There are 5 red, 4 white and 3 blue marbles in a bag. They are drawn one by one and arranged in a row. Assuming that all the 12 marbles -Maths 9th

There are 5 red, 4 white and 3 blue marbles in a bag. They are drawn one by one and arranged in a row. Assuming that all the 12 marbles -Maths 9th
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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 \(rac{5 imes4 imes3 imes2 imes1}{30 imes29 imes28 imes27 imes26}\) = \(rac{285}{5278}.\)

There are 10 orange, 2 violet and 4 purple balls in a bag. All the 16 balls are drawn one by one and arranged in a row. Find out the number of different arrangements possible. A) 25230 B) 23420 C) 120120 D) 27720

Description : There are 10 orange, 2 violet and 4 purple balls in a bag. All the 16 balls are drawn one by one and arranged in a row. Find out the number of different arrangements possible. A) 25230 B) 23420 C) 120120 D) 27720

Last Answer : Answer: C)  Number of different arrangements possible  = {16!} / {10! 2! 4!}  = {16×15×14×13×12×11×10×9×8×7×6×5×4×3×2 } /  {(10×9×8×7×6×5×4×3×2 ) (2) (4×3×2)}}  = {16×15×14×13×12×11} / {(2)(4×3×2)}  = {8×5×7×13×3×11}  = 120120

A bag contains 5 white, 7 red and 4 black balls. If four balls are drawn one by one with replacement, what is the probability that none is white. -Maths 9th

Description : A bag contains 5 white, 7 red and 4 black balls. If four balls are drawn one by one with replacement, what is the probability that none is white. -Maths 9th

Last Answer : Let A, B, C, D denote the events of not getting a white ball in first, second, third and fourth draw respectively. Since the balls are drawn with replacement, therefore, A, B, C, D are independent events such that P (A) = P (B) ... x \(rac{11}{16}\) x \(rac{11}{16}\) = \(\big(rac{11}{16}\big)^4.\)

A bag contains 7 white, 5 black and 4 red balls. Four balls are drawn without replacement. -Maths 9th

Description : A bag contains 7 white, 5 black and 4 red balls. Four balls are drawn without replacement. -Maths 9th

Last Answer : Let A : Event of getting at least 3 black balls Then n(A) = 5C3 x 11C1 + 5C4 (∵ Besides 5 black balls, there are 11 other balls)(3 black + others) (4 black)= \(rac{5 imes4}{2}\) x 11 + 5 = 115Total numbers of ways ... = 1820∴ P(A) = \(rac{n(A)}{n(S)}\) = \(rac{115}{1820}\) = \(rac{23}{364}.\)

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}\) = \(rac{33+33+40+40}{14 imes13}\) = \(rac{146}{182}\) = \(rac{73}{91}.\)

A bag contains 5 green and 11 blue balls and the second one contains 3 green and 7 blue balls. Two balls are drawn from one of the bags. -Maths 9th

Description : A bag contains 5 green and 11 blue balls and the second one contains 3 green and 7 blue balls. Two balls are drawn from one of the bags. -Maths 9th

Last Answer : (c) \(rac{111}{240}\)P(Drawing of two balls of different colours from one of the bags)= P(choosing the 1st bag) P(Drawing 1 green out 5 green and 1 out of 11 blue balls) + P(choosing the 2nd bag) P(Drawing 1 green out ... (rac{11}{48}\) + \(rac{7}{30}\) = \(rac{55+56}{240}\) = \(rac{111}{240}\).

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 bag contains x white, y red and z blue balls. -Maths 9th

Description : A bag contains x white, y red and z blue balls. -Maths 9th

Last Answer : Number of blue balls = z Total balls = x + y + z therefore P(blue ball)= z /(x+y+z )

A bag contains x white, y red and z blue balls. -Maths 9th

Description : A bag contains x white, y red and z blue balls. -Maths 9th

Last Answer : Number of blue balls = z Total balls = x + y + z therefore P(blue ball)= z /(x+y+z )

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 = \(rac{n(A)}{n(S)}\) = \(rac{35}{495}\) = \(rac{7}{99}\).

