There are n letters and n addressed envelopes. If the letters are placed in the envelopes at random, -Maths 9th

There are n letters and n addressed envelopes. If the letters are placed in the envelopes at random, -Maths 9th
by

1 Answer

Answer :

Total number of ways of placing n letters in n envelopes = n! All the letters can be placed correctly in only 1 way ∴ Probability of placing all the letters in the right envelopes = \(rac{1}{n!}\) ∴ Probability that all the letters are not placed in the right envelope = 1 – \(rac{1}{n!}\) .
by

Related questions

The letters of the word ‘SOCIETY’ are placed at random in a row. What is the probability that three vowels come together ? -Maths 9th

Description : The letters of the word ‘SOCIETY’ are placed at random in a row. What is the probability that three vowels come together ? -Maths 9th

Last Answer : There are 7 letters in the word SOCIETY. ∴ Total number of ways of arranging all the 7 letters = n(S) = 7!. When the case of three vowels being together is taken, then the three vowels are considered as one unit, so the ... = 5! 3! ∴ Required probability = \(rac{5! imes3!}{7!}\) = \(rac{1}{7}\)

The probability that in the random arrangement of the letters of the word ‘UNIVERSITY’the two I‘s do not come together is -Maths 9th

Description : The probability that in the random arrangement of the letters of the word ‘UNIVERSITY’the two I‘s do not come together is -Maths 9th

Last Answer : (b) \(rac{4}{5}\)Let S be the sample space. Then, n(S) = Total number of waysin which the letters of the word UNIVERSITY' can be arranged = \(rac{10!}{2!}\) (∵ There are 2I s) ... ! imes36}{rac{10!}{2!}}\) = \(rac{ ot8! imes36 imes2!}{10 imes9 imes ot8!}\) = \(rac{4}{5}\).

A letter is taken out at random from ‘ASSISTANT’ and another is taken out from ‘STATISTICS’. The probability that they are same letters is -Maths 9th

Description : A letter is taken out at random from ‘ASSISTANT’ and another is taken out from ‘STATISTICS’. The probability that they are same letters is -Maths 9th

Last Answer : (c) 0.8645Required probability = P(X not defective and Y not defective) = P(\(\bar{X}\)) x P(\(\bar{Y}\))= (1 – P(X)) (1 – P(Y))= \(\bigg(1-rac{9}{100}\bigg)\)\(\bigg(1-rac{5}{100}\bigg)\)= \(rac{91}{100}\) x \(rac{95}{100}\) = \(rac{8645}{10000}\) = 0.8645.

If n integers taken at random are multiplied together, then the probability that the last digit of the product is 1, 3, 7 or 9 is -Maths 9th

Description : If n integers taken at random are multiplied together, then the probability that the last digit of the product is 1, 3, 7 or 9 is -Maths 9th

Last Answer : (d) 226 × 52C26 | 104C26Since there are 52 distinct cards in a deck and each distinct card is 2 in number.∴2 decks will also contain only 52 distinct cards, two each.∴ Probability that the player gets all distinct cards = \(rac{^{52}C_{26} imes2^{26}}{^{104}C_{26}}\).

A wealthy wise old woman feared that her daughter was lazy and as a result rather stupid. When the old woman died, her will stipulated that her assets were to be liquidated and a check was to be written for the full amount. The check was to be placed in one of three envelopes. The other two envelopes would contain a blank piece of paper. If the daughter could determine from the writing on the envelope which envelope contained the check, she would inherit her mother's fortune. Otherwise, the fortune would go to the old woman's favorite charity for animals. The daughter was not allowed to touch the envelopes. Her decision had to be made based on the writing on the envelopes. The daughter was told that only one envelope had a true statement and that the other two statements were false. The en

Description : A wealthy wise old woman feared that her daughter was lazy and as a result rather stupid. When the old woman died, her will stipulated that her assets were to be liquidated and a check was to be ... only one envelope had a true statement and that the other two statements were false. The en -Riddles

Last Answer : The daughter should pick envelope 1. Unfortunately she picked envelope 3. Statements 1 and 2 were false, and the only true statement was statement 3. If the check was in envelope 1, that would make statement 1 ... both be true. If the check was in envelope 3, statements 1 and 3 would both be true.

