AP Inter 2nd Year

Referring to the AP Inter 2nd Year Maths Study Material Chapter 1 Relations and Functions Exercise 1b Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Relations and Functions Solutions Exercise 1b

I.

Question 1.
Check the injectivity and surjectivity of the function f : N → N given by f (x) = x²
Solutions:
f : N → N; f (x) = x²
∴ f is injective, (one-one)
∴ f is not surjective(onto)
(i) f(x₁) = f(x₂) ⇒ x₁² = x₂² ⇒ x₁ = x₂
(ii) For 2 ∈ N we can’t find in x in N such that f(x) = x² = 2
Aliter: We know N= {1, 2, 3, ….} and f (x) = x²
(i) f(1) = 1² = 1; f(2) = 2² = 4; f(3) = 3² = 9;
∴ f = {(1, 1), (2, 4), (3, 9), ….} ………….. (1)
Here, distinct elements of the domain N have distinct images.
∴ f is injective.

ii) From(1), Range off = {1, 4, 9, ….}
(set of all second elements in the ordered pairs)
But codomain N= {1, 2, 3, 4, ….}
Thus, Range set ≠ Codomain N.
∴ f is not surjective
Given function f is injective but not surjective.

Question 2.
Check the injectivity and surjectivity of the function f : Z → Z given by f (x) = x²
Solution:
f : Z → Z; f (x) = x²
(i) f(x₁) = f(x₂) ⇒ x₁² = x₁² ⇒ x₁= ±x₂
(ii) For 2 ∈ Z we can’t find in x in Z such that f(x) = x² = 2
Aliter: Now f(1) = 1² = 1; f(-1) = (-1)² = 1;
Thus, distinct elements 1, -1 of the domain Z have the same image 1.
∴ f is not injective,

ii) Range off = {0, 1, 4, 9, , …}
But codomain Z= {…-3, -2, -1, 0, 1, 1.5 2, 3, 4, ….};
Thus, Range set ≠ Codomain Z.
∴ f is not surjective
Given function f is not injective and not surjective.

Question 3.
Check the injectivity and surjectivity of the function f : R → R given by f (x) = x²
i) f: R → R given by f (x) = x²
f(1) = 1² = 1; f(-1) = (-1)² = 1;
Here, distinct elements 1, -1 of the domain R have the same image 1.
∴ f is not injective.

ii) For -2 ∈ R we can’t find in x in R such that f(x) = x² = -2
∴ f is not surjective
Given function f is not injective and not surjective.

Question 4.
Check the injectivity and surjectivity of the function f : N → N given by f (x) = x³
Solution:
f: N → N; f (x) = x³
(i) f (x₁) = f (x₂) ⇒ x₁³ = x₂³ ⇒ x₁ = x₂
∴ f is injective.
(ii) For 2 ∈ N we can’t find in x in N such that f(x) = x³ = 2
∴ f is not surjective
Aliter: We know N= {1, 2, 3, 4, ….} and f(x) = x³

i) f(1) = 1³ = 1; f(2) = 2³ = 8; f(3) = 3³ = 27;
∴ f{(1, 1), (2, 8), (3, 27), …………} ………….. (1)
Here, distinct elements of the domain N have distinct images.
∴ f is injective.
ii) From (1) Range of f contains only the cube values = {1, 8, 27, ….}
But codomain N = {1, 2, 3, 4, ….}
Thus, Range set ≠ Codomain N.
∴ f is not surjective
Given function f is injective but not surjective.

Question 5.
Check the injectivity and surjectivity of the function f: Z → Z given by f (x) = x³
Solution:
f : Z → Z; f(x) = x³
(i) f(x₁) = f(x₂) = x₁³ = x₂³ = x₁ = x₂
∴ f is injective.
(ii)For 2 ∈ Z we can’t find in x in Z such that f(x) = x³ = 2
∴ f is not swje1ctive

Question 6.
Prove that the Greatest Integer Function f : R → R, given by f (x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.
Solution:
f: R → R is given by f(x) = [x]= Greatest integer ≤ x

Now,f(0) = [0] = 0; f(1) = [1] = 1;
f(1.2) = [1.2] = 1, f(2) = [2] = 2;
f(-o.5) = [-0.5] = -1; f(-1.1) = [-1.1] = -2;
(i) Thus, we have f(1) = 1, f(1.2) = 1.
Thus distinct elements R have the same image.
∴ f is not one-one.

(ii) Thus, we see that the range contains only the integers f={…., -2, -1, 0, 1, 2, 3, ….}= Z
Thus, Range Z ≠ Codomain R.
∴ f is not onto.
∴ The greatest integer function is neither one-one nor onto.

Question 7.
Show that the Modulus Function f : R → R, given by f (x) = | x |, is neither one- one nor onto, where | x | is x, if x is positive or 0 and | x | is – x, if x is negative.
Solution:
f: R → R is given as f(x) = |x| =
(i) f(-1) = |-1| = 1 and f(1) = |1| = 1
Thus distinct elements -1, 1 of the domain R have the same image 1.
∴ f is not one-one.

(ii) Range of |x| is [0, ∞]; [ This range donot contain any negative reals)
Thus, range ≠ codomain R
∴ f is not onto.
∴ The modulus function is neither one-one nor onto.

Question 8.
Show that the Signum Function f : R → R, given by

Solution:
(i) From the given function we have f(1) = 1 and f(2) = 1 [∵ f(x) = 1 if x > 1]
Thus distinct elements of R have the same image.
∴ f is not one-one.

(ii) From the given function f we see that the Range of f(x)is {1, 0, -1}.But codomain is R.
Thus, range ≠ codomain R
∴ f is not onto.
∴ The signum function is neither one-one nor onto.

Question 9.
Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show that f is one-one.

Solution:
Given that A = {1, 2, 3} , B = {4, 5, 6, 7}
f: A → B is defined as f = {(1, 4), (2, 5), (3, 6)}
∴ f(1) = 4, f(2) = 5, f(3) = 6
Thus, distinct elements of A have distinct images in B under f.
∴ f is one-one.

Question 10.
Define One-one (Injective) function
Solution:
One-one (Injective) function : A function f: A → B is said to be a one one function if x₁, x₂ ∈ A
be such that f(x₁)= f(x₁) ⇒ x₁= x₂

Question 11.
Define onto(Surjective) function
Solution:
Onto(Surjective) function: A function f: A → B is said to be an onto function if its
Range equals to Codomain. [f(A)= B] (or) .
A function f: A → B is an onto function if ∀ y ∈ B ∃ x ∈ A such that f(x) = y

Question 12.
Define bijective function.
Solution:
Bijective function : A function which is both one-one and onto then f is called bijective.

II.

Question 1.
Show that the function f: R → R„ defined by f(x) = \(\frac{1}{x}\) is one-one and onto, where R_* is the set of all non-zero real numbers. Is the result true, if the domain R* is replaced by N with co-domain being same as R_*?
Solution:
a) When f: R_* → R_* is given as f(x) = —. Let x₁, x₂ ∈ R_*
(i) f(x₁) = f(x₂) ⇒ \(\frac{1}{x_1}=\frac{1}{x_2}\) ⇒ x₁ = x₂
∴ f is one-one.

(ii) Given that f(x) = \(\frac{1}{x}\) …………. (1)
Let f(x) = y ⇒ y = \(\frac{1}{x}\) ⇒ x = \(\frac{1}{y}\), exists ∀ y ≠ 0 [ as y ∈ R_*, set of non-zero reals]
∴ f is onto.
Verification: From (1), f (x) = \(\frac{1}{x}\) = \(\frac{1}{\left(\frac{1}{y}\right)}\) = y
∴ Given function f is one-one and onto.

b) We take g: N → R_* as g(x) = \(\frac{1}{x}\). Let x₁, x₂ ∈ N
(i) g(x₁) = g(x₂) ⇒ \(\frac{1}{x_1}=\frac{1}{x_2}\) ⇒ x₁ = x₂
∴ g is one-one.

(ii) For some 1.5 ∈ R_* (codomain) there is no x in N such that
g(x) = \(\frac{1}{1.5}\) {∵ \(\frac{1}{1.5}\) is not a natural}
∴ the other function g is one-one but not onto.

Question 2.
State whether the function f : R → R defined by f (x) = 3 – 4x is one-one, onto or bijective. Justify your answer.
Solution:
f: R. → R is defined as f(x) = 3 – 4x. Let x₁, x₂ ∈ R
(i) f(x₁) = f(x₂) ⇒ 3 – 4x₁ = 3 – 4x₂ ⇒ -4x₁ = -4x₂ ⇒ x₁ = x₂
∴ f is one-one.

(ii) Given that f (x) = 3 – 4x (1); Put f(x) = y
⇒ y = 3 – 4x ⇒ 4x = 3 – y ⇒ x = \(\frac{3-y}{4}\) exists ∀ y ∈ R( codomain)
∴ f is onto.
Justification: From (1), f(x) = \(\mathrm{f}\left(\frac{3-\mathrm{y}}{4}\right)=3-4\left(\frac{3-\mathrm{y}}{4}\right)\) = 3 – (3 – y) = y
Hence, f is both one-one and onto hence bijective.