A bag contains 6 black and 3 white balls. Another bag contains 5 black and 4 white balls. If one ball is drawn from each bag, -Maths 9th

Description : A bag contains 6 black and 3 white balls. Another bag contains 5 black and 4 white balls. If one ball is drawn from each bag, -Maths 9th

Last Answer : Let W1 and W2 denote the events of drawing a white ball from the first and one from the second bag respectively. Let B1 and B2 denote the events of drawing black balls from the two bags in the same order. Then P ... }\) = \(rac{14}{27}.\) (By addition theorem for mutually exclusive events.

An urn contains 3 white and 5 blue balls and a second urn contains 4 white and 4 blue balls. If one ball is drawn from each urn, -Maths 9th

Description : An urn contains 3 white and 5 blue balls and a second urn contains 4 white and 4 blue balls. If one ball is drawn from each urn, -Maths 9th

Last Answer : Let E : Event of drawing both the balls of same colour from the two urns E1 : Getting 1 white ball from the first urn and 1 white ball from the second urn E2 : Getting 1 blue ball from the first urn and 1 blue ball from ... a ball from other urn)= \(rac{12}{64}+rac{20}{64}=rac{32}{64}=rac{1}{2}.\)

Twice the number of marbles with Aman exceeds thrice the number of marbles with Vinay by 12. Assume the number of marbles with Aman and Vinay as x and y respectively .Express the statement in the form of a linear equation in two variables. -Maths 9th

Description : Twice the number of marbles with Aman exceeds thrice the number of marbles with Vinay by 12. Assume the number of marbles with Aman and Vinay as x and y respectively .Express the statement in the form of a linear equation in two variables. -Maths 9th

Last Answer : answer:

Jasmine has 50 marbles in a bag. 20% of the marbles are blue. How many marbles are blue?

Description : Jasmine has 50 marbles in a bag. 20% of the marbles are blue. How many marbles are blue?

Last Answer : 25

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) = \(rac{n(A)}{n(S)}\) = \(rac{7}{27}.\)

A bag contain 4 white & 5 red and 6blue color balls,3 balls are drawn randomly.What is the probability of all the balls are red? 1)1/22 2)3/22 3)2/90 4)2/91

Description : A bag contain 4 white & 5 red and 6blue color balls,3 balls are drawn randomly.What is the probability of all the balls are red? 1)1/22 2)3/22 3)2/90 4)2/91

Last Answer : 4)2/91 Exp: 15C3/5C2=(15×14×13)/(3×2×1)=10/455=2/91

How do you find the probability of 4 blue marbles 5 red marbles 1 green marble and 2 black marbles?

Description : How do you find the probability of 4 blue marbles 5 red marbles 1 green marble and 2 black marbles?

Last Answer : It depends what probability exactly you want to find.probability = number of successful ways / total number of waysIf the problem is:You have a bag containing 4 blue, 5 red, 1 green, 2 black marble what is the probability of ... 2 black)= 16/144 + 25/144 + 1/144 + 4/144 = 46/144 = 23/72And so on.

bassam has some marbles. one third of them are blue, one half of them are red, and the rest are green. If he has 7 green marbles, how many marbles does he have?

Description : bassam has some marbles. one third of them are blue, one half of them are red, and the rest are green. If he has 7 green marbles, how many marbles does he have?

Last Answer : Because 1/3 are blue and 1/2 are red and the rest are green we can calculate that 1/6 of the marbles are green.1/3 = 2/61/2 = 3/62/6 + 3/6 = 5/6 which leaves 1/6 green.Since there are 7 green marbles we know there must be 7 x 6 total.The answer is 42 marbles.

