• Skip to main content
  • Skip to primary sidebar
  • Skip to footer
  • NCERT Solutions
    • NCERT Books Free Download
  • TS Grewal
    • TS Grewal Class 12 Accountancy Solutions
    • TS Grewal Class 11 Accountancy Solutions
  • CBSE Sample Papers
  • GSEB Solutions
  • Maharashtra Board
  • Kerala Board
    • Kerala Syllabus 9th Standard Physics Solutions Guide
    • Kerala Syllabus 9th Standard Biology Solutions Guide
  • Goa Board

CBSE Tuts

CBSE Maths notes, CBSE physics notes, CBSE chemistry notes

NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1

NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1 are part of NCERT Solutions for Class 12 Maths. Here we have given NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1.

  • Probability Class 12 Ex 13.2
  • Probability Class 12 Ex 13.3
  • Probability Class 12 Ex 13.4
  • Probability Class 12 Ex 13.5
Board CBSE
Textbook NCERT
Class Class 12
Subject Maths
Chapter Chapter 13
Chapter Name Probability
Exercise Ex 13.1
Number of Questions Solved 17
Category NCERT Solutions

NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1

Ex 13.1 Class 12 Maths Question 1
Given that E and Fare events such that
P (E) = 0.6, P (F) = 0.3 and P(E∩F) = 0.2
find P(E|F) and P (F|E).
Solution:
Given: P (E)=0.6, P (F)=0.3, P (E ∩ F)=0.2
P(E|F)=\frac { P(E\cap F) }{ P(E) } =\frac { 0.2 }{ 0.3 } =\frac { 2 }{ 3 }
P(F|E)=\frac { P(E\cap F) }{ P(E) } =\frac { 0.2 }{ 0.6 } =\frac { 1 }{ 3 }

Ex 13.1 Class 12 Maths Question 2
Compute P(A|B) if P(B)=0.5 and P (A∩B) = 0.32.
Solution:
Given: P (B)=0.5, P(A∩B)=0.32
P(A|B)=\frac { P(A\cap B) }{ P(B) } =\frac { 0.32 }{ 0.50 } =\frac { 32 }{ 50 } =0.64

Ex 13.1 Class 12 Maths Question 3.
If P (A)=0.8, P (B)=0.5 and P(B/A)=0.4, find
(i) P(A∩B)
(ii) P(A/B)
(iii)P(A∪B)
Solution:
(i) P(B/A)=\frac { P(A\cap B) }{ P(A) } \Rightarrow 0.4=\frac { P(A\cap B) }{ 0.8 }
∴ P(A∩B) = 0.4 x 0.8 = 0.32
(ii) P(A/B) = P(A/B)=\frac { P(A\cap B) }{ P(B) } =\frac { 0.32 }{ 0.5 } =\frac { 32 }{ 50 } =\frac { 16 }{ 25 }
(iii) P(A∪B) = P(A) + P(B) – P(A∩B)
= 0.8 + 0.5 – 0.32 = 1.30 – 0.32 = 0.98

Ex 13.1 Class 12 Maths Question 4.
Evaluate P(A∪B) if 2P(A) = P(B) = \frac { 5 }{ 13 } and P(A|B) = \frac { 2 }{ 5 }.
Solution:
Given:
2P(A) = P(B) = \frac { 5 }{ 13 } and P(A|B) = \frac { 2 }{ 5 }.
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1 Q4.1

Ex 13.1 Class 12 Maths Question 5.
If P(A) = \frac { 6 }{ 11 },P(B) =\frac { 5 }{ 11 } and P(A∪B) = \frac { 7 }{ 11 },
Find
(i) P(A∩B)
(ii) P(A|B)
(iii) P(B|A)
Solution:
Given:
P(A) = \frac { 6 }{ 11 },P(B) =\frac { 5 }{ 11 } and P(A∪B) = \frac { 7 }{ 11 },
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1 Q5.1

Ex 13.1 Class 12 Maths Question 6.
Determine P(E/F) in question 6 to 9:
A coin is tossed three times, where
(i) E: head on third toss F: heads on first two tosses.
(ii) E: at least two heads F : at most two heads
(iii) E: at most two tails F: at least one tail
Solution:
(i) E = Head occurs on third toss as {HHH, HTH, THH, TTH}
F : Heads on first two tosses = {HHH, HHT} E ∩ F = {HHH}
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1 Q6.1

Ex 13.1 Class 12 Maths Question 7.
Two coins are tossed once
(i) E: tail appears on one coin F: one coin shows head
(ii) E: no tail appears F: no head appears.
Solution:
S = {HH, TH, HT, TT} n (S) = 4
(i) E : tail appears on one coin
E = {TH,HT}, P{E) = \frac { 1 }{ 2 }
F : one coin shows head,
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1 Q7.1

Ex 13.1 Class 12 Maths Question 8.
A die is thrown three times.
E: 4 appears on the third toss
F: 6 and 5 appears respectively on first two tosses.
Solution:
A die is thrown three times E : 4 appears on third toss = {(1,1,4), (1,2,4), (1,3,4), (1,4,4), (1,5,4),
(1.6.4),(2,1,4), (2,2,4), (2,3,4), (2,4,4), (2,5,4),
(2.6.4), (3,1,4), (3,2,4), (3,3,4), (3,4,4), (3,5,4),
(3.6.4), (4,1,4), (4,2,4), (4,3,4), (4,4,4), (4,5,4),
(4.6.4), (5,1,4), (5,2,4), (5,3,4), (5,4,4), (5,5,4),
(5.6.4), (6,1,4), (6,2,4), (6,3,4), (6,4,4), (6,5,4),
(6.6.4)}
These are 36 cases
F: 6 and 5 appears respectively on first two tosses = {(6,5,1), (6,5,2), (6,5,3), (6,5,4), (6, 5,5), (6,5,6)}
These are six cases E ∩ F = {6,5,4}
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1 Q8.1