Question 3.
State whether the function f : R → R defined by f (x) = 1 + x² is one-one, onto or bijective. Justify your answer.
Solution:
f: R → R is defined as f (x) = 1 + x². Let x₁, x₂ ∈ R
(i) Take two elements 1, -1 in the domain
Now f(1) = 1 + 1² = 2; f(-1) = 1 + (-1) = 1 + 1 = 2;
Two distinct elements in the domain have the same image.
∴ f is not one-one.

(ii) From the given function f (x) = 1 + x² it can be seen that Range of f is always positive.
So negative elements in the codomain R like -2 do not have any preimage in the
domain R such that f(x) = 1 + x²
∴ f is not onto.
Hence, f is neither one-one nor onto.

Question 4.
Let A and B be sets. Show that f: A × B → B × A such that f (a, b) = (b, a) is bijective function.
Solution:
f : A × B → B × A is defined as(a, b) = (b, a).
Let (a₁, b₁), (a₂, b₂) ∈ A × B
(i) f(a₁, b₁)= f(a₂, b₃) (b₁, a₁) = (b₂, a₂)
⇒ (b₂, a₂) = (b₁, a₁) (a₁, b₁)=(a₂, b₂)
∴ f is one—one.

(ii) ∀(b, a) ∈ B × A there exist (a, b) ∈ A × B such that f(a, b) = (b, a)
∴ f is onto.
Here, f is both one-one and onto hence bijective.

Question 5.
Let f : N → N be defined by

for all n ∈ N
State whether the function f is bijective. Justify your answer.
Solution:
Case i: For odd n = 1, 3, 5, …. we have f(n) = \(\frac{\mathrm{n}+1}{2}\)
∴ f(1) = \(\frac{1+1}{2}\) = 1; f(3) = \(\frac{3+1}{2}\) = 2; f(5) = \(\frac{5+1}{2}\) = 3;
⇒ f = {(1, 1), (3, 2), (5, 3), …………} ……. (1)

Case ii: For even n = 2, 4, 6, …. we have f(n) = \(\frac{\mathrm{n}}{2}\)
∴ f(2) = \(\frac{2}{2}\) = 1; f(4) = \(\frac{4}{2}\) = 2; f(6) = \(\frac{6}{2}\) = 3; ⇒ {(2, 1), (4, 2), (6, 3), }….. (2)
From (1) and (2) we get f= {(1, 1), (2, 1), (3, 2), (4, 2), (5, 3), (6, 3), ….} (3)
a) From (3), we have f(1) = 1 and f(2) = 1
Thus distinct elements 1, 2 of the domain N have the same image 1.
∴ f is not one-one

b) From (3) the range set = {1, 1, 2, 2, 3, 3, } = {1, 2, 3, ….}
⇒ Range of = {1, 2, 3, }= codomain N
∴ f is onto.
∴ f is onto but not one-one. Hence, f is not a bijective function.

Question 6.
Let A = R – {3} and B = R – {1}. Consider the function f: A → B defined by f(x) = \(\left(\frac{x-2}{x-3}\right)\) Is f one-one and onto? Justify your answer.
Solution:
Given that f: A-+B is defined as f (x) = \(\left(\frac{x-2}{x-3}\right)\) for A = R -{3}, B = R -{1}
a) Take x₁, x₂ ∈ A such that f(x₁) = f(x₂) ⇒ \(\frac{x_1-2}{x_1-3}=\frac{x_2-2}{x_2-3}\) ⇒ (x₁ -2)(x₂ – 3) = (x₂ – 2)(x₁ – 3)
⇒ x₁x₂ – 3x₁ – 2x₂ + 6 = x₁x₂ – 3x₂ – 2x₁ + 6
⇒ -3x₁ – 2x₂ = -3x₂ – 2x₁
⇒ 3x₁ – 2x₁ = 3x₂ – 2x₂
⇒ x₁ = x₂
∴ f is one-one.

b) Given that f(x) = \(\left(\frac{x-2}{x-3}\right)\) …………. (1). Put f(x) = y
⇒ y = \(\left(\frac{x-2}{x-3}\right)\) ⇒ x – 2 = xy – 3y ⇒ x(1 – y) = -3y + 2 ⇒ x = \(\frac{2-3 y}{1-y}\) exists ∀ y ∈ B = R-{1}
∴ f is onto.

Referring to the AP Inter 2nd Year Maths Study Material Chapter 1 Relations and Functions Exercise 1a Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Relations and Functions Solutions Exercise 1a

I. Give an example of a relation. Which is

Question 1.
Symmetric but neither reflexive nor transitive:
Solution:
Let A = {1, 2}. On A × A consider a relation R = {(1, 2),(2, 1)}
(i) (1, 1),(2, 2) ∈ R . So, R is not reflexive.
(ii) (1, 2) ∈ R and (2, 1) ∈ R. So, R is symmetric
(iii) (1, 2), (2, 1) ∈ R,but (1, 1) ∉ R. So, R is not transitive.
∴ This relation R is symmetric but not reflexive or transitive.

Question 2.
Transitive but neither reflexive nor symmetric:
Solution:
Let A = {1, 2, 3}. On A × A consider a relation R = {(a, b): a < b}
(i) (1, 1) ∉ R. So, R is not reflexive, [v 1 cannot be less than itself ]
(ii) (1, 2) ∈ R. But (2, 1) ∉ R.So, R is not symmetric. [∵ 2 is not less than 1]
(iii) (1, 2),(2, 3) ∈ R ⇒ (1, 3) ∈ R. So, R is transitive. [∵ 1 < 2 and 2 < 3 ⇒ 1 < 3]
∴ This relation R is transitive but not reflexive and symmetric.

Question 3.
Reflexive and symmetric but not transitive:
Solution:
LetA = {1, 2, 3} and R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)}
(i) (a, a) ∈ R ∀ a ∈ A, . So, R is reflexive.
(ii) (1, 2) ∈ R ⇒ (2, 1) ∈ R for a, b ∈ R. So, R is symmetric.
(iii) (1, 2), (2, 3) ∈ R ⇒ (1, 3) ∉ R. So, R is not transitive.
∴ This relation R is reflexive and symmetric but not transitive.

Question 4.
Reflexive and transitive hut not symmetric.
Solution:
Let A = {1, 2, 3} and R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}
(i) (a, a) ∈ R ∀ a ∈ A, . So, R is reflexive.
(ii) (1, 2) ∈ R ⇒ (2, 1) ∉ R . So, R is not symmetric.
(iii) (1, 2) ∈ R, (2, 3) ∈ R ⇒ (1, 3) ∈ R . So, R is transitive.
∴ This relation R is reflexive and transitive but not symmetric.

Question 5.
Symmetric and transitive but not reflexive.
Solution:
Let A = {1, 2} and R = {(1, 2), (2, 1), (1, 1)}
(i) (2, 2) ∉ R . So, R is not reflexive.
(ii) (1, 2), (2, 1) ∈ R .So, R is symmetric.
(iii) (1, 2), (2, 1) ∈ R ⇒ (1, 1) ∈ R . So, R is transitive.
∴ This relation R is not reflexive but symmetric and transitive.

Question 6.
Define reflexive relation
Solution:
Reflexive Relation: A relat ion R on a set A is reflexive if (a, a) ∈ R for all a ∈ A.
Ex : The relation equality (=) on reals.

Question 7.
Define symmetric relation
Solution:
Symmetric Relation: A relation R on a set A is symmetric if (a, b) ∈ R ⇒ (b, a) ∈ R
Ex: The relation ‘parallel of lines (||) in a plane.

Question 8.
Define transitive relation
Solution:
Transitive Relation: A relation R on a set A is transitive if(a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R
Ex: The relation greater than ( > ) on reals.

Question 9.
Define an equivalence relation
Solution:
Equivalence Relation:
A relation R on a set A is called an equivalence relation if R is reflexive, symmetric and transitive.
Ex: The relation equality (=) on reals.

II.

Question 1.
Determine whether the Relation R in the set A = {I, 2, 3, 13, 14} defined as R = {(x, y): 3x – y = 0} is reflexive, symmetric and transitive.
Solution:
Given that A = {1, 2, 3, …, 13, 14} andR= {(x, y): 3x – y = 0}.
∴ R = {(1, 3), (2, 6), (3, 9), (4, 12), (14, 52)}
(i) (1, 1), (2, 2), …. ∉ R. So, R is not reflexive.
(ii) (1, 3) ∈ R, but (3, 1) ∉ R. So, R is not symmetric.
(iii) (1, 3), (3, 9) ∈ R, but(1, 9) ∉ R. So, R is not transitive.

