A box contains 5 different red and 6 different white balls. In how many ways can 6 balls be selected so that there are at least -Maths 9th

A box contains 5 different red and 6 different white balls. In how many ways can 6 balls be selected so that there are at least -Maths 9th
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Answer :

The selection of 6 balls, consisting of at least two balls of each color from 5 red and 6 white balls can be made in the following ways: Red balls (5) White balls(6) Number of ways 2 4 5 C 2 ​ × 6 C 4 ​ =150 3 3 5 C 3 ​ × 6 C 3 ​ =200 4 2 5 C 4 ​ × 6 C 2 ​ =75 Total 425
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Related questions

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)

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.

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

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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) = \(rac{n(A)}{n(S)}\) = \(rac{115}{1820}\) = \(rac{23}{364}.\)

A box contains 4 green, 5 red and 6 white balls. Three balls are drawn randomly. What is the probability that the balls drawn are of different colours? a) 24/91 b) 67/91 c) 21/91 d) 70/91 e) 3/13

Description : A box contains 4 green, 5 red and 6 white balls. Three balls are drawn randomly. What is the probability that the balls drawn are of different colours? a) 24/91 b) 67/91 c) 21/91 d) 70/91 e) 3/13

Last Answer : Answer is: a)

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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 )

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

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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.

A box contains 2 pink balls, 3 brown balls and 4 blue balls. In how many ways can 3 balls be drawn from the box, if at least one brown ball is to be included in the draw? A) 32 B) 48 C) 64 D) 96 E) None

Description : A box contains 2 pink balls, 3 brown balls and 4 blue balls. In how many ways can 3 balls be drawn from the box, if at least one brown ball is to be included in the draw? A) 32 B) 48 C) 64 D) 96 E) None

Last Answer : Answer: C) 

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 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 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 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)= \(rac{12}{64}+rac{20}{64}=rac{32}{64}=rac{1}{2}.\)

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) = \(rac{n(A)}{n(S)}\) = \(rac{24}{84}\) = \(rac{2}{7}\).

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

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

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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

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A box contains 15bolts of which 5 are defective. If 5 bolts are selected at random from the box, what is the probability that at least one of them is defective? a) 91/143 b) 101/143 c) 111/143 d) 121/143 e) 131/143

Description : A box contains 15bolts of which 5 are defective. If 5 bolts are selected at random from the box, what is the probability that at least one of them is defective? a) 91/143 b) 101/143 c) 111/143 d) 121/143 e) 131/143

Last Answer : Probability that at least one is defective = 1 – probability that none is defective Probability that none is defective = 10C5 / 15C5 = 12/143 Required Probability = 1-12/143 = 131/143 Answer: e)

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

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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 = \(rac{n(A)}{n(S)}\) = \(rac{7}{10}\).

In how many ways can 8 different balls be distributed in 6 different boxes can contain any number of balls except that ball 4 can only be put into box 4 or 5 ? A) 2×5^6 B) 2×6^7 C) 2×5^4 D) 2×4^7

Description : In how many ways can 8 different balls be distributed in 6 different boxes can contain any number of balls except that ball 4 can only be put into box 4 or 5 ? A) 2×5^6 B) 2×6^7 C) 2×5^4 D) 2×4^7

Last Answer : Answer: B)  1st ball can be put in any of the 6 boxes.  2nd ball can be put in any of the 6 boxes.  3rd ball can be put in any of the 6 boxes.  Ball 4 can only be put into box 4 or box 5. Hence, 4th ball ... put in any of the 6 boxes.  Hence, required number of ways = 6 6 6 2 6 6 6 6  = 2 6^7

In how many ways can a team of 11 players be selected from 14 players when two of them play as goalkeepers only? -Maths 9th

Description : In how many ways can a team of 11 players be selected from 14 players when two of them play as goalkeepers only? -Maths 9th

Last Answer : As each team of 11 players has one goalkeeper and 10 team members, and out of 14 players there are 2 goalkeepers and 12 team members. = 12×112×2 = 132.

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}\) + \(rac{2}{64}\) = \(rac{26}{64}\) = \(rac{13}{32}\).

A bowl contains 8 violet, 6 purple and 4 magenta balls. Three balls are drawn at random. Find out the number of ways of selecting the balls of different colours? A) 362 B) 2 48 C) 122 D) 192

Description : A bowl contains 8 violet, 6 purple and 4 magenta balls. Three balls are drawn at random. Find out the number of ways of selecting the balls of different colours? A) 362 B) 2 48 C) 122 D) 192

Last Answer : Answer: D)  1 violet ball can be selected is 8C1 ways.  1 purple ball can be selected in 6C1 ways.  1 magenta ball can be selected in 4C1 ways.  Total number of ways = 8C1 × 6C1 × 4C1  = 8×6×4  = 192

A box contains 50 bolts and 150 nuts. On checking the box, it was found that half of the bolts and half of the nuts are rusted. -Maths 9th

Description : A box contains 50 bolts and 150 nuts. On checking the box, it was found that half of the bolts and half of the nuts are rusted. -Maths 9th

Last Answer : Total number of nuts and bolts in the box = 150 + 50 = 200 Number of nuts and bolts rusted = 1 / 2 × 200 = 100 P(a rusted nut or bolt) = 100 / 200 = 1 / 2

A box contains 50 bolts and 150 nuts. On checking the box, it was found that half of the bolts and half of the nuts are rusted. -Maths 9th

Description : A box contains 50 bolts and 150 nuts. On checking the box, it was found that half of the bolts and half of the nuts are rusted. -Maths 9th

