Let A = {2, 3, 5, 6}. Then, which of the following relations is transitive only? -Maths 9th

Let A = {2, 3, 5, 6}. Then, which of the following relations is transitive only? -Maths 9th
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1 Answer

Answer :

(d) R = {(2, 3), (3, 5), (2, 5)}.
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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. -Maths 9th

Description : 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. -Maths 9th

Last Answer : Reflexive: R = {(a, b) : b = a +1} = {(a, a + l) : a, a + 1∈{l, 2, 3, 4, 5, 6}} = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)} ⇒ R is not reflexive since (a, a) ∉R for all a. Symmetric: R is not symmetric as (a ... as (a, b) ∈ R and (b, c) ∈ R but (a, c) ∉ R e.g., (1, 2) ∈ R (2, 3) ∈ R but (1, 3) ∉R

Let A = {a, b, c, d} and B = {x, y, z}. Which of the following are relations from A to B ? -Maths 9th

Description : Let A = {a, b, c, d} and B = {x, y, z}. Which of the following are relations from A to B ? -Maths 9th

Last Answer : (i) Yes. (ii) No, because in the ordered pair (a, d), a ∈ A and d ∉ B. (iii) No, because in (y, d), y ∈ B. (iv) No. because here the first entries in all the ordered pairs are in ... ) No, because the element z is not an ordered pair. (vii) No, because the elements of the set are not ordered pairs.

If n(A) = 5 and n(B) = 7, then the number of relations on A × B is -Maths 9th

Description : If n(A) = 5 and n(B) = 7, then the number of relations on A × B is -Maths 9th

Last Answer : (b) 235n(A) = 5 and n(B) = 7 ∴ n(A × B) = 5 × 7 = 35 Total number of relations from A to B = Total number of subsets of (A × B) = 235.

If R is a relation on a finite set A having n elements, then the number of relations on A is -Maths 9th

Description : If R is a relation on a finite set A having n elements, then the number of relations on A is -Maths 9th

Last Answer : (d) \(2^{n^2}\)Set A has n elements ⇒ n(A) = n ⇒ A × A has n × n = n2 elements ∴ Number of relations on A = Number of subsets of A × A = \(2^{n^2}\)

Let A = {1, 2, 3, 4}, B = {5, 6, 7, 8}. Then R = {(1, 5), (1, 7), (2, 6)} is a relation from set A to B defined as : -Maths 9th

Description : Let A = {1, 2, 3, 4}, B = {5, 6, 7, 8}. Then R = {(1, 5), (1, 7), (2, 6)} is a relation from set A to B defined as : -Maths 9th

Last Answer : (d) R = {(a, b) : b/a is odd} Since (2, 6) ∈R, the relation a and b are odd does not exist. Since (1, 5) and (1, 7) ∈R, the relation a and b are even does not exist. None of the ... \(rac{7}{1}\) = 7, \(rac{6}{2}\) = 3, quotients being all odd numbers, the relation b/a is odd exists.

Let A = {(2, 5, 11)}, B = {3, 6, 10} and R be a relation from A to B defined by R = {(a, b) : a and b are co-prime}. Then R is -Maths 9th

Description : Let A = {(2, 5, 11)}, B = {3, 6, 10} and R be a relation from A to B defined by R = {(a, b) : a and b are co-prime}. Then R is -Maths 9th

Last Answer : (c) {(2, 3), (5, 3), (5, 6), (11, 3), (11, 6), (11, 10)}

Which of the following relations is only symmetric? -Maths 9th

Description : Which of the following relations is only symmetric? -Maths 9th

Last Answer : (c) “is perpendicular to” on a set of a coplanar lines(a) Let a, b, c ∈ A where A is a set of real numbers. Then R = {(a, b) : a ≤ b, a, b ∈ A} is : Reflexive: a ≤ a ⇒ (a, a) ∈ R (Yes) Symmetric: a ≤ b ⇒ (a, b) ∈ R, but a ≤ b ⇒ a < b or a = b ⇒ b = a but b \( ot

Let u = (log2x)^2 – 6(log2x) + 12, where x is a real number. Then the equation x^u = 256 has : -Maths 9th

Description : Let u = (log2x)^2 – 6(log2x) + 12, where x is a real number. Then the equation x^u = 256 has : -Maths 9th