Question 2.
Determine whether the Relation R in the set N of natural numbers defined as,
R = {(x, y): y = x + 5 and x < 4} is reflexive, symmetric and transitive.
Solution:
Given N = {1, 2, 3, …} and R = {(x, y): y = x + 5 and x < 4}
∴ R = {(1, 6), (2, 7), (3, 8)}
(i) (1, 1) ∉ R. So, R is not reflexive.
(ii) (1, 6) ∈ R but(6, 1) ∉ R So, R is not symmetric.
(iii) We find no 3 ordered pairs such that (x, y), (y, z) ∈ R ⇒ (x, z) ∈ R .
Among the three ordered pairs in R, no chaining is possible vacously true. So, R is transitive.

Question 3.
Determine whether the Relation R in the set A = {1, 2, 3, 4, 5, 6} as R = {(x, y): y is divisible by x} is reflexive, symmetric and transitive.
Solution:
Given that A = {1, 2, 3, 4, 5, 6} as R = {(x, y): y is divisible by x}
(i) (x, x) ∈ R ∀ x ∈ A. So, R is reflexive.
[∵ Any number other than 0 is divisible by itself. ]

(ii) (x, y) ∈ R then (y, x)need not belong to R. So, R is not symmetric.
Let (2, 4) ∈ R [because 4 is divisible by 2]
But (4, 2)∉ R[since 2 is not divisible by 4]

(iii) (x, y) and (y, z) ∈ R ⇒(x, z) ∈ R. So, R is transitive.
Let (1, 3) ∈ R [because 3 is divisible by 1]
and (3, 6) ∈ R[since 6 is divisible by 3 ]
Then (1, 6) ∈ R[since 6 is divisible by 1]

Question 4.
Determine whether, the Relation R in the set Z of all integers defined as R = {(x, y): x – y is an integer} is reflexive, symmetric and transitive.
Solution:
Given that x, y ∈ Z and R = {(x, y): x – y is an integer)}
(i) (x, x) ∈R x ∈ Z. So, R is reflexive.
[∵ x – x = 0 and 0 is an integer.]

(ii) (x, y) ∈ R, =(y, x) ∈ R. So, R is symmetric.
[∵ if (x – y) is an integer then (y – x) is also an integer.]

(iii) (x, y), (y, z) ∈ R =(x, z) ∈ R where x, y, z ∈ Z. So, R is transitive.
[∵ if (x – y) and (y – z) are integers then (x – z) = (x – y) + (y – z) is also an integer]

Question 5.
Show that the relation R in the set R of real numbers, defined as
R = {(a, b): a ≤ b²) is neither reflexive nor symmetric nor transitive.
Solution:
Given that a, b ∈ R and R =(a, b): a ≤ b²}
(i) (\(\frac{1}{2}\), \(\frac{1}{2}\)) ∉ R. So, R is not reflexive. [∵ \(\frac{1}{2}\) > (\(\frac{1}{2}\))²]
(ii) (1, 2) ∈ R ⇒ (2, 1) ∉ R. So, R is not symmetric.
[∵ (1, 2) ∈ R ⇒ 1< 2² = 4, (2, 1) ∈ R ⇒ 2 < 1² = 1, which is absurd.]

(iii) (3, 2), (2, 1.5) ∈ R ⇒ (3, 1.5) ∉ R So, R is not transitive.
[(3, 2) ∈ R⇒ 3 < 2² = 4 and also 2 < (1.5)² = 2.25 .But 3 < (1.5)² = 2.25, which is absurd.]
∴ R is not reflexive, not symmetric, not transitive.

Question 6.
Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.
Solution:
Given that A = {1, 2, 3, 4, 5, 6} and R = {(a, b):b = a + 1}
∴ R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}
(i) (1, 1), (2, 2) ∉ R . So, R is not reflexive.
(ii) (1, 2) ∈ R, but (2, 1) ∉ R. So, R is not symmetric.
(iii) (1, 2), (2, 3) ∈ R but (1, 3) ∉ R. So, R is not transitive.
∴ R is not reflexive, not symmetric, not transitive.

Question 7.
Show that the relation R in R defined as R = {(a, b): a ≤ b} is reflexive and transitive but not symmetric.
Solution:
Let a, b ∈ R and R = {(a, b): a ≤ b}
(i) (a, a)∈ R. So, R is reflexive [∵ a ≤ a]
(ii) (a, b) ∈ R . But (b, a) ∉ R. So, R is not symmetric.
[∵ (2, 4) ∈ R(as 2 < 4) and(4, 2) ∉ R(as 4 > 2)]
(iii) (a, b), (b, c) ∈ R ⇒ (a, c) ∈ R . So, R is transitive, [∵ a ≤ b and b ≤ c ⇒ a ≤ c]
∴ R is reflexive and transitive but not symmetric.

Question 8.
Check whether the relation R in R defined by R = {(a, b): a ≤ b³} is reflexive, symmetric or transitive.
Solution:
Given that a, b ∈ R and R = {(a, b):a ≤ b³}
(i) (\(\frac{1}{2}\), \(\frac{1}{2}\)) ∉ R. So, R is not reflexive. [∵ \(\frac{1}{2}\) > (\(\frac{1}{2}\))³]
(ii) (1, 2) ∈ R ⇒ (2, 1) ∉ R. So, R is not symmetric.
[∵ (1, 2) ∈ R ⇒ 1 < 2³ = 8.(2, 1) ∈ R ⇒ 2 < 1³ = 1, which is absurd.]
(iii) Consider (3, 1.5), (1.5, 1.2), (3, 1.2) . Here R is not transitive [ (1.2)³ = 1.728, (1.5)³ = 3.75]
[∵ 3 ≤ (1.5)³ ⇒ (3, 1.5) ∈ R; 1.5 ≤ (1.2)³ ⇒ (1.5, 1.2) ∈ R. But(1.5)³ >3 ⇒ (3, 1.5) ∉ R]
∴ R is not reflexive, not symmetric and not transitive.

Question 9.
Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.
Solution:
Given that A = {1, 2, 3}; R = {(1, 2), (2, 1)}
(i) (1, 1), (2, 2), (3, 3) ∉ R. So, R is not reflexive
(ii) (1, 2) ∈ R and (2, 1) ∈ R. So, R is symmetric.
(iii) (1, 2) ∈ R and (2, 1) ∈ R, (1, 1) ∉ R. So, R is not transitive.
∴ R is not reflexive, not transitive but R is symmetric.

Question 10.
Show that the relation R in the set A of all the books in a library of a college, given by R = {(x, y): x and y have same number of pages} is an equivalence relation. .
Solution:
Set A contain various library books and x, y related to the number of pages in any two books.
Given that R = {(x, y): x and y have same number of pages}.
(i) (x, x) ∈ R. So, R is reflexive
∵ x and x have same number of pages.]

(ii) (x, y) ∈ R ⇒ (y, x) ∈ R. So, R is symmetric.
[ ∵ If x and y have same number of pages then y and x also have same number of pages.]

(iii) (x, y) ∈ R, (y, z) ∈ R ⇒ (z, x) ∈ R. So, R is transitive.
[∵ If x and y have same number of pages, and y and z have same number of pages, then x and z have same number of pages. ]
Thus, R is reflexive, symmetric, transitive. Hence, R is an equivalence relation.

Question 11.
Show that the relation R in the set A = {1, 2, 3, 4, 5} given by R = {(a, b): |a – b| is even} is an equivalence relation. Show that all the elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are related to each other. But no element of {1, 3, 5} is related to ally element of {2, 4}.
Solution:
GiyenthatA={1, 2, 3, 4, 5}givenbyR={(a, b):|a-b| is even},
(i) (a, a) ∈ R ∀ a ∈ A. So, R is reflexive. [∵ |a-a| = 0 (which is even)]
(ii) (a, b) ∈ R ⇒ (b, a) ∈ R. So, R is symmetric. [∵ if |a-b| is even then |-(a-b)| =|b-a| is also even]
(iii) (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R. So, R is transitive.
[∵ if |a—b| , |b—c| are even then (a-b), (b-c) are also even .
Now (a-c) = (a+b) +(b-c) is even ⇒ |a-c| is even ]
Thus, R is reflexive, symmetric, transitive. Hence, R is an equivalence relation.
Further, all elements of {1, 3, 5} are related to each other because they are all odd.
So, the modulus of the difference between any two elements is even.
Similarly, all elements {2, 4} are related to each other because they are all even.
No element of {1, 3, 5}is related to any elements of {2, 4} as all elements of {1, 3, 5}are odd and all elements of {2, 4} are even. So, the modulus of the difference between the two elements will not be even.