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

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

Last Answer : Let S : Drawing 2 cards out of 52 card A : Drawing 2 red cards B : Drawing 2 kings A ∪ B : Drawing 2 red cards or 2 kings ∴ n(S) = 52C2 n(A) = 26C2 (∵ There are 26 red cards) n(B) = 4C2 ... \(rac{4 imes3}{52 imes51}\) - \(rac{2}{52 imes51}\) = \(rac{660}{2652}\) = \(rac{55}{221}.\)

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}\)= \(rac{44}{52}\) + \(rac{16}{52}\) = \(rac{28}{52}\) = \(rac{7}{13}\) .

There are 20 marbles in a box which are marked with distinct numbers from 1 to 20. If a marble is drawn, then find the probability that the marble bei

Description : There are 20 marbles in a box which are marked with distinct numbers from 1 to 20. If a marble is drawn, then find the probability that the marble bei

Last Answer : There are 20 marbles in a box which are marked with distinct numbers from 1 to 20. If a marble is drawn, then find the ... `2//5` C. `1//5` D. `4//5`

Abag contains 4 red and 3 black balls.Asecond bag contains 2 red and 4 black balls. -Maths 9th

Description : Abag contains 4 red and 3 black balls.Asecond bag contains 2 red and 4 black balls. -Maths 9th

Last Answer : (b) \(rac{19}{42} \)A red ball can be selected in two mutually exclusive ways. (i) Selecting bag I and then drawing a red ball from it (ii) Selecting bag II and them drawing a red ball from it ∴ P(red ball) = P(Selecting bag I) ... \(rac{2}{6}\) = \(rac{2}{7}\) + \(rac{1}{6}\) = \(rac{19}{42} \).

One bag contains 3 black and 4 white balls and the other bag contains 4 black and 3 white balls. A die is rolled. -Maths 9th

Description : One bag contains 3 black and 4 white balls and the other bag contains 4 black and 3 white balls. A die is rolled. -Maths 9th

Last Answer : Let A : Getting 2 or 5 B : Getting white ball from first bag C : Getting white ball from second bag.∴ P(A) = \(rac{2}{6}\) = \(rac{1}{3}\) ⇒ P(\(\bar{A}\)) = 1 - P(A) = 1 - \(rac{1}{3}\) = \(rac{2}{3}\)∴ Required ... \(rac{4}{7}\) + \(rac{2}{3}\) x \(rac{3}{7}\) = \(rac{4+6}{21}\) = \(rac{10}{21}.\)

A bag contains a white and b black balls. Two players A and B alternately draw a ball from the bag -Maths 9th

Description : A bag contains a white and b black balls. Two players A and B alternately draw a ball from the bag -Maths 9th

Last Answer : (c) 2 : 1Let W denote the event of drawing a white ball at any draw and B that of drawing a black ball. Then, P (W) = \(rac{a}{a+b},\) P(B) = \(rac{b}{a+b}\)∴ P (A wins the game) = P (W or BBW or BBBBW or ... ... the given condition,\(rac{a+b}{a+2b}\) = 3. \(rac{b}{a+2b}\) ⇒ a = 2b ⇒ a : b = 2 : 1.

There are 55 marbles and 100 bags. You need to put every marble in a way that the 1st bag has 1 marbles, the 2nd bag has 2 marbles and so on. How many bags will you need to use 55 marbles? -Riddles

Description : There are 55 marbles and 100 bags. You need to put every marble in a way that the 1st bag has 1 marbles, the 2nd bag has 2 marbles and so on. How many bags will you need to use 55 marbles? -Riddles

Last Answer : 55 bags. Put a marble in the first bag then the bag goes in the second bag with 1 marble, making the second have 2 marbles. Repeat this until use 55 marbles, in total,the 55th bag has 55 marbles.