Last Answer : ⇒ u=(log2 x)2−log2 x+12 ⇒ We can take log2 x=y ⇒ Then equation becomes y2−6y+12=u ⇒ Given that xu=256 Applying log we get, ⇒ ulog2 x=8 ∴ u=log2 x8 =y8 So our equation becomes, ⇒ y2−6y+12=y8 ⇒ ... We get y=2 ⇒ So, log2 x=2 ⇒ x=22 ∴ x=4 ∴ The given equation has exactly one solution for x

Define the types of relations and their examples. -Maths 9th

Description : Define the types of relations and their examples. -Maths 9th

Last Answer : There are various types of relations:Let A be a non-empty set. Then, a relation R on A is said to be Reflexive if (a, a) ∈ R for each a ∈ A, i.e., if a R a for each a ∈ A. For example, the relation is as strong as is reflexive ... a || b ⇒ b || a a || b, b || c ⇒ a || c.

Determine the domain and range of the following relations: (i) {(–3, 1), (–1, 1), (1, 0), (3, 0)} -Maths 9th

Description : Determine the domain and range of the following relations: (i) {(–3, 1), (–1, 1), (1, 0), (3, 0)} -Maths 9th

Last Answer : (i) Domain = {-3, -1, 1, 3}, Range = {0, 1} (ii) Domain = {x : x is a multiple of 3} = {3n : n ∈ Z} Range = {y : y is a multiple of 5} = {5n : n ∈ Z} (iii) Relation = {(x, x2) : x is a prime number ... (7, 49), (11, 121), (13, 169)} Domain = {2, 3, 5, 7, 11, 13}, Range = {4, 9, 25, 49, 121, 169}

I is the set of integers. Describe the following relations in words, giving its domain and range. -Maths 9th

Description : I is the set of integers. Describe the following relations in words, giving its domain and range. -Maths 9th

Last Answer : R = {(0, 0), (1, – 1), (2, – 2), (3, – 3) ...} = {(x, y) : y = – x, x ∈ W} Domain = {0, 1, 2, 3, ....} = W, Range = {...,– 3, – 2, – 1, 0}

Write ‘yes’ if each of the following relations is an equivalence relation. -Maths 9th

Description : Write ‘yes’ if each of the following relations is an equivalence relation. -Maths 9th

Last Answer : (i) is parallel to' is reflexive, because any line is parallel to itself. Symmetric, because if line l is parallel to line m, then line m is parallel to line l. Transitive, because if l || m ... y cannot be a factor of x. (v) No, because the relation is reflexive and transitive but not symmetric.

Which of the following are relations from B to A where A = {a, b, c, d} and B = {x, y, z}? -Maths 9th

Description : Which of the following are relations from B to A where A = {a, b, c, d} and B = {x, y, z}? -Maths 9th

Last Answer : (c) (i), (ii) and (iv)The set of ordered pairs {(b, y), (z, a), (x, c)} does not state a relation from B to A as the ordered pair (b, y) has the first element ‘b’ from set A, whereas it should be from set B.

Given A = {–2, –1, 0, 1, 2}, which of the following relations on A have both domain and range equal to A? -Maths 9th

Description : Given A = {–2, –1, 0, 1, 2}, which of the following relations on A have both domain and range equal to A? -Maths 9th

Last Answer : (d) (i) and (iii)Let us examine the domain and range of the each relation individually: (i) R : is equal to means R = {(a, b) : a = b, a ∈ A, b ∈A} ∴ R = {(-2, -2), (-1, -1), (0, 0), (1, ... = {-2, -1, 0, 1} and range = {-1, 0, 1, 2}. ∴ Relations given in (i) and (iii) satisfy the given condition.

Consider the following relations R = {(x, y) | x, y are real numbers and x = wy for some rational number w}; -Maths 9th

Description : Consider the following relations R = {(x, y) | x, y are real numbers and x = wy for some rational number w}; -Maths 9th

Last Answer : (c) S is an equivalence relation but R is not an equivalence relationR = {(x, y) | x, y ∈ R, x = wy, w is a rational number} Reflexive: x R x ⇒ x = wx ⇒ w = 1, (a rational number) Hence R is reflexive. Symmetric ... \(rac{r}{s}\) ⇒ \(rac{m}{n}\) S \(rac{r}{s}\) (True)∴ S is an equivalence relation.