………………… is a tool in MS_Word using which customized letters, mailing labels and envelopes can be printed in bulk. -Technology

Description : ………………… is a tool in MS_Word using which customized letters, mailing labels and envelopes can be printed in bulk. -Technology

Last Answer : Mail Mergeis a tool in MS-Word using which customized letters, mailing labels and envelopes can be printed in bulk.

A coin is tossed 500 times and we get Heads : 285 and tails : 215 times. When a coin is tossed at random, what is the probability of getting a. head? b. tail? -Maths 9th

Description : A coin is tossed 500 times and we get Heads : 285 and tails : 215 times. When a coin is tossed at random, what is the probability of getting a. head? b. tail? -Maths 9th

Last Answer : Given, Total number of events = 500 No. of times heads occur = 285 Probability of getting head when coin is tossed at random = 285/500 = 57/100 No. of times tails occur = 215 Probability of getting tails when coin is tossed at random = 215/500 = 43/100

Two coin are tossed 400 times and we get a. Two Heads : 112 times b. One Head : 160 times c. No Head : 128 times. When two coins are tossed at random, what is the probability of getting a. Two Heads b. One Head c. No Head -Maths 9th

Description : Two coin are tossed 400 times and we get a. Two Heads : 112 times b. One Head : 160 times c. No Head : 128 times. When two coins are tossed at random, what is the probability of getting a. Two Heads b. One Head c. No Head -Maths 9th

Last Answer : Given, Total number of events = 400 (a) No. of times two heads occur = 112 Probability of getting two heads = 112/400 = 7/25 (b) No. of times one heads occur = 160 Probability of getting one heads = 160/400 = 2/5 (c) No. of times no heads occur = 128 Probability of getting no heads = 128/400 = 8/25

In a class, there are x girls and y boys, a student is selected at random, then find the probability of selecting a boy. -Maths 9th

Description : In a class, there are x girls and y boys, a student is selected at random, then find the probability of selecting a boy. -Maths 9th

Last Answer : Number of boys = y Total students = (x + y) Thus,P(a boy)= y/(x+y)

750 families with 3 children were selected randomly and the following data recorded If a family member is chosen at random, compute the probability that it has : -Maths 9th

Description : 750 families with 3 children were selected randomly and the following data recorded If a family member is chosen at random, compute the probability that it has : -Maths 9th

Last Answer : (i) P(no boy child) =100 / 750 = 2 / 15 (ii) P (no girl child) = 120 /750 =4 / 25

A coin is tossed 500 times and we get Heads : 285 and tails : 215 times. When a coin is tossed at random, what is the probability of getting a. head? b. tail? -Maths 9th

Description : A coin is tossed 500 times and we get Heads : 285 and tails : 215 times. When a coin is tossed at random, what is the probability of getting a. head? b. tail? -Maths 9th

Last Answer : Given, Total number of events = 500 No. of times heads occur = 285 Probability of getting head when coin is tossed at random = 285/500 = 57/100 No. of times tails occur = 215 Probability of getting tails when coin is tossed at random = 215/500 = 43/100

Two coin are tossed 400 times and we get a. Two Heads : 112 times b. One Head : 160 times c. No Head : 128 times. When two coins are tossed at random, what is the probability of getting a. Two Heads b. One Head c. No Head -Maths 9th

Description : Two coin are tossed 400 times and we get a. Two Heads : 112 times b. One Head : 160 times c. No Head : 128 times. When two coins are tossed at random, what is the probability of getting a. Two Heads b. One Head c. No Head -Maths 9th

Last Answer : Given, Total number of events = 400 (a) No. of times two heads occur = 112 Probability of getting two heads = 112/400 = 7/25 (b) No. of times one heads occur = 160 Probability of getting one heads = 160/400 = 2/5 (c) No. of times no heads occur = 128 Probability of getting no heads = 128/400 = 8/25

In a class, there are x girls and y boys, a student is selected at random, then find the probability of selecting a boy. -Maths 9th

Description : In a class, there are x girls and y boys, a student is selected at random, then find the probability of selecting a boy. -Maths 9th

Last Answer : Number of boys = y Total students = (x + y) Thus,P(a boy)= y/(x+y)