Question 12.
Show that each of the relation R in the set A = {x ∈ Z : 0 ≤, x ≤ 12}, given by R = {(a, b): |a – b| is a multiple of 4} is an equivalence relation. Find the set of all elements related to 1 .
Solution:
Given that A = {x ∈ Z: 0 ≤ x ≤ 12} ⇒ A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
R = {(a, b):|a – b|is a mutiple of 4}
(i) (a, a) ∈ R ∀ a ∈ A. So, R is reflexive, [∵ |a – a| = 0 and 0 is a multiple of 4]

(ii) (a, b) ∈ R ⇒ (b, a) ∈ R . So, Ris symmetric.
[∵ |a – b| is a multiple of 4 ⇒ |-(a -b)| = |b – a| is also a multiple of 4]

(iii) (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R So, Ris transitive.
[∵If |a-b|, |b-c| are multiples of 4 ⇒ (a-b), (b-c) are multiples of 4.
Now (a-c) = (a-b) + (b-c) is a multiple of 4 ⇒ |a – c| is a multiple of 4 ]
Thus, R is reflexive, symmetric, transitive. Hence, R is an equivalence relation. Further, the set of elements related to 1 is {1, 5, 9}
∵ |1 – 1| = 0 is a multiple of 4; |5 – 1| = 4 is a multiple of 4., |9 -1| = 8 is a multiple of 4

Question 13.
Show that each of the relation R in the set A = {x ∈ Z : 0≤ x ≤ 12}, given by R = {(a, b): a = b} is an equivalence relation. Find the set of all elements related to 1.
Solution:
Given that R = {(a, b):a = b}
(i) (a, a) ∈ R ∀ a ∈ A . So, R is reflexive, [∵ a = a ∀a]
(ii) (a, b) ∈ R ⇒ a = b => b = a ⇒ (b, a)e R . So, R is symmetric.
(iii) (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R. So, R is transitive. [∵ a = b and b = c ⇒ a = c]
Thus, R is reflexive, symmetric, transitive. Hence, R is an equivalence relation.
Further, the set of elements related to 1 are those elements of A which are equal to 1.
∴ the required set is {1}.

III.

Question 1.
Show that the relation R in the set A of points in a plane given by R = {(P, Q): distance of the point P from the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, show that the set of all points related to a point P ≠ (0, 0) is the circle passing through P with origin as center.

Solution:
Given that A is the set of points in a plane and
R = {(P, Q): Distance of the point P from the origin = the distance of the point Q from the origin}
(i) Clearly (P, P) ∈ R. So, R is reflexive. [∵ OP = OP]
(ii) (P, Q) ∈ R ⇒ (Q, P) ∈ R . So, R is symmetric. [∵ PQ = QP]
(iii) (P, Q), (Q, S) ∈ R ⇒ (P, S) ∈ R . So, R is transitive
[∵ (P, Q), (Q, S) ∈ R ⇒ OP = OQ and OQ = OS then OP = OS ⇒ (P, S) ∈ R]
Thus, R is reflexive, symmetric, transitive. Hence, R is an equivalence relation.
Further, the set of points related to P ≠ (0, 0) will be those points whose distance from origin is same as distance of P from the origin.
Set of points forms a circle with the centre as brigin and this circle passes through P.

Question 2.
Show that the relation R defined in the set A of all triangles as
R = {(T₁, T₂): T₁ is similar to T₂}, is an equivalence relation. Consider three right angle triangles T₁ with sides 3, 4, 5, T₂ with sides 5, 12, 13 and T₃ with sides 6, 8, 10. Which triangles among T₁, T₂ and T₃ are related?

Solution:
Given that R = {(T₁, T₂):T₁ is similar to T₂}
(i) (T₁, T₁) ∈ R ∀ T₁ ∈ A . So, R is reflexive.
[∵ Every triangle is similar to itself. ]

(ii) If (T₁, T₂) ∈ R ⇒ (T₂, T₁) ∈ R. So, R is symmetric.
[∵ If T₁ is similar to T₂ then T₂ is similar to T₁]

(iii) (T₁, T₂), (T₂, T₃) ∈ R ⇒ (T₁, T₃) ∈ R. So, R is transitive.
[∵ If T₁ is similar to T₂ and T₂ is similar to T₃ then T₁ is similar to T₃]
The given 3 triangles with sides (3, 4, 5), (5, 12, 13) and (6, 8, 10) are right angled triangles.
But among T₁, T₂, T₃ the Corresponding sides of T₁ and T₃ are in the same ratio.
∵ \(\frac{3}{6}=\frac{4}{8}=\frac{5}{10}=\left(\frac{1}{2}\right)\)
So, triangle T₁ is similar to triangle T₃. Hence, T₁ is related to T₃.

Question 3.
Show that the relation R defined in the set A of all polygons as R = {(P₁, P₂): P₁ and P₂ have same number of sides}, is an equivalence relation. What is the set of all elements in A related to the right-angle triangle T with sides 3, 4 and 5?
Solution:
Given that A is the set of all polygons and R = {(P₁, P₂):P₁ and P₂ have same number of sides}
(i) (P₁, P₁) ∈ R. So, R is reflexive.
[∵ Same polygon has same number of sides.]

(ii) (P₁, P₂) ∈ R ⇒ (P₂, P₁) ∈ R. So, R is symmetric.
[∵ if P₁ and P₂ have same number of sides then P₂ and P₁) have same number of sides.]

(iii) (P₁, P₂), (P₂, P₃) ∈ R ⇒ (P₁, P₃) ∈ R. So, R is transitive.
[∵ (P₁, P₂), (P₂, P₃) ∈ R ⇒ P₁, P₂ and P₂ P₃ have same number of sides.
Then P₁, P₃ have same number of sides ⇒ (P₁, P₃) ∈ R. ]
Thus, R is reflexive, symmetric, transitive. Hence, R is an equivalence relation.
Further, the elements in A related to right-angled triangle (T) with sides 3, 4, 5 are those polygons which have three sides.
Set of all elements in a related to triangle T is the set of all triangles.

Question 4.
Let L be the set of all lines in XY plane and R be the relation in L defined as R = {(L₁, L₂): L₁ is parallel to L₂}. Show that R is an equivalence relation. Find the set of all lines related to the line y = 2x + 4.
Solution:
Given that L is the set of all lines and R = {(L₁, L₂): L₁ is parallel to L₂}.
(i) (L₁, L₁) ∈ R. So, R is reflexive ‘
[∵ Any line L₁ is parallel to itself. ]

(ii) If (L₁, L₂) ∈ R ⇒ (L₂, L₁) ∈ R. So, R is symmetric.
[∵ If L₁ is parallel to L₂ then L₂ is parallel to L₁]

(iii) (L₁, L₂), (L₂, L₃) ∈ R ⇒ (L₁, L₃) ∈ R. So, R is transitive.
[∵ If L₁ is parallel to L₂ and L₂ is parallel to L₃ then L₁ is parallel to L₃.]
Thus, R is reflexive, symmetric, transitive. Hence, R is an equivalence relation.
Also, set of all lines related to the line y = 2x + 4 is the set of all lines parallel to y = 2x + 4
Slope of the line y = 2x + 4 is m = 2.
Line parallel to the given line is in the form y = 2x + c, where c ∈ R .
Set of all lines related to the given line is given by y = 2x + c, where c ∈ R

Question 5.
Determine whether each of the following relations are reflexive, symmetric and transitive. Relation R in the set A of human beings in a town at a particular time given by a) R = {(x, y): x and y work at the same place)
b) R {(x, y): x and y live in the same locality)
Solution:
Given that A is the set of all human beings in a town
a) Given that R = {(x,y): x and y work at the same place)
(i) (x, x) ∈ R ∀ x ∈ A. So, R is reflexive.
[∵ Every person x, he himself works at the same place]

(ii) (x, y) ∈ R ⇒ (y, x) ∈ R. So, R is symmetric.
[∵ If x and y work at the same place then y and x also work at the same place.]

(iii) If (x, y) ∈ R and (y, z) ∈ R then (x, z) ∈ R. So, R is transitive.
[∵ If x and y work at the same place, and y and z also work at the same place, then x and z also work at the same place] .

b) Given that R={(x, y): x and y live in the same locality)
(i) (x, x) ∈ R. So, R is reflexive.[∵ Every person x, he himself lives at the same locality]
(ii) (x,y) ∈ R ⇒ (y, x) ∈ R. So, R is symmetric.
[∵ If x and y live in the same locality then y and x also live in the same locality.]
(iii) If(x, y) ∈ R and (y, z) ∈ R then (x, z) ∈ R. So, R is transitive
∵ If x and y live in the same locality, and y and z also live in the same locality, then x and z also live in the same locality.]

Question 6.
Determine whether each of the following relations are reflexive, symmetric and transitive. Relation R in the set A of human beings in a town at a particular time given by
a) R = {(x, y): x is exactly 7 cm taller than y}
b) R = {(x, y): x is wife of y} .
c) R = {(x, y): x is father of y}
Solution:
a) Given that R = {(x, y): x is exactly 7 cm taller than y}
(i) (x, x) ∉ R. So, R is not reflexive
[∵ A person cannot be 7 cm taller than himself.]

(ii) (x, y) ∈ R ⇒ (y, x) ∉ R. So, R is not symmetric .
[∵ If x is exactly 7cm taller than y then y is clearly not taller than x.]