Let y varies directly as x. If y = 12 when x = 4, then write a linear equation. What is value of y when x = 5 ? -Maths 9th

Description : Let y varies directly as x. If y = 12 when x = 4, then write a linear equation. What is value of y when x = 5 ? -Maths 9th

Last Answer : Linear equation

Let y varies directly as x. If y = 12 when x = 4, then write a linear equation. What is value of y when x = 5 ? -Maths 9th

Description : Let y varies directly as x. If y = 12 when x = 4, then write a linear equation. What is value of y when x = 5 ? -Maths 9th

Last Answer : Linear equation

Let ABC be a triangle of area 16 cm^2 . XY is drawn parallel to BC dividing AB in the ratio 3 : 5. If BY is joined, then the area of triangle BXY is -Maths 9th

Description : Let ABC be a triangle of area 16 cm^2 . XY is drawn parallel to BC dividing AB in the ratio 3 : 5. If BY is joined, then the area of triangle BXY is -Maths 9th

Last Answer : answer:

Let P(–3, 2), Q(–5, –5), R(2, –3) and S(4, 4) be four points in a plane. Then show that PQRS is a rhombus. Is it a square ? -Maths 9th

Description : Let P(–3, 2), Q(–5, –5), R(2, –3) and S(4, 4) be four points in a plane. Then show that PQRS is a rhombus. Is it a square ? -Maths 9th

Last Answer : Let P(1, -1), Q \(\big(rac{-1}{2},rac{1}{2}\big)\) and R(1,2) be the vertices of the ΔPQR.Then, PQ = \(\sqrt{\big(rac{-1}{2}-1\big)^2+\big(rac{1}{2}+1\big)^2}\) = \(\sqrt{rac{9}{4}+rac{9}{4}} ... {3\sqrt2}{2}\)PR = \(\sqrt{(1-1)^2+(2+1)^2}\) = \(\sqrt9\) = 3∵ PQ = QR, the triangle PQR is isosceles.

Let R be a relation from A = {1, 2, 3, 4, 5, 6} to B = {1, 3, 5} which is defined as “x is less than y”. -Maths 9th

Description : Let R be a relation from A = {1, 2, 3, 4, 5, 6} to B = {1, 3, 5} which is defined as “x is less than y”. -Maths 9th

Last Answer : R = {a, b : a < b, a ∈ A, b ∈ B}, where A = {1, 2, 3, 4, 5, 6} and B = {1, 3, 5}. ∴ R = {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)} Domain of R = {1, 2, 3, 4} Range of R = {3, 5} Codomain of R = {1, 3, 5}.

Let x be the mean of x1, x2,….,xn and y be the mean of y1, y2, ……,yn the mean of z is x1, x2,….,xn , y1, y2, ……,yn then z is equal to -Maths 9th

Description : Let x be the mean of x1, x2,….,xn and y be the mean of y1, y2, ……,yn the mean of z is x1, x2,….,xn , y1, y2, ……,yn then z is equal to -Maths 9th

Last Answer : NEED ANSWER

Let x be the mean of x1, x2,….,xn and y be the mean of y1, y2, ……,yn the mean of z is x1, x2,….,xn , y1, y2, ……,yn then z is equal to -Maths 9th

Description : Let x be the mean of x1, x2,….,xn and y be the mean of y1, y2, ……,yn the mean of z is x1, x2,….,xn , y1, y2, ……,yn then z is equal to -Maths 9th

Last Answer : According to question find the value of z

Let R be a relation defined on the set A of all triangles such that R = {(T1, T2) : T1 is similar to T2}. Then R is -Maths 9th

Description : Let R be a relation defined on the set A of all triangles such that R = {(T1, T2) : T1 is similar to T2}. Then R is -Maths 9th

Last Answer : (d) An equivalence relation.Every triangle is similar to itself, so (T1, T1) ∈ R ⇒ R is reflexive. (T1, T2) ∈ R ⇒ T1 ~ T2 ⇒T2 ~ T1, ⇒ (T2, T1) ∈ R ⇒ R is symmetrictransitive. ∴ R is an equivalence relation.