750 families with 3 children were selected randomly and the following data recorded If a family member is chosen at random, compute the probability that it has : -Maths 9th

Description : 750 families with 3 children were selected randomly and the following data recorded If a family member is chosen at random, compute the probability that it has : -Maths 9th

Last Answer : (i) P(no boy child) =100 / 750 = 2 / 15 (ii) P (no girl child) = 120 /750 =4 / 25

80 bulbs are selected at random from a lot -Maths 9th

Description : 80 bulbs are selected at random from a lot -Maths 9th

Last Answer : Number of bulbs having life less than 900 hours = 10 + 12 + 23 = 45 P (a bulb has life less than 900 hours) = 45/80 = 9/16

Two balls are drawn at random from a bag containing 3 white, 3 red, 4 green and 4 black balls, one by one without replacement. -Maths 9th

Description : Two balls are drawn at random from a bag containing 3 white, 3 red, 4 green and 4 black balls, one by one without replacement. -Maths 9th

Last Answer : Given, 3 white (3 W), 3 red (3 R), 4 green (4 G), 4 black (4 B) balls Total no. of balls = 3 + 3 + 4 + 4 = 14 Two balls are to be drawn, one by one without replacement. There are 4 possibilities.First BallSecond ... }{13}\) = \(rac{33+33+40+40}{14 imes13}\) = \(rac{146}{182}\) = \(rac{73}{91}.\)

An integer is chosen at random from the first two hundred positive integers. What is the probability that the integer chosen is divisible by 6 or 8 ? -Maths 9th

Description : An integer is chosen at random from the first two hundred positive integers. What is the probability that the integer chosen is divisible by 6 or 8 ? -Maths 9th

Last Answer : As there are 200 integers, total number of exhaustive, mutually exclusive and equally likely cases, i.e, n(S) = 200 Let A : Event of integer chosen from 1 to 200 being divisible by 6⇒ n(A) = 33 \(\bigg(rac{200}{6}=33rac{1}{3}\ ... (rac{25}{200}\) - \(rac{8}{200}\) = \(rac{50}{200}\) = \(rac{1}{4}\).

A bag contains 30 tickets numbered from 1 to 30. Five tickets are drawn at random and arranged in ascending order -Maths 9th

Description : A bag contains 30 tickets numbered from 1 to 30. Five tickets are drawn at random and arranged in ascending order -Maths 9th

Last Answer : Total number of ways in which 5 tickets can be drawn = n(S) = 30C5. The tickets are arranged in the form T1, T2, T3 (= 20), T4, T5 Where T1, T2 ∈{1, 2, 3, , 19} and T4, T5 ∈{21, 22, , 30 ... {10 imes9}{2}\) x \(rac{5 imes4 imes3 imes2 imes1}{30 imes29 imes28 imes27 imes26}\) = \(rac{285}{5278}.\)

In a group there are 3 women and 3 men. 4 people are selected at random from this group -Maths 9th

Description : In a group there are 3 women and 3 men. 4 people are selected at random from this group -Maths 9th

Last Answer : A : Selected 3 women and 1 man B : Selected 1 women and 3 men S : Selected 4 people from 6 people (3 + 3) Then n(A) = 3C3 3C1, n(B) = 3C1 3C3, n(S) = 6C4∴ Required probability = P(A) + P(B) = \(rac{ ... 3C_1}{^6C_4}\) + \(rac{^3C_1 imes^3C_3}{^6C_4}\)= \(rac{2 imes1 imes3}{15}\) = \(rac{2}{5}.\)

What is the probability that a number selected at random from the set of numbers {1, 2, 3, …, 100} is a perfect cube? -Maths 9th

Description : What is the probability that a number selected at random from the set of numbers {1, 2, 3, …, 100} is a perfect cube? -Maths 9th

Last Answer : (a) \(rac{1}{25}\) Let us assume S as the sample space in all questions. S means the set denoting the total number of outcomes possible. Let S = {1, 2, 3, , 100} be the sample space. Then, n(S) = 100 Let A : ... ∴Required probability P(A) = \(rac{n(A)}{n(S)}\) = \(rac{4}{100}\) = \(rac{1}{25}\)

From a group of 3 man and 2 women, two person are selected at random. Find the probability that at least one women is selected. -Maths 9th

Description : From a group of 3 man and 2 women, two person are selected at random. Find the probability that at least one women is selected. -Maths 9th

Last Answer : (b) \(rac{7}{10}\)Total number of ways of selecting 2 persons at random out of 5 persons = 5C2 ∴ n(S) = 5C2 = \(rac{|\underline5}{|\underline3|\underline2}\) = \(rac{5 imes4}{2 imes1}\) = 10Let A : Event of selecting ... = 2 3 + 1 = 7 ∴ Required probability = \(rac{n(A)}{n(S)}\) = \(rac{7}{10}\).

A bag contains 7 red and 5 green balls. The probability of drawing all four balls asred balls, when four balls are drawn at random is -Maths 9th

Description : A bag contains 7 red and 5 green balls. The probability of drawing all four balls asred balls, when four balls are drawn at random is -Maths 9th

Last Answer : (b) \(rac{7}{99}\)There are (7 + 5) = 12 balls in the bag. 4 balls can be drawn at random from 12 balls in 12C4 ways. ∴ n(S) = 12C4 = \(rac{|\underline{7}}{|\underline3|\underline4}\) = \(rac{7 imes6 imes5}{3 ... ) = 35∴ Required probability = \(rac{n(A)}{n(S)}\) = \(rac{35}{495}\) = \(rac{7}{99}\).

Five horses are in a race. Mr A. Selects two of the horses at random and bets on them. The probability that Mr A selected the winning horse is -Maths 9th

Description : Five horses are in a race. Mr A. Selects two of the horses at random and bets on them. The probability that Mr A selected the winning horse is -Maths 9th

Last Answer : (b) \(rac{2}{5}\)As each horse has equal chance of winning the race, Number of ways in which one of the five horses wins the race = 5C1 ∴ n(S) =5C1 = \(rac{|\underline5}{|\underline4|\underline1}\) 5To find the chance ... n(E) = 2C1 = 2 ∴ Required probability = \(rac{n(E)}{n(S)}\) = \(rac{2}{5}\).

Two cards are drawn at random from a well-shuffled pack of 52 cards. What is the probability that either both are black or both are kings ? -Maths 9th

Description : Two cards are drawn at random from a well-shuffled pack of 52 cards. What is the probability that either both are black or both are kings ? -Maths 9th

Last Answer : (b) \(rac{55}{221}\)S : Drawing 2 cards out of 52 cards ⇒ n(S) = 52C2 = \(rac{|\underline{52}}{|\underline{52}|\underline2}\) = \(rac{52 imes51}{2}\) = 1326A : Event of drawing 2 black cards out of 26 black cards⇒ n ... ) + \(rac{6}{1326}\) - \(rac{1}{1326}\) = \(rac{330}{1326}\) = \(rac{55}{221}\).

A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability of getting a red card or a diamond or a jack ? -Maths 9th

Description : A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability of getting a red card or a diamond or a jack ? -Maths 9th

Last Answer : (d) \(rac{7}{13}\)Here n(S) = 52 Let A, B, C be the events of getting a red card, a diamond and a jack respectively. ∵ There are 26 red cards, 13 diamonds and 4 jacks, n(A) = 26, n(B) = 13, n(C) = 4 ⇒ n(A ∩ B) = ... rac{1}{52}\)= \(rac{44}{52}\) + \(rac{16}{52}\) = \(rac{28}{52}\) = \(rac{7}{13}\) .

A natural number is chosen at random from amongst the first 300. What is the probability that the number chosen is a multiple of 2 or 3 or 5 ? -Maths 9th

Description : A natural number is chosen at random from amongst the first 300. What is the probability that the number chosen is a multiple of 2 or 3 or 5 ? -Maths 9th

Last Answer : (b) \(rac{11}{15}\)n(S) = 300 Let A : Event of getting a number divisible by 2 B : Event of getting a number divisible by 3 C : Event of getting a number divisible by 5 ∴ A ∩ B : Event of getting a number divisible by ... \(rac{320}{300}\) - \(rac{100}{300}\) = \(rac{220}{300}\) = \(rac{11}{15}\).