(iii) If(x, y), (y, z) ∈ R then (x, z) ∉ R. So, R is not transitive
[∵ If x is exactly 7 cm taller than y and y is.exactly 7cm taller than z, then x is not exactly 7 cm taller than z.]

b) Given that R = {(x, y): x is wife of y}
(i) (x, x) ∉ R. So, R is not reflexive
[∵ A women can never be a wife to herself]
(ii) (x, y) ∈ R ⇒ (y, x) ∉ R. So, R is not symmetric.
[∵ If x is the wife of y then y can never be the wife of x.]
(iii) If (x, y) ∈ R and (y, z)e R then (x, z) ∉ R. So, R is not transitive
[∵ If x is wife of y and y is wife of z (this itself is a contradiction, so we need not go for the next step)]

c) Given that R = {(x, y): x is father of y}
(i) (x, x) ∉ R So, R is not reflexive
[∵ A man can never be a father to himself.]
(ii) (x, y) ∈ R ⇒ (y, x) ∉ R. So, R is not symmetric .
[∵ if x is the father of y then y can never be a father of x]
(iii) If (x, y) ∈ R and (y, z) ∈ R then (x, z) ∉ R. So, R is not transitive
[∵ if x is father of y and y is father of z (this itself is a contradiction, so we need not go for the next step)]

Question 7.
If R₁ and R₂ are equivalence relations in a set A, show that R₁ ∩ R₂ is also an equivalence relation.
Solution:
Given that R₁ and R₂ are equivalence relations .
(i) ∀ a ∈ A, (a, a) ∈ R₁, and (a, a) ∈ R₂ ⇒ (a, a) ∈ R₁ ∩ R₂. So R₁ ∩ R₂ is reflexive.
(ii) (a, b) ∈ R₁ ∩ R₂ ⇒ (a, b) ∈ R₁ and (a, b) ∈ R ₂ ⇒ (b, a) ∈ R₁ and (b, a) ∈ R₂
⇒ (b, a) ∈ R₁ ∩ R₂. So; R₁ ∩ R₂ is symmetric.
(iii) (a, b) ∈ R₁ ∩ R₂ and (b, c) ∈ R₁ ∩ R₂ ⇒ (a, c) ∈ R₁ and (a, c) ∈ R₂ ⇒ (a, c) ∈ R₁ ∩ R₂.
So, R₁ ∩ R₂ is transitive.
Thus, R₁ ∩ R₂ is reflexive, symmetric, transitive. Hence it is an equivalence relation.

Question 8.
Let R be a relation on the set A of ordered pairs of positive integers defined by (x, y)R(u, v) if and only if xv = yu. Show that R is an equivalence relation.
Solution:
(i) (x, y) R(x, y), [∵ xy = yx, ∀ (x, y) ∈ A]. So, R is reflexive.
(ii) (x, y) R (u, v) ⇒ xv = yu ⇒ uy = vx ⇒ (u, v) R (x, y). So, R is symmetric.
(iii) (x, y) R (u, v) and (u, v) R (a, b) ⇒ xv = yu and ub = va; Now we show that xb = ya
xv = yu ⇒ y = \(\frac{x v}{u}\) and ub = va ⇒ a = \(\frac{u b}{v}\)
Now ya = \(\left(\frac{\mathrm{xv}}{\mathrm{u}}\right)\left(\frac{\mathrm{ub}}{\mathrm{v}}\right)\). Hence, we have xb = ya
⇒ (x, y) R (a, b). So, R is transitive.
Thus, R is reflexive, symmetric, transitive. Hence R is an equivalence relation.

Question 9.
Let X = {1, 2, 3, 4, 5, 6, 7, 8, 9}. Let R₁ be a relation in X given by
R₁ = {(x, y): x – y is divisible by 3} and R₂ be another relation on X given by R₂ = {(x, y): {x, y} ⊂ {1, 4, 7}} or {x, y} ⊂ {2, 5, 8} or {x, y} ⊂ {3, 6, 9}}. Show that R₁ = R₂.
Solution:
R₁ Contains (x, y) such that x – y should be divisible by 3
So, elements of X are grouped according to their remainders when divided by 3:
Remainder 1: {1, 4, 7}; Remainder2: {2, 5, 8}; Remainder 0: {3, 6, 9}
Thus, R₁ relates all ordered pairs where both elements come from the same group.
Step 1: Ordered pairs in R₁
Frpm {1, 4, 7}: (1, 1), (1, 4), (1, 7), .(4, 1), (4, 4), (4, 7), (7, 1), (7, 4), (7, 7)
From {2, 5, 8}: (2, 2), (2, 5), (2, 8), (5, 2), (5, 5), (5, 8), (8, 2), (8, 5), (8, 8)
From {3, 6, 9}: (3, 3), (3, 6), (3, 9), (6, 3), (6, 6), (6, 9), (9, 3), (9, 6), (9, 9)

Step 2: Ordered pairs in R₂
By definition, R₂ consists of all ordered pairs (x, y) such that: both x, y ∈ {1, 4, 7} or both x, y ∈ {2, 5, 8} or both x, y ∈ (3, 6, 9}
This gives exactly the same ordered pairs listed above for R₁. Hence R₁ = R₂

Question 10.
Let f : X → Y be a function. Define a relation R in X given by
R = {(a, b): f(a) = f(b)}. Examine whether R is an equivalence relation or not.
Solution:
(i) ∀a ∈ X, (a, a) ∈ R, since f(a) = f(a). So, R is reflexive.
(ii) (a, b) ∈ R ⇒ f(a) = f(b) ⇒ f(b) = f(a) ⇒ (b, a) ∈ R. So, R is symmetric.
(iii) (a, b) ∈ R and (b, c) ∈ R ⇒ f(a) = f(b) and f(b) = f(c)
⇒ f(a) = f(c) ⇒ (a, c) ∈ R, So, R is transitive.
Thus, R is reflexive, symmetric, transitive. Hence R is an equivalence relation.

Question 11.
Find the number of all one-one functions from set A = {1, 2, 3} to itself.
Solution:
Total number of one-one function from {1, 2, 3} to itself is
= Total number of permutations on{1, 2, 3} = ³P₃ = 3! = 3 × 2 × 1 = 6.

Question 12.
Let A = {1, 2, 3}. Then show that the number of relations containing (1, 2) and (2, 3) which are reflexive and transitive but not symmetric is three.
Solution:
The smallest relation R₁ containing (1, 2) and (2, 3) which is reflexive and transitive but not
symmetric is R₁ = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}.
Now, if we add the pair (2, 1) to R₁ we get R₂, which is reflexive, transitive but not symmetric. Similarly, if we add the pair (3, 2) we get R₃ by adding (3, 2) to R₁ to get the desired relation. Thus, the total number of desired relations is three.(R₁, R₂, R₃)
[Note: We can not add two pairs (2, 1), (3, 2) or single pair (3, 1) to R₁ at a time, as by doing so, we need to add the remaining pair in order to maintain transitivity. Then the relation will become symmetric which is not required.]

Question 13.
Show that the number of equivalence relation in the set {1, 2, 3} containing (1, 2) and (2, 1) is two.
Solution:
The smallest equivalence relation R₁ containing (1, 2) and (2, 1) is
R₁ = {(1, 1), (2, 2), (1, 2), (2, 1)}.
Now we are left with only 4 pairs namely (2, 3), (3, 2), (1, 3) and (3, 1).
If we add any one pair (2, 3) to R₁, then for symmetry we must add (3, 2) and for transitivity we need to add (1, 3) and (3, 1). So R₂ = {(1, 1), (2, 2), (1, 2), (2, 1), (2, 3), (3, 2), (1, 3) , (3, 1), (3, 3)}.
Thus, the only equivalence relation bigger than R₁ is this universal relation.
This shows that the total number of equivalence relations containing (1, 2) and (2, 1) is two.

Question 14.
Given a non empty set X, consider P(X) which is the set of all subsets of X. Define the relation R in P(X) as follows: For subsets A, B in P(X), ARB if and only if A ⊂ B, Is R an equivalence relation on P(X)? Justify your answer.
Solution:
(i) ARA for all A e P(X) [ every set is a subset of itself]. So, R is reflexive.
(ii) Let ARB ⇒ A ⊂ B . This does not imply that B ⊂ A. So, R is not symmetric.
(iii) If ARB and BRC, then A ⊂ B and B ⊂ C ⇒ A ⊂ C ⇒ ARC . So, R is transitive.
R is not an equivalence relation as it is not symmetric.

AP Intermediate 2nd Year Maths Textbook Solutions 2026 2027

AP Inter 2nd Year Maths Study Material

AP Board Solutions Class 12 Maths

Chapter 1 Relations and Functions

• Chapter 1 Relations and Functions Exercise 1a
• Chapter 1 Relations and Functions Exercise 1b
• Chapter 1 Relations and Functions Exercise 1c
• Chapter 1 Relations and Functions MCQ

Chapter 2 Inverse Trigonometric Functions

• Chapter 2 Inverse Trigonometric Functions Exercise 2a
• Chapter 2 Inverse Trigonometric Functions Exercise 2b
• Chapter 2 Inverse Trigonometric Functions Exercise 2c
• Chapter 2 Inverse Trigonometric Functions MCQ

Chapter 3 Matrices

• Chapter 3 Matrices Exercise 3a
• Chapter 3 Matrices Exercise 3b
• Chapter 3 Matrices Exercise 3c
• Chapter 3 Matrices Exercise 3d
• Chapter 3 Matrices MCQ

Chapter 4 Determinants

• Chapter 4 Determinants Exercise 4a
• Chapter 4 Determinants Exercise 4b
• Chapter 4 Determinants Exercise 4c
• Chapter 4 Determinants Exercise 4d
• Chapter 4 Determinants Exercise 4e
• Chapter 4 Determinants Exercise 4f
• Chapter 4 Determinants MCQ

Chapter 5 Continuity and Differentiability

• Chapter 5 Continuity and Differentiability Exercise 5a
• Chapter 5 Continuity and Differentiability Exercise 5b
• Chapter 5 Continuity and Differentiability Exercise 5c
• Chapter 5 Continuity and Differentiability Exercise 5d
• Chapter 5 Continuity and Differentiability Exercise 5e
• Chapter 5 Continuity and Differentiability Exercise 5f
• Chapter 5 Continuity and Differentiability Exercise 5g
• Chapter 5 Continuity and Differentiability Exercise 5h
• Chapter 5 Continuity and Differentiability MCQ

Chapter 6 Application of Derivatives

• Chapter 6 Application of Derivatives Exercise 6a
• Chapter 6 Application of Derivatives Exercise 6b
• Chapter 6 Application of Derivatives Exercise 6c
• Chapter 6 Application of Derivatives Exercise 6d
• Chapter 6 Application of Derivatives MCQ

Chapter 7 Integrals

• Chapter 7 Integrals Exercise 7a
• Chapter 7 Integrals Exercise 7b
• Chapter 7 Integrals Exercise 7c
• Chapter 7 Integrals Exercise 7d
• Chapter 7 Integrals Exercise 7e
• Chapter 7 Integrals Exercise 7f
• Chapter 7 Integrals Exercise 7g
• Chapter 7 Integrals Exercise 7h
• Chapter 7 Integrals Exercise 7i
• Chapter 7 Integrals Exercise 7j
• Chapter 7 Integrals MCQ

Chapter 8 Application of Integrals

• Chapter 8 Application of Integrals Exercise 8a
• Chapter 8 Application of Integrals Exercise 8b
• Chapter 8 Application of Integrals MCQ

Chapter 9 Differential Equations

• Chapter 9 Differential Equations Exercise 9a
• Chapter 9 Differential Equations Exercise 9b
• Chapter 9 Differential Equations Exercise 9c
• Chapter 9 Differential Equations Exercise 9d
• Chapter 9 Differential Equations Exercise 9e
• Chapter 9 Differential Equations Exercise 9f
• Chapter 9 Differential Equations MCQ

Chapter 10 Vector Algebra

• Chapter 10 Vector Algebra Exercise 10a
• Chapter 10 Vector Algebra Exercise 10b
• Chapter 10 Vector Algebra Exercise 10c
• Chapter 10 Vector Algebra Exercise 10d
• Chapter 10 Vector Algebra Exercise 10e
• Chapter 10 Vector Algebra MCQ

Chapter 11 Three Dimensional Geometry

• Chapter 11 Three Dimensional Geometry Exercise 11a
• Chapter 11 Three Dimensional Geometry Exercise 11b
• Chapter 11 Three Dimensional Geometry MCQ

Chapter 12 Linear Programming

• Chapter 12 Linear Programming Exercise 12a
• Chapter 12 Linear Programming MCQ

Chapter 13 Probability

• Chapter 13 Probability Exercise 13a
• Chapter 13 Probability Exercise 13b
• Chapter 13 Probability Exercise 13c
• Chapter 13 Probability MCQ

AP Inter 2nd Year Maths Weightage Blue Print 2026-2027

Referring to the AP Inter 2nd Year Maths Study Material Chapter 13 Probability Exercise 13a Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Probability Solutions Exercise 13a

I.

Question 1.
Given that E and F are 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 that P(E) = 0.6, P(F) = 0.3, P(E ∩ F) = 0.2
∴ P(E/F) = \(\frac{P(E \cap F)}{P(F)}=\frac{0.2}{0.3}=\frac{2}{3}\) and
P(F/E) = \(\frac{P(F \cap E)}{P(E)}=\frac{0.2}{0.6}=\frac{1}{3}\)

Question 2.
Compute P(A|B), if P(B) = 0.5 and P(A ∩ B) = 0.32
Solution:
Given that P(B) = 0.5 and P(A ∩ B) = 0.32
∴ P(A|B) = \(\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}\)
= \(\frac{0.32}{0.5}\) = \(\frac{32}{50}\)
= \(\frac{16}{25}\)

Question 3.
A and B are two events such that P (A) ≠ 0. Find P(B|A), if
(i) A is a subset of B
(ii) A ∩ B = Φ
Solution:
Given,P(A) ≠ 0
(i) A is a subset of B ⇒ A ∩ B = A – Then P(A ∩ B) = P(B ∩ A) = P(A)
∴P(B|A) = \(\frac{\mathrm{P}(\mathrm{B} \cap \mathrm{A})}{\mathrm{P}(\mathrm{A})}=\frac{\mathrm{P}(\mathrm{A})}{\mathrm{P}(\mathrm{A})}\) = 1
(ii) A ∩ B Φ ⇒ P (A ∩ B) = 0
∴ P(B|A) = \(\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}\) = 0

Question 4.
If a leap year is selected at random, what is the chance that it will contain 53 tuesdays?
Solution:
In a leap year, there are 366 days i.e., 52 weeks and 2 days.
In 52 weeks, there are 52 Tuesdays.
∴ the probability that the leap year will contain 53 Tuesdays is equal to the probability that the remaining 2 days will be Tuesdays.
The remaining 2 days can be any of the following:
Monday and Tuesday or Tuesday and Wednesday or Wednesday and Thursday or Thursday and Friday or Friday and Saturday or Saturday and Sunday or Sunday and Monday.
Total number of cases = 7
Favourable cases = 2
Probability that a leap year will have 53 Tuesdays = 2/7

II.

Question 1.
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:
Given that P (A) = 0.8, P (B) = 0.5 and P (B|A) = 0.4
(i) Now P (B|A) = 0.4 ⇒ \(\frac{\mathrm{P}(\mathrm{B} \cap \mathrm{A})}{\mathrm{P}(\mathrm{A})}\) = 0.4
⇒ \(\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{0.8}\) = 0.4
⇒ P(A ∩ B) = 0.8 × 0.4 = 0.32

(ii) P(A|B) = \(\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}=\frac{0.32}{0.5}\)
= \(\frac{32}{100} \times \frac{10}{5}\) = \(\frac{64}{100}\) = 0.64

(iii) P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = 0.8 + 0.5 – 0.32 = 1.3- 0.32 = 0.98.

Question 2.
Evaluate P(A ∪ B), if 2P(A) = P(B) = \(\frac{5}{13}\) and P(A | B) = \(\frac{5}{5}\)
Solution:
Given that 2P(A)= P(B) = \(\frac{5}{13}\) =>P(A) = \(\frac{5}{26}\), P(B) = \(\frac{5}{13}\)
Now P(A|B) = \(\frac{2}{5}\) ⇒ \(\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}=\frac{2}{5}\)
⇒ P(A ∩ B) = \(\frac{2}{5}\)P(B) = \(\frac{2}{5}\) × \(\frac{5}{13}\) = \(\frac{2}{13}\)
∴ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{5}{26}\) + \(\frac{5}{13}\) – \(\frac{2}{13}\) = \(\frac{5+10-4}{26}\)
= \(\frac{11}{26}\)

Question 3.
If P(A) = \(\frac{6}{11}\), P(B) = \(\frac{5}{11}\) andP(A ∪ B) = \(\frac{7}{11}\), find
(i) P(A ∩ B)
(ii) P(A|B)
(iii) P(B|A)
Solution:
(i) Given that P(A ∪ B) = \(\frac{7}{11}\)
⇒ P(A) + P(B) – P(A ∩ B) = \(\frac{7}{11}\)
⇒ \(\frac{6}{11}\) + \(\frac{5}{11}\) – P(A ∩ B) = \(\frac{7}{11}\)
⇒ P(A ∩ B) = 1 – \(\frac{7}{11}\)= \(\frac{4}{11}\)

(ii) P(A/B) = \(\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}\)
= \(\frac{\frac{4}{11}}{\frac{5}{11}}\) = \(\frac{4}{5}\)

(iii) P(B/A) = \(\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}\)
= \(\frac{\frac{4}{11}}{\frac{6}{11}}\) = \(\frac{2}{3}\)