Let R be a relation defined as a Rb if | a – b | > 0, then the relation is -Maths 9th

Description : Let R be a relation defined as a Rb if | a – b | > 0, then the relation is -Maths 9th

Last Answer : (d) Symmetric and transitive| a - a | = | 0 | = 0 so (a, a) ∉R ⇒ R is not reflexive(a, b) ∈ R ⇒ | a - b | > 0 ⇒ | b - a | > 0 ⇒ (b, a) ∈R (∵ | a - b | = | b - a |) ⇒ R is symmetric (a, b) ∈ R ... a, b, c. ∴ | a - b | > 0 and | b - c | > 0 ⇒ | a - c | > 0 ⇒ (a, c) ∈ R ⇒ R is transitive.

Let R be a relation on the set of integers given by a = 2^k .b for some integer k. Then R is -Maths 9th

Description : Let R be a relation on the set of integers given by a = 2^k .b for some integer k. Then R is -Maths 9th

Last Answer : (c) equivalence relationGiven, a R b = a = 2k .b for some integer. Reflexive: a R a ⇒ a = 20.a for k = 0 (an integer). True Symmetric: a R b ⇒ a = 2k b ⇒ b = 2-k . a ⇒ b R a as k, -k are both ... = 2k1 + k2 c, k1 + k2 is an integer. ∴ a R b, b R c ⇒ a R c True ∴ R is an equivalence relation.

Let R be a relation on the set N, defined by {(x, y) : 2x – y = 10} then R is -Maths 9th

Description : Let R be a relation on the set N, defined by {(x, y) : 2x – y = 10} then R is -Maths 9th

Last Answer : (a) ReflexiveGiven, {(\(x\), y) : 2\(x\) – y = 10} Reflexive, \(x\) R \(x\) = 2\(x\) – \(x\) = 10 ⇒ \(x\) = 10 ⇒ y = 10 ∴ Point (10, 10) ∈ N ⇒ R is reflexive.

Let R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}. Then R is -Maths 9th

Description : Let R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}. Then R is -Maths 9th

Last Answer : (b) Reflexive and transitive but not symmetricLet A = {1, 2, 3, 4} • ∵ (1, 1), (2, 2), (3, 3) and (4, 4) ∈R ⇒ R is reflexive • ∵ (1, 2) ∈ R but (2, 1) ∉ R ; (1, 3) ∈ R and (3, 1) ∉ R ; (3, 2) ∈R and (2, 3) ∉ R ⇒ R is not symmetric • (1, 3) ∈ R and (3, 2) ∈ R and (1, 2) ∈ R ⇒ R is transitive.

Let a, b, c be the lengths of the sides of a right angled triangle, the hypotenuse having the length c, then a + b is -Maths 9th

Description : Let a, b, c be the lengths of the sides of a right angled triangle, the hypotenuse having the length c, then a + b is -Maths 9th

Last Answer : answer:

Let a, b, c be positive numbers, then a/(b+c) + b/(c+a) + c/(a+b) is -Maths 9th

Description : Let a, b, c be positive numbers, then a/(b+c) + b/(c+a) + c/(a+b) is -Maths 9th

Last Answer : answer:

Let a, b, c be positive numbers lying in the interval (0, 1], then a/(1+b+ca)+b/(a+c+ab)+c/(1+a+bc) is -Maths 9th

Description : Let a, b, c be positive numbers lying in the interval (0, 1], then a/(1+b+ca)+b/(a+c+ab)+c/(1+a+bc) is -Maths 9th

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Let a1, a2, ..... an be positive real numbers such that a1a2a3 ...... an = 1. Then (1 + a1) (1 + a2) ..... (1 + an) is -Maths 9th

Description : Let a1, a2, ..... an be positive real numbers such that a1a2a3 ...... an = 1. Then (1 + a1) (1 + a2) ..... (1 + an) is -Maths 9th

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Let A, B, C be the angles of a plain triangle (A/2) = (1/3 ), tan (B/2) = (2/3). Then tan (C/2) is equal to -Maths 9th

Description : Let A, B, C be the angles of a plain triangle (A/2) = (1/3 ), tan (B/2) = (2/3). Then tan (C/2) is equal to -Maths 9th

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Let ABC be a right angled triangle with AC as its hypotenuse. Then, -Maths 9th