A bag contains 5 green and 7 red balls, out of which two balls are drawn at random. What is the probability that they are of the same colour ? -Maths 9th

Description : A bag contains 5 green and 7 red balls, out of which two balls are drawn at random. What is the probability that they are of the same colour ? -Maths 9th

Last Answer : (d) \(rac{31}{66}\)Total number of balls in the bag = 12 (5 Green + 7 Red) Let S be the sample space of drawing 2 balls out of 12 balls.Thenn(S) = 12C2 = \(rac{12 imes11}{2}\) = 66∴ Let A : Event of drawing two red balls⇒ ... \(rac{n(B)}{n(S)}\) = \(rac{21}{66}\) + \(rac{10}{66}\) = \(rac{31}{66}\).

A management institute has six senior professors and four junior professors. Three professors are selected at random -Maths 9th

Description : A management institute has six senior professors and four junior professors. Three professors are selected at random -Maths 9th

Last Answer : (a) \(rac{5}{6}\)P(At least one junior professor is selected) = P(Selecting 1 Junior) P(Selecting 2 Seniors) + P(Selecting 2 Junior) P(Selecting 1 Senior) + P(Selecting all 3 Juniors)∴ Required probability = \(rac{^4C_1 imes^ ... }{30}\) = \(rac{15+9+1}{30}\) = \(rac{25}{30}\) = \(rac{5}{6}\).

A card is drawn at random from a well shuffled pack of 52 cards -Maths 9th

Description : A card is drawn at random from a well shuffled pack of 52 cards -Maths 9th

Last Answer : (c) P(X) = P(Y) > P(Z) P(X) = \(rac{26}{52}\) + \(rac{4}{52}\) - \(rac{2}{52}\) = \(rac{28}{52}\) (∵ There are 26 black cards, 4 kings and 2 black kings)P(Y) = \(rac{13}{52}\) + \(rac{ ... }{52}\)(∵ There are 4 aces, 13 diamonds, 4 queens, 1 ace of diamond, 1 queen of diamond) ∴ P(X) = P(Y) > P(Z).

A group of 2n boys and 2n girls is divided at random into two equal batches. -Maths 9th

Description : A group of 2n boys and 2n girls is divided at random into two equal batches. -Maths 9th

Last Answer : (c) \(rac{(^{2n}C_n)^2}{^{4n}C_{2n}}\)Total number of boys and girls = 2n + 2n = 4n Since, there are two equal batches, each batch has 2n members ∴ Let S (Sample space) : Selecting one batch out of 2 ⇒ S : ... )2∴ Required probability = \(rac{n(E)}{n(S)}\) = \(rac{(^{2n}C_n)^2}{^{4n}C_{2n}}\)

A box contains 100 balls numbers from 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability -Maths 9th

Description : A box contains 100 balls numbers from 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability -Maths 9th

Last Answer : (d) \(rac{1}{4}\)The box contains 100 balls numbered from 1 to 100. Therefore, there are 50 even and 50 odd numbered balls. The sum of the three numbers drawn will be odd, if all three are odd or one is even and 2 are odd. ∴ Required probability = P(odd) × P(odd) × P(odd) + P(even) × P(odd) × P(odd)

Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle with these vertices is equilateral equals : -Maths 9th

Description : Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle with these vertices is equilateral equals : -Maths 9th

Last Answer : (c) \(rac{1}{10}\)Let S be the sample space.Then n(S) = Number of triangles formed by selecting any three vertices of 6 vertices of a regular hexagon= 6C3 = \(rac{6 imes5 imes4}{3 imes2}\) = 20.Let A : Event that the ... Required probability = \(rac{n(A)}{n(S)}\) = \(rac{2}{20}\) = \(rac{1}{10}\).