Ex 13.1 Class 12 Maths Question 9.
Mother, father and son line up at random for a family picture:
E: son on one end, F: father in middle
Solution:
Mother (m), Father (f) and son (s) line up at random E : son on one end: {(s, m, f), (s, f, m), (f, m, s), (m, f, s)}
F: Father in middle: {(m, f,s), (s, f, m), (s, f, m)}
E∩F = {(m, f, s), (s, f, m)}
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1 Q9.1

Ex 13.1 Class 12 Maths Question 10.
A Mack and a red die 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:
(a) n(S) = 6 x 6 = 36
Let A represent obtaining a sum greater than 9 and B represents black die resulted in a 5.
A= {46,64,55,36,63,45,54,65,56,66}
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1 Q10.1

Ex 13.1 Class 12 Maths Question 11
A fair die is rolled. Consider events E = {1,3,5} F = {2,3} and G = {2,3,4,5}, Find
(i) P(E|F) and P(F|E)
(ii) P(E|G) and P(G|E)
(iii) P((E ∪ F) |G) and P (E ∩ F)|G)
Solution:
(i) E = {1,3,5}, F = {2,3},E∩F = {3}
P(E) = \frac { 3 }{ 6 } ,P(F) = \frac { 2 }{ 6 },P(E∩F) = \frac { 1 }{ 6 },
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1 Q11.1
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1 Q11.2

Ex 13.1 Class 12 Maths Question 12.
Assume that each child born 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 girl?
Solution:
Let first and second girls are denoted by G1 and G2 and Boys by B1 and B2.
Sample space
S = {(G1G2),(G1B2),(G2B1),(B1B2)}
Let A = Both the children are girls = {G1G2}
B = youngest child is a girls = {G1G2, B1G2}
C = at least one is a girl = {G1B2, G1G2, B1G2}
A∩B = {G1G2},
A∩C = {G1G2}
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1 Q12.1

Ex 13.1 Class 12 Maths Question 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:
The given data may be tabulated as
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1 Q13.1

Ex 13.1 Class 12 Maths Question 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:
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1 Q14.1
Let A represents the event “the sum of numbers on the dice is 4”
and B represents the event “the two numbers appearing on throwing two dice are different.”
A = {13,22,31}, n (A) = 3
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1 Q14.2

Ex 13.1 Class 12 Maths Question 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:
Let there be n throws in which a multiple of 3 occurs every time.
Probability of getting a multiple of 3 (i.e. 3 or 6)
in one throw = \frac { 2 }{ 6 } = \frac { 1 }{ 3 }
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1 Q15.1
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1 Q15.2

In each of the following choose the correct answer:

Ex 13.1 Class 12 Maths Question 16.
If P(A) = \frac { 1 }{ 2 }, P (B) = 0 then P (A | B) is
(a) 0
(b) \frac { 1 }{ 2 }
(c) not defined
(d) 1
Solution:
P(A) = P(B) = 0
∴ P(A∩B) = 0
\therefore P(A|B)=\frac { P(A\cap B) }{ P(B) } =\frac { 0 }{ 0 }
Thus option C is correct

Question 17.
If A and B are events such that P(A | B) = P(B | A) then
(a) A⊂B but A≠B
(b) A = B
(c) A∩B = φ
(d) P(A) = P(B)
Solution:
P(A | B) = P(B | A)
Thus option (d) is correct.
\frac { P(A\cap B) }{ P(B) } =\frac { P(B\cap A) }{ P(A) }
⇒ P(A) = P(B)

We hope the NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1 help you. If you have any query regarding NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1, drop a comment below and we will get back to you at the earliest.

 

Primary Sidebar

NCERT Exemplar problems With Solutions CBSE Previous Year Questions with Solutoins CBSE Sample Papers

Recent Posts

  • Political Science Class 12 Important Questions Chapter 1 The Cold War Era
  • MCQ Questions for Class 6 Science with Answers PDF Download Chapter Wise
  • MCQ Questions for Class 7 Science with Answers PDF Download Chapter Wise
  • NCERT Solutions for Class 6 Sanskrit Ruchira Bhag 1
  • MCQ Questions for Class 8 Science with Answers PDF Download Chapter Wise
  • NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes InText Questions
  • NCERT Solutions for Class 7 Maths Chapter 14 Symmetry InText Questions
  • NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers InText Questions
  • NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions InText Questions
  • NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area InText Questions
  • NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers InText Questions
  • NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities InText Questions
  • NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles InText Questions
  • NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties InText Questions
  • NCERT Solutions for Class 7 Maths Chapter 5 Lines and Angles InText Questions

Footer

Maths NCERT Solutions

NCERT Solutions for Class 12 Maths
NCERT Solutions for Class 11 Maths
NCERT Solutions for Class 10 Maths
NCERT Solutions for Class 9 Maths
NCERT Solutions for Class 8 Maths
NCERT Solutions for Class 7 Maths
NCERT Solutions for Class 6 Maths

SCIENCE NCERT SOLUTIONS

NCERT Solutions for Class 12 Physics
NCERT Solutions for Class 12 Chemistry
NCERT Solutions for Class 11 Physics
NCERT Solutions for Class 11 Chemistry
NCERT Solutions for Class 10 Science
NCERT Solutions for Class 9 Science
NCERT Solutions for Class 7 Science
MCQ Questions NCERT Solutions
CBSE Sample Papers
NCERT Exemplar Solutions LCM and GCF Calculator
TS Grewal Accountancy Class 12 Solutions
TS Grewal Accountancy Class 11 Solutions