Description : Let ABC be a right angled triangle with AC as its hypotenuse. Then, -Maths 9th

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Let R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 12), (3, 6)} be a relation on set A = {3, 6, 9, 12}. The relation is -Maths 9th

Description : Let R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 12), (3, 6)} be a relation on set A = {3, 6, 9, 12}. The relation is -Maths 9th

Last Answer : (c) Reflexive and transitive onlyHere A = {3, 6, 9, 12} and R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 12), (3, 6)} ∵ (3, 3), (6, 6), (9, 9), (12, 12) all belong to R ⇒ {(a, a): ∈R V a ∈A} ... 12) ∈ R and (3,12) ∈ R ⇒ (3, 12) ∈ R (x, y) ∈ R, (y, z) ∈ R ⇒ (x, z) ∈ R Hence R is transitive.

Let R1 and R2 be the remainders when the polynomials x^3 + 2x^2 – 5ax – 7 and x^2 + ax^2 – 12x + 6 are divided by (x + 1) and (x – 2) respectively. -Maths 9th

Description : Let R1 and R2 be the remainders when the polynomials x^3 + 2x^2 – 5ax – 7 and x^2 + ax^2 – 12x + 6 are divided by (x + 1) and (x – 2) respectively. -Maths 9th

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Let PS be the median of the triangle with vertices P(2, 2), Q(6, –1) and R(7, 3). -Maths 9th

Description : Let PS be the median of the triangle with vertices P(2, 2), Q(6, –1) and R(7, 3). -Maths 9th

Last Answer : (b) a = √2b Let D be the mid-point of BC. Then D ≡ \(\bigg(rac{a+0}{2},rac{0}{2}\bigg)\)i.e. \(\bigg(rac{a}{2},0\bigg)\)Let E be the mid-point of AC, thenE = \(\bigg(rac{a+0}{2},rac{0+b}{2}\bigg)\) = \(\bigg ... \(rac{b}{a}.\)∴ From (i), \(rac{-2b}{a}\) x \(rac{b}{a}\) = -1⇒ 2b2 = a2 ⇒ a = √2 .

The relation "divides" on a set of positive integers is .................. (A) Symmetric and transitive (B) Anti symmetric and transitive (C) Symmetric only (D) Transitive only

Description : The relation "divides" on a set of positive integers is .................. (A) Symmetric and transitive (B) Anti symmetric and transitive (C) Symmetric only (D) Transitive only

Last Answer : (B) Anti symmetric and transitive Explanation: The ‘divide’ operation is antisymmetric because if a divides b does not necessarily implies that b divides a. If a divides b and b divides c then a divides c. So, it is transitive as well.

Let R and S be two fuzzy relations defined as: Image Then, the resulting relation, T, which relates elements of universe x to elements of universe z using max-min composition is given by

Description : Let R and S be two fuzzy relations defined as: Then, the resulting relation, T, which relates elements of universe x to elements of universe z using max-min composition is given by

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Let x and y be rational and irrational numbers, respectively. -Maths 9th

Description : Let x and y be rational and irrational numbers, respectively. -Maths 9th

Last Answer : Yes, (x + y) is necessarily an irrational number.

Let x be rational and y be irrational. Is xy necessarily irrational? Justify your answer by an example. -Maths 9th

Description : Let x be rational and y be irrational. Is xy necessarily irrational? Justify your answer by an example. -Maths 9th

Last Answer : No, (xy) is necessarily an irrational only when x ≠0. Let x be a non-zero rational and y be an irrational. Then, we have to show that xy be an irrational. If possible, let xy be a rational number. Since ... wrong. Hence, xy is an irrational number. But, when x = 0, then xy = 0, a rational number.

Let x and y be rational and irrational numbers, respectively. -Maths 9th

Description : Let x and y be rational and irrational numbers, respectively. -Maths 9th

Last Answer : Yes, (x + y) is necessarily an irrational number.

Let x be rational and y be irrational. Is xy necessarily irrational? Justify your answer by an example. -Maths 9th

Description : Let x be rational and y be irrational. Is xy necessarily irrational? Justify your answer by an example. -Maths 9th

Last Answer : No, (xy) is necessarily an irrational only when x ≠0. Let x be a non-zero rational and y be an irrational. Then, we have to show that xy be an irrational. If possible, let xy be a rational number. Since ... wrong. Hence, xy is an irrational number. But, when x = 0, then xy = 0, a rational number.