Out of 3n consecutive natural numbers 3 natural numbers are chosen at random without replacement. -Maths 9th

Description : Out of 3n consecutive natural numbers 3 natural numbers are chosen at random without replacement. -Maths 9th

Last Answer : (c) \(rac{3n^2-3n+2}{(3n-1)(3n-2)}\)In 3n consecutive natural numbers, (i) n numbers are of the form 3p (ii) n numbers are of the form 3p + 1 (iii) n numbers are of the form 3p + 2 For the ... ) We can select one number from each set.∴ Favourable number of cases = nC3 + nC3 + nC3 + (nC1 nC1 nC1)

A point is selected at random inside an equilateral triangle. From this point a perpendicular is dropped to each side. -Maths 9th

Description : A point is selected at random inside an equilateral triangle. From this point a perpendicular is dropped to each side. -Maths 9th

Last Answer : answer:

If an integer P is chosen at random in the interval 0 ≤ p ≤ 5, the probability that the roots of the equation x^2 + px -Maths 9th

Description : If an integer P is chosen at random in the interval 0 ≤ p ≤ 5, the probability that the roots of the equation x^2 + px -Maths 9th

Last Answer : answer:

A basket contains 2 blue, 4 red, 3 green and 5 black balls. If 4 balls are picked at random, what is the probability that -Maths 9th

Description : A basket contains 2 blue, 4 red, 3 green and 5 black balls. If 4 balls are picked at random, what is the probability that -Maths 9th

Last Answer : (d) None of theseThe month having 3 days less than 31 days has 28 days, i.e, it is the month of February. P(Choosing February) = \(rac{1}{12}\).

Three equal cubes are placed adjacently in a row. Find the ratio of total surface area of the new cuboid to that of the sum of the surface areas of the three cubes. -Maths 9th

Description : Three equal cubes are placed adjacently in a row. Find the ratio of total surface area of the new cuboid to that of the sum of the surface areas of the three cubes. -Maths 9th

Last Answer : Let each side of a cube = a cm Then surface area = 6a² cm² and surface area of 3 such cubes = 3 x 6a² = 18a² cm² By placing three cubes side by side we get a cuboid whose ... + 3a²] = 14 a² ∴ Ratio between their surface areas = 14a² : 18a² = 7 : 9

ABCD is a square. Another square EFGH with the same area is placed on the square ABCD such that the point of intersection of diagonals of square -Maths 9th

Description : ABCD is a square. Another square EFGH with the same area is placed on the square ABCD such that the point of intersection of diagonals of square -Maths 9th

Last Answer : (a) 32 (2 - √2)As is seen in the given figure, the sides of one square are parallel to the diagonals of another square. Also, square ABCD and EFGH have same area.⇒ Sides of square ABCD and square EFGH are 4 cm each. Let ... four Δs outside ABCD= 16 +16 (3 -2√2)= 16 + (4 - 2√2) = 32 (2 - √2) cm2.

Counters marked 1, 2, 3 are placed in a bag and one is drawn and replaced. -Maths 9th

Description : Counters marked 1, 2, 3 are placed in a bag and one is drawn and replaced. -Maths 9th

Last Answer : (b) \(rac{7}{27}.\)Let S be the sample space of drawing a counter three times and replacing it each time. Then, n(S) = 3 3 3 = 27 Let A : Event of obtaining a total of 6 in the three draws of counters. Then, A = {(1, 2, 3), ... (2, 2, 2)} ⇒ n(A) = 7 ∴ P(A) = \(rac{n(A)}{n(S)}\) = \(rac{7}{27}.\)

There are 6 numbered chairs placed around a circular table. 3 boys and 3 girls want to sit on them such that neither of two boys nor two girls -Maths 9th

Description : There are 6 numbered chairs placed around a circular table. 3 boys and 3 girls want to sit on them such that neither of two boys nor two girls -Maths 9th

Last Answer : Since the chairs are numbered, so they are distinguishable. Therefore 3 boys can be arranged on 3 alternate chairs in 3! ways. 3 girls can be arrenged in 3! ways Also, the girls can be seated before the boys. Total number of required ways = 3! × 3! + 3! × 3! = 2 × (3!)2

Five balls of different colours are to be placed in three boxes of different sizes. Each box can hold all five. In how many different ways -Maths 9th

Description : Five balls of different colours are to be placed in three boxes of different sizes. Each box can hold all five. In how many different ways -Maths 9th

Last Answer : According to the question, we have 5 balls to be placed in 3 boxes where no box remains emptyHence, we can have the following kinds of distribution firstly, where the distribution will be (3,1,1) that is, one box gets three ... go in second box is = 4 C 2 . Total no. of ways =90 Total:60+90=150.