Last Answer : Total number of nuts and bolts in the box = 150 + 50 = 200 Number of nuts and bolts rusted = 1 / 2 × 200 = 100 P(a rusted nut or bolt) = 100 / 200 = 1 / 2

Out of 5 men and 2 women, a committee of 3 is to be formed. In how many ways can it be formed if at least one woman is to be included? -Maths 9th

Description : Out of 5 men and 2 women, a committee of 3 is to be formed. In how many ways can it be formed if at least one woman is to be included? -Maths 9th

Last Answer : Number of selections = Number of ways of selecting 2 men out of 5 men x number of ways of selecting l woman out of 3 women. = 5C2×3C1=5×41×2×3=30

A box contains 50 tickets. Each ticket is numbered from 1 to 50. One ticket is selected at random, find the probability that the number on the ticket

Description : A box contains 50 tickets. Each ticket is numbered from 1 to 50. One ticket is selected at random, find the probability that the number on the ticket

Last Answer : A box contains 50 tickets. Each ticket is numbered from 1 to 50. One ticket is selected at random ... number on the ticket is not a perfect square.

From 5 different green balls, four different blue balls and three different red balls, how many combinations of balls can be chosen taking at least one green and one blue ball?

Description : From 5 different green balls, four different blue balls and three different red balls, how many combinations of balls can be chosen taking at least one green and one blue ball?

Last Answer : Answer: 3720.

How many balls, each of radius 2 cm can be made from a solid sphere of lead of radius 8 cm ? -Maths 9th

Description : How many balls, each of radius 2 cm can be made from a solid sphere of lead of radius 8 cm ? -Maths 9th

Last Answer : No.of balls = Volume of share / Volume of each ball = 4 / 3π × 8 × 8 × 8 / 4 / 3π × 2 × 2 × 2 = 64

How many balls, each of radius 2 cm can be made from a solid sphere of lead of radius 8 cm ? -Maths 9th

Description : How many balls, each of radius 2 cm can be made from a solid sphere of lead of radius 8 cm ? -Maths 9th

Last Answer : No.of balls = Volume of share / Volume of each ball = 4 / 3π × 8 × 8 × 8 / 4 / 3π × 2 × 2 × 2 = 64

The cost of a toy horse is same as that of cost of 3 balls. Express this statement as a linear equation in two variables. Also draw its graph. -Maths 9th

Description : The cost of a toy horse is same as that of cost of 3 balls. Express this statement as a linear equation in two variables. Also draw its graph. -Maths 9th

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A cylindrical rod of iron whose height is eight times its radius is melted and cast into spherical balls each of half the radius of the cylinder. -Maths 9th

Description : A cylindrical rod of iron whose height is eight times its radius is melted and cast into spherical balls each of half the radius of the cylinder. -Maths 9th

Last Answer : Let radius of iron rod = r ∴∴ Height = 8r ∴∴ Volume of iron rod =π×(r)2×8r⇒8πr3=π×(r)2×8r⇒8πr3 ⇒⇒ Radius of spherical ball =r2=r2 Volume of spherical ball =43π(r2)3=43π(r2)3 Let n balls are casted ∴n×43π(r38)=8πr3∴n×43π(r38)=8πr3 ⇒n6=8⇒n=48

A cylindrical rod of iron whose radius is one-fourth of its height is melted and cast into spherical balls of the same radius as that of the cylinder. -Maths 9th

Description : A cylindrical rod of iron whose radius is one-fourth of its height is melted and cast into spherical balls of the same radius as that of the cylinder. -Maths 9th

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Three identical balls fit snugly into a cylindrical can. The radius of the spheres is equal to the radius of the can and the balls just touch the -Maths 9th

Description : Three identical balls fit snugly into a cylindrical can. The radius of the spheres is equal to the radius of the can and the balls just touch the -Maths 9th

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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

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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

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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

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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

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

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Description : 1500 families with 2 children were selected -Maths 9th

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Description : An Insurance company selected 2000 drivers -Maths 9th

Last Answer : Total number of drivers = 2000 (i) Number of drivers who are 18-29 years old and have exactly 3 accidents in one year is 61 So, P (driver is 18-29 years old with exactly 3 accidents) = 61/2000 = 0.0305 ~ 0. ... = 440 + 505 + 360 = 1305 So, P (drivers with no accident) = 1305/2000 = 0.6525 = 0.653

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) = \(rac{n(A)}{n(S)}\) = \(rac{4}{100}\) = \(rac{1}{25}\)

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 = \(rac{n(E)}{n(S)}\) = \(rac{2}{5}\).

If three natural numbersfrom 1 to 100 are selected randomly, then the probability that all are divisible by both 2 and 3 is -Maths 9th

Description : If three natural numbersfrom 1 to 100 are selected randomly, then the probability that all are divisible by both 2 and 3 is -Maths 9th

Last Answer : (c) \(rac{4}{1155}\)Let n(S) = Number of ways of selecting 3 numbers from 100 numbers = 100C3 Let E : Event of selecting three numbers divisible by both 2 and 3 from numbers 1 to 100 = Event of selecting three ... C_3}{^{100}C_3}\) = \(rac{16 imes15 imes14}{100 imes99 imes98}\) = \(rac{4}{1155}\).

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 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:

Three letters are randomly selected from the 26 capital letters of the English alphabet. -Maths 9th

Description : Three letters are randomly selected from the 26 capital letters of the English alphabet. -Maths 9th

Last Answer : (d) \(rac{36}{1001}\)Required probability = \(rac{P( ext{2 blue balls}) imes{P}( ext{2 red balls})}{P( ext{4 balls out of 14 balls})}\) + \(rac{P( ext{2 green balls}) imes{P}( ext{2 black balls})}{P( ext{4 out of 14 balls})}\)