Question 4.
Determine P(E|F) when a coin is tossed three times, where
(i) E : head on third toss , F : heads on first two tosses
Solution:
We know that the sample space for the random experiment ‘a coin is tossed three times’ is
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} ⇒ n(S) = 8
(i) E: Head on third toss ⇒ E = {HHH, HTH, THH, TTH} ⇒ n(E) = 4
F: Heads on first two tosses ⇒ F = {HHH, HHT} ⇒ n(F) = 2
Hence E ∩ F = {HHH} ⇒ n(E ∩ T) = 1
Now P(E) = \(\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}\) = \(\frac{4}{8}\) = \(\frac{1}{2}\)
P(F) = \(\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}\) = \(\frac{2}{8}\) = \(\frac{1}{4}\)
P(E ∩ F) = \(\frac{1}{8}\)
∴ P(E/F) = \(\frac{P(E \cap F)}{P(F)}\)
= \(\frac{\frac{1}{8}}{\frac{1}{4}}\) = \(\frac{1}{2}\)

Question 5.
Determine P(E|F) when a coin Is tossed three times, where
E : at least two heads , F: at most two heads
Solution:
E : at least two heads ⇒ E = {HHII, HHT, HTH, THH} = n(E) = 4
F: at most two beads ⇒ F = {HI-IT, HTH, THH, HTT, THT, TTH, TTT} ⇒ n(F) 7
Hence E ∩ F = {HHT, HTH, THH} ⇒ n(E ∩ F) = 3
Now P(E) = \(\frac{4}{8}\) = \(\frac{1}{2}\) ,
P(F) = \(\frac{7}{8}\),
P(E ∩ F) = \(\frac{\mathrm{n}(\mathrm{E} \cap \mathrm{F})}{\mathrm{n}(\mathrm{S})}\) = \(\frac{3}{8}\)
∴ P(E/F) = \(\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}\)
= \(\frac{\frac{3}{8}}{\frac{7}{8}}\) = \(\frac{3}{7}\)

Question 6.
Determine P(EIF) when a coin is tossed three times, where
E : at most two tails , F : at least one tail.
Solution:
E : at moat two tails ⇒ E = {TTH, THT, HTT, THH, HTH, HHT, HHH} ∴ n(E) = 7
F: at least one tail ⇒ F = {THH, HTH, HHT, TTH, THT, HTT, TTT) ∴ n(F) = 7
Hence E ∩ F = {TTH, THT, HTT, THH, HTH, HHT} ⇒ n(E ∩ F) = 6
Now P(E) = \(\frac{7}{8}\), P(F) = \(\frac{\mathrm{n}(\mathrm{F})}{\mathrm{n}(\mathrm{S})}\) = \(\frac{7}{8}\),
P(E ∩ F) = \(\frac{6}{8}\) = \(\frac{3}{4}\)
∴ P(E/F) = \(\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}\) = \(\frac{\frac{3}{4}}{\frac{7}{8}}\)
= \(\frac{3}{4} \times \frac{8}{7}\) = \(\frac{6}{7}\)

Question 7.
Find P(E|F) when two coins are tossed once, where
(i) E : tail appears on one coin, F : one coin shows head
(ii) E : no tail appears, F : no head appears.
Solution:
When ‘two coins are tossed once’ in the sample space S = {HH, HT, TH, TT} ⇒ n(S) = 4
(i) E : tail appears on one coin ⇒ E = {HT, TH} ⇒ n(E) = 2
F : one coin shows head ⇒ F = {HT, TH} ⇒ n(F) = 2
Hence E ∩ F = {HT, TH} ⇒ n(E ∩ F) = 2
Now P(E) = \(\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\)
P(F) = \(\frac{2}{4}\) = \(\frac{1}{2}\)
P(E ∩ F) = \(\frac{2}{4}\) = \(\frac{1}{2}\)
∴ P(E/F) = \(\frac{P(E \cap F)}{P(F)}\)
= \(\frac{\frac{1}{2}}{\frac{1}{2}}\) = 1

(ii) E: no tail = {HH}; F: no head = {TT}
∴ E ∩ F = Φ; P(F) = 1 and P (E ∩ F) = 0
∴ P(E|F) = \(\frac{P(E \cap F)}{P(F)}\)
= \(\frac{0}{1}\) = 0

Question 8.
Find P(E|F) when a die is thrown three times,
E : 4 appears on the third toss,
F : 6 and 5 appears respectively on first two tosses.
Solution:
When a die is thrown 3 times, n(S) = 6³ = 216
E: 4 appears on third toss = {(1, 1, 4) (1, 2, 4)… (1, 6, 4) (2, 1, 4) (2, 2, 4)…
(2, 6, 4)(3, 1, 4) (3, 2, 4) … (3 6 4) (4, 1, 4)(4, 2, 4)… (4, 6, 4)(5, 1, 4) (5, 2, 4)… (5, 6, 4) (6, 1, 4) (6, 2, 4)…
F: 6 and 5 appear respectively on first two tossess
= {(6, 5, 1) (6, 5, 2) (6, 5, 3) (6, 5, 4) (6, 5, 5) (6, 5, 6)}
Hence E ∩ F = {(6, 5, 4)}
Now P(F) = \(\frac{6}{216}\) and P(E ∩ F) = \(\frac{1}{216}\)
∴ P(E|F) = \(\frac{P(E \cap F)}{P(F)}=\frac{\left(\frac{1}{2 \times 6}\right)}{\left(\frac{6}{2 \times 6}\right)}\)
= \(\frac{1}{6}\)

Question 9.
Determine P(E|F) when Mother, father and son line up at random for a family picture E : son on one end, F : father in middle. Find P(E|F)
Solution:
Let m, f and s denote the mother, father and son respectively.
The sample space is S = {mfs, msf, fms, fsm, smf, sfm} ∴ n(S) = 6
E: son on one end ⇒ E = {mfs, fins, smf, sftn} ⇒ n(E) = 4 ⇒ P(E) = \(\frac{4}{6}\) = \(\frac{2}{3}\)
F: father in middle F= {mfs, sfm} ⇒ n(F) = 2 ⇒ P(F) = \(\frac{2}{6}\) = \(\frac{1}{3}\)
Hence, E ∩ F = {mfs, sfm} ⇒ n(E ∩ F) = 2 ⇒ P(E ∩ F) = \(\frac{2}{6}\) = \(\frac{1}{3}\)
∴ P(E|F) = \(\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}=\frac{1 / 3}{1 / 3}\) = 1.

Question 10.
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:
Let the first (elder) child be denoted by capital letter and the second (younger) by a small
letter. The sample space is S = {Bb, Bg, Gb, Gg} ⇒ n(S) = 4
Let E: both children are girls, then E = {Gg.} ⇒ n(E) = 1 ⇒ P(E) = \(\frac{1}{4}\)
(i) Let F: the youngest child is a girl, then
F = {Bg, Gg} ⇒ n(F) = 2 ⇒ P(F) = \(\frac{\mathrm{n}(\mathrm{F})}{\mathrm{n}(\mathrm{S})}\) = \(\frac{2}{4}\)
Hence E ∩ F = {Gg} ⇒ n(E ∩ F) = 1 ⇒ P(E ∩ F) = \(\frac{1}{4}\)
∴ P(E|F) = \(\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}=\frac{\frac{1}{4}}{\frac{2}{4}}\) = \(\frac{1}{2}\)

(ii) Let F : at least one child is a girl then F {Bg, Gb, Gg} ⇒ n(F) = 3
⇒ P(F) = \(\frac{\mathrm{n}(\mathrm{F})}{\mathrm{n}(\mathrm{S})}\) = \(\frac{3}{4}\)
Hence, E ∩ F = {Gg}:. n(E ∩ F) = 1 ⇒ P(E ∩ F) = \(\frac{1}{4}\)
∴ Required probability is P(E|F) = \(\frac{P(E \cap F)}{P(F)}=\frac{\frac{1}{4}}{\frac{3}{4}}\) = \(\frac{1}{3}\)

Question 11.
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:
Total number of questions = 300 + 200 + 500 + 400 = 1400
∴ n(S) = 1400
Let E : selected question is easy
F : selected question is a multiple choice question then
E ∩ F : selected question is an easy multiple choice question
Thus n(E ∩ F) = 500, n(F) = 500 + 400 = 900
∴ P(E ∩ F) = \(\frac{n(\mathrm{E} \cap \mathrm{F})}{\mathrm{n}(\mathrm{S})}=\frac{500}{1400}\) and P(F) = \(\frac{\mathrm{n}(\mathrm{F})}{\mathrm{n}(\mathrm{S})}=\frac{900}{1400}\)
∴ Required probability is P(E/F) = \(\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}=\)
= \(\frac{\frac{500}{1400}}{\frac{900}{1400}}\) = \(\frac{500}{1400} \times \frac{1400}{900}\)
= \(\frac{5}{9}\)

Question 12.
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:
Sample space of throwing two dice is
S = {( x, y): x, y e {1, 2, 3, 4, 5, 6}} ⇒ n(S) = 6 × 6 = 36
Let E : the sum of numbers on the dice is 4 ⇒ E = {(1, 3), (2, 2), (3,1)}
⇒ n(E) = 3 ⇒ P(E) = \(\frac{n(E)}{n(S)}=\frac{3}{36}\)
Let F : numbers appearing on the dice are different
⇒ F = S – {(1, 1),(2, 2),(3, 3),(4, 4),(5, 5),(6, 6)}
∴ n(F) = 36 – 6 = 30 ⇒ P(F) = \(\frac{30}{36}\)
Also, E ∩ F = {(1,3), (3,1)} ⇒ n(E ∩ F) = 2
∴ P(E ∩ F) = \(\frac{n(E \cap F)}{n(S)}=\frac{2}{36}\)
∴ Required probability is P(E|F) = \(\frac{P(E \cap F)}{P(F)}\)
= \(\frac{\frac{2}{36}}{\frac{30}{36}}=\) = \(\frac{2}{36} \times \frac{36}{30}\)
= \(\frac{1}{15}\)

Question 13.
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:
The sample space of given random experiment is
S = {(1, H), (1, T), (2, H), (2, T), (3,1), (3,2), (3,3), (3, 4) (3, 5), (3, 6), (4, H), (4, T), (5, H), (5, T), (6, 1), (6, 2),(6, 3), (6, 4), (6, 5), (6, 6)}
⇒ n(S) = 20
Let E: the coin shows a tail ⇒ E= {(1, T), (2, T), (4, T), (5, T)} ⇒ n(E) = 4
Let F: at least one die shows a 3
⇒ F = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (6, 3)} ⇒ n(F) = 7
Hence, E ∩ F = Φ ⇒ n(E ∩ F) = 6 ⇒ P(E ∩ F) = \(\frac{n(E \cap F)}{n(S)}=\frac{0}{20}\) = 0.
∴ Required probability is P(E|F) = \(\frac{P(E \cap F)}{P(F)}=\frac{0}{P(F)}\) = 0.

Question 14.
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:
If a couple has two children, then the sample space is S = {(B,B), (B,G), (QB), (G,G)}
(i) Let E and F respectively denote the events that both children are male and atleast onechildren is a male. .
E ∩ F ={(G,G)} ⇒ P(E ∩ F) = \(\frac{1}{4}\)
P(E) = \(\frac{1}{4}\)
P(F) = \(\frac{3}{4}\)
⇒ P(E|F) = \(\frac{P(E \cap F)}{P(F)}\) = \(\frac{1 / 4}{3 / 4}=\frac{1}{4} \times \frac{4}{3}\) = \(\frac{1}{3}\)

(ii) Let C and D respectively denote the events that both children are females and the elder child is a female.
C = {(G, G)} ⇒ P(C) = \(\frac{1}{4}\); D = {(G, B),(G, G)} ⇒ P(D) = \(\frac{2}{4}\)
C ∩ D = {(G, G)} ⇒ P(C ∩ D) = \(\frac{1}{4}\)
∴ P(C|D) = \(\frac{\mathrm{P}(\mathrm{C} \cap \mathrm{D})}{\mathrm{P}(\mathrm{D})}=\frac{1 / 4}{2 / 4}\) = \(\frac{1}{2}\)

Question 15.
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 \(\frac{1}{2}\)).
Solution:
The total number of determinants of second order with each element being 0 or 1 is (2)⁴ = 16.
The value of determinant is positive in the following cases. \(\left|\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right|,\left|\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right|,\left|\begin{array}{ll}
1 & 0 \\
1 & 1
\end{array}\right|\)
∴ Required probability = \(\frac{3}{16}\)

Question 16.
An electronic assembly consists of two sub systems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known:
P(A fails) = 0.2, P(B fails alone) = 0.15, P(A and B fail) = 0.15
Evaluate the following probabilities (i) P(A fails|B has failed)
(ii) P(A fails alone)
Solution:
Let the event in which A fails and B fails he denote by E_A and E_B
P(E_A) = 0.2,P(E_A ∩ E_B) = 0.15 .
P(B fails alone) = P(E_B) – P(E_A ∩ E_B)
∴ 0.15 = P(E_B) – 0.15
∴ P(E_B) = 0.3
(i) P(E_A |E_B) = \(\frac{P\left(E_A \cap E_B\right)}{P\left(E_B\right)}\) = \(\frac{0.15}{0.3}\) = 0.5
(ii) P(A fails alone) = P(E_A) – P(E_A and E_B) = 0.2 – 0.15 = 0.05

III.

Question 1.
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:
Let x denote the outcome on black die and y denote the outcome on red die.
The sample space is S = {(x, y) : x, y ∈ {1, 2, 3, 4, 5, 6}} ⇒ n(S) = 6 × 6 = 36
(a) LetE : sum x + y > 9 ⇒ x + y = 10, 11, 12
⇒ E = {(6,4), (6,5), (6,6), (5, 5), (5,6), (4,6)}
F : black die resulted in a 5. ⇒ F = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
∴ n(F) = 6 ⇒ P(F) = \(\frac{6}{36}\)
Hence, E ∩ F = {(5, 5), (5, 6)}
∴ n(E ∩ F) = 2 ⇒ P(E ∩ F) = \(\frac{2}{36}\)
∴ Required probability is P(E|F) = \(\frac{P(E \cap F)}{P(F)}=\frac{\frac{2}{36}}{\frac{6}{36}}\) = \(\frac{6}{3}\)

(b) Let E : sum x + y = 8 ⇒ E = {(2, 6), (3, 5), (4,4), (5, 3), (6, 2)}
F : red die resulted in a number less than 4
⇒ F = {(x, y) : x ∈ {1, 2, 3, 4, 5, 6} and y ∈ {1, 2, 3}}
= {(1,1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1),(3, 2), (3, 3), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3), (6, 1), (6, 2), (6, 3)}
∴ n(F) = 6 × 3 = 18 ⇒ P(F) = \(\frac{18}{36}\)
Also, E ∩ F = {(5, 3), (6, 2)}
∴ n(E ∩ F) = 2 ⇒ P(E ∩ F) ⇒ P(E ∩ F) = \(\frac{2}{36}\)
∴ Required probability is P(E/F) = \(\frac{P(E \cap F)}{P(F)}=\frac{\frac{2}{36}}{\frac{18}{36}}\) = \(\frac{1}{9}\)

Question 2.
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:
Sample space S = {1, 2, 3, 4, 5, 6} ⇒ n(S) = 6
Given: Event E = {1, 3, 5}, F = {2, 3}, G = {2, 3, 4, 5}
(i) E ∩ F = {3}, n(E) = 3, n(F) = 2, n(G) = 4, n(E ∩ F) = 1
∴ P(E) = \(\frac{n(E)}{n(S)}=\frac{3}{6}\), P(F) = \(\frac{2}{6}\), P(G) = \(\frac{4}{6}\), P(E ∩ F) = \(\frac{1}{6}\)
∴ P(E|F) = \(\frac{P(E \cap F)}{P(F)}=\frac{\frac{1}{6}}{\frac{2}{6}}\) = \(\frac{1}{2}\) and P(F|E) = \(\frac{P(F \cap E)}{P(E)}=\frac{\frac{1}{6}}{\frac{3}{6}}\) = \(\frac{1}{3}\)

(ii) Now E ∩ G = {3, 5} ⇒ n(E ∩ G) = 2 ⇒ P(E ∩ G) = \(\frac{2}{6}\)
∴ P(E|G) = \(\frac{P(E \cap G)}{P(G)}=\frac{\frac{2}{6}}{\frac{4}{6}}\) = \(\frac{1}{2}\) and P(G|E) = \(\frac{P(E \cap G)}{P(E)}=\frac{\frac{2}{6}}{\frac{3}{6}}\) = \(\frac{2}{3}\)

(iii) Also E ∪ F = {1, 2, 3, 5}
⇒ E ∩ F = {3} (E ∪ F) ∩ G = {1, 2, 3, 5} ∩ {2, 3, 4, 5} = {2, 3, 5}
⇒ n((E ∪ F) ∩ G) = 3 ⇒ P((E ∪ F) ∩ G) = \(\frac{3}{6}\)
Now (E ∩ F) ∩ G = {3} ∩ {2, 3, 4, 5} = {3}
⇒ n((E ∩ F) ∩ G) = 1 ⇒ P((E ∩ F) ∩ G) = \(\frac{1}{6}\)
∴ P((E ∪ F)|G) = \(\frac{P((E \cup F) \cap G)}{P(G)}=\frac{\frac{3}{6}}{\frac{4}{6}}\) = \(\frac{3}{4}\)
P((E ∩ F)|G) = \(\frac{P((E \cap F) \cap G)}{P(G)}=\frac{\frac{1}{6}}{\frac{4}{6}}\) = \(\frac{1}{4}\)