Let cost of a pen and a pencil be “x” and “y” respectively. A girl pays ₹16 for 2 pens and 3 pencils. Write the given data in the form of a linear equation in two variables. Also represent it graphically. -Maths 9th

Description : Let cost of a pen and a pencil be “x” and “y” respectively. A girl pays ₹16 for 2 pens and 3 pencils. Write the given data in the form of a linear equation in two variables. Also represent it graphically. -Maths 9th

Last Answer : Solution :-

Let the vertex of an angle ABC be located outside a circle -Maths 9th

Description : Let the vertex of an angle ABC be located outside a circle -Maths 9th

Last Answer : Given: AD = CE To prove: ∠ABC = 1/2(∠DOE - ∠AOC) In △AOD and △COE AD = CE (Given) AO = OC and DO = OE (Radii of same circle) ∴ △AOD ≅ △COE (By SSS congruence criterion) ⇒ ∠1 = ∠3, ∠2 = ∠4 (CPCT) ... = 1/2(4z + 4x - 360°) ....(vii) From (iv) and (vii), we have ∠BAC = 1/2(∠DOE - ∠AOC)

Let N be the set of natural numbers. Describe the following relation in words giving its domain -Maths 9th

Description : Let N be the set of natural numbers. Describe the following relation in words giving its domain -Maths 9th

Last Answer : The given relation stated in words is R = {(x, y) : x is the fourth power of y; x ∈ N, y ∈ {1, 2, 3, 4}}.

Let X be the set of all graduates in India. Elements x and y in X are said to be related, if they are graduates of the same university. -Maths 9th

Description : Let X be the set of all graduates in India. Elements x and y in X are said to be related, if they are graduates of the same university. -Maths 9th

Last Answer : R = {(x, y)}: x and y are graduates of same university, x, y ∈ {All graduates of India}. R is reflexive as (x, x) ∈ R, since x and x are graduates from the same university. R ... graduates from the same university ⇒ (x, z) ∈ R. R being reflexive, symmetric and transitive is an equivalence relation.

Let A = {1, 2, 3} and R = {(1, 2), (1, 1), (2, 3)} be a relation on A. -Maths 9th

Description : Let A = {1, 2, 3} and R = {(1, 2), (1, 1), (2, 3)} be a relation on A. -Maths 9th

Last Answer : (a) 1For the relation R to become transitive: (1, 2) ∈R and (2, 3) ∈R should imply (1, 3) ∈R ∴ Minimum one ordered pair (1, 3) should be added to R.

A fair coin is tossed three times. Let A, B and C be defined as follows: -Maths 9th

Description : A fair coin is tossed three times. Let A, B and C be defined as follows: -Maths 9th

Last Answer : The sample space is S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} A = {HHH, HHT, HTH, HTT}, B = {HHH, HHT, THH, THT} and C = {HHT, THH} Also, A ∩ B = {HHH, HHT}, B ∩ C = {HHT, THH}, C ∩ A = {HHT}P (A ... (C), i.e., if the events are pairwise independent and (ii) P (A ∩ B ∩ C) = P (A) . P (B) . P (C)

Let a pair of fair coins be tossed. Here S = {HH, HT, TH, TT}. Consider the events A = {heads on the first coin} = {HH, HT}, -Maths 9th

Description : Let a pair of fair coins be tossed. Here S = {HH, HT, TH, TT}. Consider the events A = {heads on the first coin} = {HH, HT}, -Maths 9th

Last Answer : ThenP (A) = P (B) = P (C) = \(rac{2}{4}\) = \(rac{1}{2}\) andP (A ∩ B) = P ({HH}) = \(rac{1}{4}\), P (A ∩ C) = P ({HT}) = \(rac{1}{4}\)P ( ... C)Thus condition (i) is satisfied, i.e., the events are pairwise independent. But condition (ii) is not satisfied and so the three events are not independent

Let O be any point inside a triangle ABC. Let L, M and N be the points on AB, BC and CA respectively, -Maths 9th

Description : Let O be any point inside a triangle ABC. Let L, M and N be the points on AB, BC and CA respectively, -Maths 9th

Last Answer : answer: