap F)}{P(F)}
Substituting the values, we get
\displaystyle P(E\mid F)=\frac{\frac14}{\frac34}=\frac13
Hence,
\displaystyle \boxed{P(E\mid F)=\frac13}
Question 13: Probability Exercise 13.1
13. An instructor has a question bank consisting of 300 easy True / False questions, 200 difficult True / False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple choice question?
Solution:
Think First… The phrase “given that it is a multiple choice question” is the key. Once we know that the selected question is an MCQ, the True / False questions are no longer part of the relevant sample space. We only need to compare the number of easy MCQs with the total number of MCQs.
Let \displaystyle E = the event that the selected question is an easy question
and \displaystyle F = the event that the selected question is a multiple choice question.
We are required to find
\displaystyle P(E\mid F)
Using the formula for conditional probability,
\displaystyle P(E\mid F)=\frac{P(E\cap F)}{P(F)}
There are 500 easy multiple choice questions and 400 difficult multiple choice questions.
Therefore, the total number of multiple choice questions is
\displaystyle 500+400=900
The event \displaystyle E\cap F means that the selected question is both easy and a multiple choice question. There are 500 such questions.
Thus, out of the 900 multiple choice questions, 500 are easy.
Therefore,
\displaystyle P(E\mid F)=\frac{500}{900}=\frac59
Hence, the required probability is
\displaystyle \boxed{P(E\mid F)=\frac59}
Question 14: Conditional Probability
14. Given that the two numbers appearing on throwing two dice are different. Find the probability of the event ‘the sum of numbers on the dice is 4’.
Solution:
Think First… The phrase “given that the two numbers are different” tells us to exclude all outcomes in which both dice show the same number. We then find the probability of getting a sum of 4 among the remaining outcomes.
Let \displaystyle E = the event that the sum of the numbers on the two dice is 4
and \displaystyle F = the event that the numbers appearing on the two dice are different.
We have to find
\displaystyle P(E\mid F)
Using the formula for conditional probability,
\displaystyle P(E\mid F)=\frac{P(E\cap F)}{P(F)}
When two dice are thrown, the total number of equally likely outcomes is
\displaystyle 6\times6=36
The event F means that the numbers on the two dice are different. The outcomes with equal numbers are
\displaystyle \{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\}
There are 6 such outcomes. Therefore, the number of outcomes favourable to F is
\displaystyle 36-6=30
Hence,
\displaystyle P(F)=\frac{30}{36}=\frac56
Now, for the sum to be 4, the possible outcomes are
\displaystyle E=\{(1,3),(2,2),(3,1)\}
But the condition F requires the two numbers to be different. Therefore, the outcome \displaystyle (2,2) is not allowed.
Thus, \displaystyle E\cap F=\{(1,3),(3,1)\}
Therefore,
\displaystyle P(E\cap F)=\frac{2}{36}=\frac1{18}
Using the formula for conditional probability,
\displaystyle P(E\mid F)=\frac{\frac1{18}}{\frac56}=\frac1{18}\times\frac65=\frac1{15}
Therefore, the required probability is
\displaystyle \boxed{P(E\mid F)=\frac1{15}}
Question 15: Probability 13.1
15. Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’.
Solution:
Think First… The multiples of 3 on a die are \displaystyle 3 and \displaystyle 6. If the first throw gives 3 or 6, the die is thrown one more time. If the first throw gives 1, 2, 4 or 5, a coin is tossed. The experiment ends after this second die throw or the coin toss.
Let \displaystyle E = the event that the coin shows a tail
and \displaystyle F = the event that at least one die shows a 3.
We have to find
\displaystyle P(E\mid F)
Using the formula for conditional probability,
\displaystyle P(E\mid F)=\frac{P(E\cap F)}{P(F)}
Let us first write the sample space of the experiment.
If the first throw gives 1, 2, 4 or 5, a coin is tossed. Thus, we get the outcomes
\displaystyle (1,H),(1,T),(2,H),(2,T)\\[2pt](4,H),(4,T),(5,H),(5,T)
If the first throw gives 3 or 6, the die is thrown once more. Hence, the corresponding outcomes are
\displaystyle (3,1),(3,2),(3,3),(3,4),(3,5),(3,6)
and
\displaystyle (6,1),(6,2),(6,3),(6,4),(6,5),(6,6)
Therefore, the sample space contains
\displaystyle 8+6+6=20
equally likely outcomes.
Now, event E is the event that the coin shows a tail. Therefore,
\displaystyle E=\{(1,T),(2,T),(4,T),(5,T)\}
Event F is the event that at least one die shows 3.
The first die shows 3 in the six outcomes
\displaystyle (3,1),(3,2),(3,3),(3,4),(3,5),(3,6)
In addition, the first die can show 6 and the second die can show 3, giving the outcome \displaystyle(6,3).
Hence,
\displaystyle F=\{(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(6,3)\}
Clearly, none of the outcomes in E is present in F. Therefore, the two events are mutually exclusive:
\displaystyle E\cap F=\varnothing
Thus, \displaystyle P(E\cap F)=0
Also, there are 7 outcomes favourable to F out of 20 equally likely outcomes. Hence,
\displaystyle P(F)=\frac7{20}
Therefore,
\displaystyle P(E\mid F)=\frac{P(E\cap F)}{P(F)} =\frac{0}{\frac7{20}}=0
Hence, the required conditional probability is
\displaystyle \boxed{P(E\mid F)=0}
Maths Better Tip! Remember, the probability of any event always lies between 0 and 1. An impossible event has probability 0, while a certain event has probability 1. Therefore, a probability can never be negative or greater than 1.
Question 16: Probability 13.1 – MCQ
16. If \displaystyle P(A)=\frac12,\quad P(B)=0, then \displaystyle P(A\mid B) is
- (A) 0
- (B) \displaystyle\frac12
- (C) not defined
- (D) 1
Solution
Think First… The formula for conditional probability has \displaystyle P(B) in the denominator. Therefore, \displaystyle P(B)\ne0 is necessary for \displaystyle P(A\mid B) to be defined.
Using the formula for conditional probability,
\displaystyle P(A\mid B)=\frac{P(A\cap B)}{P(B)},\quad P(B)\ne0
Here, we are given
\displaystyle P(B)=0
Therefore, the denominator in the formula becomes zero. Hence, \displaystyle P(A\mid B) is not defined.
The value of \displaystyle P(A)=\frac12 does not change this conclusion.
✅️ Hence, the correct answer is (C).
Question 17: Probability 13.1 – MCQ
17. If \displaystyle P(A\mid B)=P(B\mid A) , then
- (A) \displaystyle A\subset B but \displaystyle A\ne B
- (B) \displaystyle A=B
- (C) \displaystyle A\cap B=\varnothing
- (D) \displaystyle P(A)=P(B)
Solution
Think First… Write both conditional probabilities using the formula \displaystyle P(A\mid B)=\frac{P(A\cap B)}{P(B)} and compare them.
Given that
\displaystyle P(A\mid B)=P(B\mid A)
Using the formula for conditional probability,
\displaystyle P(A\mid B)=\frac{P(A\cap B)}{P(B)}
and
\displaystyle P(B\mid A)=\frac{P(A\cap B)}{P(A)}
Therefore,
\displaystyle \frac{P(A\cap B)}{P(B)} = \frac{P(A\cap B)}{P(A)}
For the usual non-zero intersection case, \displaystyle P(A\cap B)\ne0, so we can cancel \displaystyle P(A\cap B) from both sides.
Thus,
\displaystyle \frac1{P(B)}=\frac1{P(A)}
Hence,
\displaystyle P(A)=P(B)
✅️ Hence, the correct answer is (D).
Common Mistakes to Avoid
- Using the wrong denominator: In \displaystyle P(E\mid F), the given event F determines the relevant sample space. Always use \displaystyle P(F) in the denominator.
- Forgetting the condition: When a question says “given that”, first identify the given event and restrict your attention to outcomes satisfying that condition before finding the required probability.
- Confusing \displaystyle P(E\mid F) with \displaystyle P(F\mid E): These are generally different probabilities. Read carefully which event is being given and which event is being found.
- Using the formula without checking the condition: The formula \displaystyle P(E\mid F)=\frac{P(E\cap F)}{P(F)} is valid only when \displaystyle P(F)\ne0. If \displaystyle P(F)=0, the conditional probability is not defined.
- Finding the intersection incorrectly: The event \displaystyle E\cap F means that both E and F occur. Carefully list the common outcomes when working with sample spaces.
- Keeping outcomes that violate the given condition: In questions involving dice, coins or other experiments, remove outcomes that do not satisfy the given event before counting favourable outcomes.
- Ignoring the order of outcomes: In experiments involving two or more dice or coins, outcomes such as \displaystyle(1,3) and \displaystyle(3,1) are different ordered outcomes when the experiment distinguishes the first and second results.
- Assuming the answer is always \displaystyle\frac12 for a fair coin: A fair coin has probability \displaystyle\frac12 of showing a tail only when the coin is actually tossed. Check the given condition and the sequence of the experiment first.
- Missing mutually exclusive events: If \displaystyle E\cap F=\varnothing, then \displaystyle P(E\cap F)=0. Hence, provided \displaystyle P(F)\ne0, \displaystyle P(E\mid F)=0.
- Canceling a common factor without checking whether it is zero: While simplifying conditional-probability equations, do not cancel \displaystyle P(E\cap F) unless it is known to be non-zero. This is especially important in conceptual MCQs.
- Stopping at the formula: Conditional-probability questions often involve a change in the sample space. Clearly explain what the given condition does to the possible outcomes before calculating the final probability.
Continue Learning
With the completion of Probability 13.1, you have now covered the basic ideas of Conditional Probability. In this exercise, you learned how additional information changes the probability of an event, how to use the conditional probability formula, and how to identify and work with events such as \displaystyle E\cap F under a given condition.
Before moving on to Exercise 13.2, make sure you are comfortable with the following concepts, as they will be used frequently in the next exercise:
- Understanding the meaning of conditional probability and interpreting the phrase “given that”.
- Using the formula \displaystyle P(E\mid F)=\frac{P(E\cap F)}{P(F)}, provided \displaystyle P(F)\ne0.
- Finding \displaystyle E\cap F correctly by identifying outcomes common to both events.
- Understanding how a given condition can reduce the relevant sample space.
- Distinguishing between \displaystyle P(E\mid F) and \displaystyle P(F\mid E).
- Recognising when two events are mutually exclusive and using \displaystyle E\cap F=\varnothing accordingly.
- Carefully analysing multi-step experiments involving coins, dice and other random experiments before applying a probability formula.
- Checking the given condition before counting favourable outcomes or applying conditional probability.
These ideas form the foundation for the next part of the chapter, where we will study Independent Events and understand when the occurrence of one event does not affect the probability of another event.
Explore More
A strong understanding of Conditional Probability forms the foundation for studying the concepts that follow in this chapter. While solving problems, first identify the given event and the required event, and then use the appropriate conditional probability formula. Pay particular attention to the meaning of “given that”, as it changes the set of outcomes that should be considered.
While solving questions, carefully identify the relevant sample space and common outcomes, especially in multi-step experiments involving coins, dice and other random experiments. Check whether the given condition makes two events mutually exclusive or affects the possible outcomes, and always verify that the denominator in a conditional probability is non-zero. With regular practice, you will be able to recognise conditional probability situations quickly and solve them accurately and confidently as you move through the remaining concepts of Probability.
All the best and keep learning 👍