On planet A , each alien has 3 eyes and on planet B , each alien has 4 eyes. A group of aliens from planets A and B landed on earth. They had 30 eyes in all. Assuming number of aliens from planets A and B respectively , express the situation in the form of linear equation in two variables. -Maths 9th

Description : On planet A , each alien has 3 eyes and on planet B , each alien has 4 eyes. A group of aliens from planets A and B landed on earth. They had 30 eyes in all. Assuming number of ... planets A and B respectively , express the situation in the form of linear equation in two variables. -Maths 9th

Last Answer : answer:

The probability of student A passing examination is 3/7 and of student B passing is 5/7 Assuming the two events “A passes”, -Maths 9th

Description : The probability of student A passing examination is 3/7 and of student B passing is 5/7 Assuming the two events “A passes”, -Maths 9th

Last Answer : p1 = P(A) = \(rac{3}{7}\), p2 = P(B) = \(rac{5}{7}\) ∴ q1 = P(\(\bar{A}\)) = 1 - P(A) = 1 - \(rac{3}{7}\) = \(rac{4}{7}\). q2 = P(\(\bar{B}\)) = 1 - P(B) = 1 - \(rac{5}{7}\) = \(rac{2}{7}\) ... passes) = p1 q2 + q1 p2 = \(rac{3}{7}\) x \(rac{2}{7}\) + \(rac{4}{7}\) x \(rac{5}{7}\) = \(rac{26}{49}.\)

Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in fig. These pieces are arranged in a way that AB||HC || GD || FE. Also BC=CD=DE, and GF=6 cm... -Maths 9th

Description : Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in fig. These pieces are arranged in a way that AB||HC || GD || FE. Also BC=CD=DE, and GF=6 cm... -Maths 9th

Last Answer : Given : Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in the fig. These pieces are arranged ... length of coloured tape required = 30 cm (b) The values are : Happiness, beauty, Knowledge.

If the median of data (arranged in ascending order) 31, 33, 35, x, x+10, 48, 48, 50 is 40, then find value of x. -Maths 9th

Description : If the median of data (arranged in ascending order) 31, 33, 35, x, x+10, 48, 48, 50 is 40, then find value of x. -Maths 9th

Last Answer : Given data is 31, 33, 35, x, x+10, 48, 48, 50 Number of observation = 8 (even) Median = Value of (8/2)th observation + Value of (8/2+1)th observation / 2 Value of 4th observation + Value of 5th observation / 2 = x + x + 10 / 2 = x + 5 ∴ x + 5 = 40 ⇒ x = 35

If the median of data (arranged in ascending order) 31, 33, 35, x, x+10, 48, 48, 50 is 40, then find value of x. -Maths 9th

Description : If the median of data (arranged in ascending order) 31, 33, 35, x, x+10, 48, 48, 50 is 40, then find value of x. -Maths 9th

Last Answer : Given data is 31, 33, 35, x, x+10, 48, 48, 50 Number of observation = 8 (even) Median = Value of (8/2)th observation + Value of (8/2+1)th observation / 2 Value of 4th observation + Value of 5th observation / 2 = x + x + 10 / 2 = x + 5 ∴ x + 5 = 40 ⇒ x = 35

Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in fig. These pieces are arranged in a way that AB||HC || GD || FE. Also BC=CD=DE, and GF=6 cm. He wants to decorate the card by putting up a colored tape on the non-parallel sides of the trapezium.. -Maths 9th

Description : Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in fig. These pieces are arranged in a way that ... the card by putting up a colored tape on the non-parallel sides of the trapezium.. -Maths 9th

Last Answer : Let us consider the following lay out of the greeting card. Trapeziums are arranged in such a way that AB || HC || GD || FE. Also BC=CD=DE and GF=6 cm and DE = 4cm. If three parallel lines make equal ... HG+GF+BC+CD+DE = 6+6+6+4+4+4=30 cm. (b) The values are: Happiness, beauty, Knowledge.

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!}}\) = \(rac{7!}{8!}\) = \(rac{7!}{8 imes7!}\) = \(rac{1}{8}\).

The letters of the word ‘COCHIN’ are permuted and all the permutations are arranged in alphabetical order as in English dictionary. -Maths 9th

Description : The letters of the word ‘COCHIN’ are permuted and all the permutations are arranged in alphabetical order as in English dictionary. -Maths 9th

Last Answer : answer:

In how many ways can the letters of the word “AFLATOON” be arranged if the consonants and vowels must occupy alternate places? -Maths 9th

Description : In how many ways can the letters of the word “AFLATOON” be arranged if the consonants and vowels must occupy alternate places? -Maths 9th

Last Answer : 24 ways is the answer

In how many ways can a mixed doubles game be arranged from amongst 8 married couples if no husband and wife play in the same game? -Maths 9th

Description : In how many ways can a mixed doubles game be arranged from amongst 8 married couples if no husband and wife play in the same game? -Maths 9th

Last Answer : For mixed doubles, we have 2 men and 2 women. 2 men out of 9 can be selected in 9C8 ways Out of 9 women, 2 will be there wife 30 only 7 are remaining for selecting 2 Now out of them 2 possible combinations can be made. So, ... =3024 Hence the answer is 3024.

There are 5 boys and 3 girls. In how many ways can they stand in a row so that no two girls are together? -Maths 9th

Description : There are 5 boys and 3 girls. In how many ways can they stand in a row so that no two girls are together? -Maths 9th

Last Answer : Have the 55 boys stand in a line. This can be done in 5!5! ways. For the moment, add a boy at each end, who will be removed when we're done. Now send the 33 girls, one at a time, to ... that each girl can either wedge herself between two boys or else stand next to a boy at one end or the other.

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) = \(rac{1}{12}\).

An urn contains nine balls, of which three are red, four are blue and two are green. -Maths 9th

Description : An urn contains nine balls, of which three are red, four are blue and two are green. -Maths 9th

Last Answer : (b) \(rac{2}{7}\)Let S be the sample space having 9 balls (3R + 4B + 2G) Then n(S) = Total number of ways in which 3 balls can be drawn out of the 9 balls= \(rac{9 imes8 imes7}{3 imes2}\) = 84Let A : Event of drawing three ... 2 = 24.∴ P(A) = \(rac{n(A)}{n(S)}\) = \(rac{24}{84}\) = \(rac{2}{7}\).

If from each of the three boxes containing 3 blue and 1 red balls, 2 blue and 2 red balls, -Maths 9th

Description : If from each of the three boxes containing 3 blue and 1 red balls, 2 blue and 2 red balls, -Maths 9th

Last Answer : (a) \(rac{13}{32}\)The three boxes B1, B2 and B2 contain the different coloured balls as follows :BlueRedBox 131Box 222Box 313There can be three mutually exclusive cases of drawing 2 blue balls and 1 red balls in the ways as given :Box ... {64}\) + \(rac{2}{64}\) = \(rac{26}{64}\) = \(rac{13}{32}\).

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

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 = \(rac{5! imes3!}{7!}\) = \(rac{1}{7}\)

6 boys and 6 girls are sitting in a row randomly. The probability that boys and girls sit alternately is -Maths 9th

Description : 6 boys and 6 girls are sitting in a row randomly. The probability that boys and girls sit alternately is -Maths 9th

Last Answer : (b) \(rac{1}{462}\)Let S be the sample space. Then, n(S) = Number of ways in which 6 boys and 6 girls can sit in a row = 12! Let E : Event of 6 girls and 6 boys sitting alternately. Then, the ... )= \(rac{2 imes6 imes5 imes4 imes3 imes2 imes1}{12 imes11 imes10 imes9 imes8 imes7}\) = \(rac{1}{462}\).

6 boys and 6 girls are seated in a row. Probability that all the boys sit together is -Maths 9th

Description : 6 boys and 6 girls are seated in a row. Probability that all the boys sit together is -Maths 9th

Last Answer : (d) \(rac{1}{132}\)Let S be the sample space for arranging 6 boys and 6 girls in a row. Then, n(S) = 12! If all 6 boys are to sit together, then consider the 6 boys as one entity. Now the ... ) = \(rac{7 imes6 imes5 imes4 imes3 imes2}{12 imes11 imes10 imes9 imes8 imes7}\) = \(rac{1}{132}\).

A bag contain 7red, 12white and 4green balls .what is the probability that ... 1. 3 balls are drwan all are white 2. 3 balls drawn on one of each colour

Description : A bag contain 7red, 12white and 4green balls .what is the probability that ... 1. 3 balls are drwan all are white 2. 3 balls drawn on one of each colour

Last Answer : A bag contain 7red, 12white and 4green balls .what is the probability that ... 1. 3 balls are drwan ... white 2. 3 balls drawn on one of each colour

A king decided to let a prisoner try to escape the prison with his life. The king placed 2 marbles in a jar that was glued to a table. One of the marbles was supposed to be black, and one was supposed to be blue. If the prisoner could pick the blue marble, he would escape the prison with his life. If he picked the black marble, he would be executed. However, the king was very mean, and he wickedly placed 2 black marbles in the jars and no blue marbles. The prisoner witnessed the king only putting 2 black marbles in the jars. If the jar was not see-through and the jar was glued to the table and that the prisoner was mute so he could not say anything, how did he escape with his life? -Riddles

Description : A king decided to let a prisoner try to escape the prison with his life. The king placed 2 marbles in a jar that was glued to a table. One of the marbles was supposed to be black, and one was ... and that the prisoner was mute so he could not say anything, how did he escape with his life? -Riddles

Last Answer : The prisoner grabbed one of the marbles from the jar and concealed it in his hand. He then swallowed it, and picked up the other marble and showed everyone. The marble was black, and since the other marble was swallowed, it was assumed to be the blue one. So the mean king had to set him free

What are the least numbers of packs of yellow marbles and blue marbles a person would have to buy t ohave the same number of each color of marble?

Description : What are the least numbers of packs of yellow marbles and blue marbles a person would have to buy t ohave the same number of each color of marble?

Last Answer : That would depend on how many yellow and blue marbles are in apack. If yellow and blue marbles are sold separately and there arethe same number of marbles in a pack, buy one of each. That'sprobably not the case.

Three girls Reshma, Salma and Mandeep are playing a game by standing on a circle of radius 5 m drawn in a park. -Maths 9th

Description : Three girls Reshma, Salma and Mandeep are playing a game by standing on a circle of radius 5 m drawn in a park. -Maths 9th

Last Answer : Solution :- Let R, S and M represent the position of Reshma, Salma and Mandeep respectively. Clearly △RSM is an isosceles triangle as RS = SM = 6m Join OS which intersect RM at A. In △ROS and △MOS OR = OM ( ... . ∴ RM = 2RA RM = 2 x 4.8 = 9.6m Hence, distance between Reshma and Mandeep is 9.6m.

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:

Lights of different colours are arranged below in order of increasing wavelengths: 1. Violet, indigo, blue 2. Green, orange, red 3. Red, green, indigo 4. Blue, green, yellow Which of the above are in proper sequence? (a) 1 only (b) 1 and 2 (c) 2 and 3 (d) 1 and 4

Description : Lights of different colours are arranged below in order of increasing wavelengths: 1. Violet, indigo, blue 2. Green, orange, red 3. Red, green, indigo 4. Blue, green, yellow Which of the above are in proper sequence? (a) 1 only (b) 1 and 2 (c) 2 and 3 (d) 1 and 4

Last Answer : Ans:(d)

If Chris Bromham took off on his Yamaha and jumped a horizontal distance of 74.0 m across a row of cars and assuming that he started and landed at the same level and was airborne for 1.3 s, what was his initial (angled) velocity as he left the ground?

Description : If Chris Bromham took off on his Yamaha and jumped a horizontal distance of 74.0 m across a row of cars and assuming that he started and landed at the same level and was airborne for 1.3 s, what was his initial (angled) velocity as he left the ground?

Last Answer : I really need help. If someone can explain this to me, it'll be great!