If three cylinders of radius r and height h are placed vertically such that the curved surface of each cylinder touches the curved surfaces -Maths 9th

Description : If three cylinders of radius r and height h are placed vertically such that the curved surface of each cylinder touches the curved surfaces -Maths 9th

Last Answer : hr2 (3-√−π2)(3−π2) The bases of the three cylinders when placed as given are as shown in the figure : Let the radius of the base of each cylinder = r cm. We are required to find the volume of air. ... ∠C = 60º) = 3 x 60o360o πr2=πr2260o360o πr2=πr22 ∴ Required volume = (3-√r2−π2r2)h=(3-√−π2)r2h.

A tank internally measuring 150 cm × 120 cm × 100 cm has 1281600 cm^3 of water in it. Porous bricks are placed in the water until the tank -Maths 9th

Description : A tank internally measuring 150 cm × 120 cm × 100 cm has 1281600 cm^3 of water in it. Porous bricks are placed in the water until the tank -Maths 9th

Last Answer : Volume of a brick = 20 cm 6 cm 4 cm = 480 cm3. Water absorbed by one brick = (110 480)(110 480) cm3 = 48 cm3. Let x bricks be placed in the water. Then, x bricks absorb 48x cm3 of water. ... tank ⇒ 1281600 + 480xx - 48xx = 150 x 120 x 100 ⇒ 432xx = 1800000 - 1281600 = 518400 ⇒ xx = 1200.

There are two congruent triangles each with area 198 cm^2. Triangle DEF is placed over triangle ABC in such a way that the centroid of -Maths 9th

Description : There are two congruent triangles each with area 198 cm^2. Triangle DEF is placed over triangle ABC in such a way that the centroid of -Maths 9th

Last Answer : answer:

The letters ofthe word ‘NATIONAL’are arranged atrandom. What is the probability that the last letter will be T ? -Maths 9th

Description : The letters ofthe word ‘NATIONAL’are arranged atrandom. What is the probability that the last letter will be T ? -Maths 9th

Last Answer : (c) \(rac{1}{8}\)Let S be the sample space.Then n(S) = Total number of ways in which the letters of word NATIONAL can be arranged= \(rac{8!}{2!\,2!}\) (∵There are 2A s and 2N's in 8 letters)Let E : Event of ... !}{2!\,2!}}\) = \(rac{7!}{8!}\) = \(rac{7!}{8 imes7!}\) = \(rac{1}{8}\).

How many words can be formed from the letters of the word “DAUGHTER” so that the vowels always come together? -Maths 9th

Description : How many words can be formed from the letters of the word “DAUGHTER” so that the vowels always come together? -Maths 9th

Last Answer : The number of words formed from 'DAUGHTER' such that all vowels are together is 4320.

Find how many arrangements can be made with the letters of the word “MATHEMATICS” in which the vowels occur together? -Maths 9th

Description : Find how many arrangements can be made with the letters of the word “MATHEMATICS” in which the vowels occur together? -Maths 9th

Last Answer : (i) There are 11 letters in the word 'MATHEMATICS' . Out of these letters M occurs twice, A occurs twice, T occurs twice and the rest are all different. Hence, the total number of arrangements of ... 4!2!=12. Hence, the number of arrangement in which 4 vowels are together =(10080×12)=120960.

How many words can be formed from the letters of the word “SUNDAY” so that the vowels never come together? -Maths 9th

Description : How many words can be formed from the letters of the word “SUNDAY” so that the vowels never come together? -Maths 9th

Last Answer : Given: The word ‘SUNDAY’ Total number of letters in the word ‘SUNDAY’ is 6. So, number of arrangements of 6 things, taken all at a time is 6P6 = 6! = 6 ... of words using letters of ‘SUNDAY’ starting with ‘N’ and ending with ‘Y’ is 24

All the words that can be formed using the letters A, H, L, U, R are written as in a dictionary -Maths 9th

Description : All the words that can be formed using the letters A, H, L, U, R are written as in a dictionary -Maths 9th

Last Answer : No. of words starting with A are 4!=24 No. of words starting with H are 4!=24 No. of words starting with L are 4!=24 These account for 72 words Next word is RAHLU and the 74th word RAHUL.