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

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

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 1Box 2Box 3Case I1 Blue1 Blue1 RedCase II1 Blue1 Red1 BlueCase III1 Red1 Blue1 Blue∴ P(Drawing 2 blue and 1 red ball)= P(B1 (Blue), B2 (Blue), B3 (Red)) + P(B1 (Blue), B2 (Red), B3 (Blue)) + P(B1 (Red), B2 (Blue), B3 (Blue)) = \(rac{3}{4}\) x \(rac{2}{4}\) x \(rac{3}{4}\) + \(rac{3}{4}\) x \(rac{2}{4}\) x \(rac{1}{4}\) + \(rac{1}{4}\) x \(rac{2}{4}\) x \(rac{1}{4}\)= \(rac{18}{64}\) + \(rac{6}{64}\) + \(rac{2}{64}\) = \(rac{26}{64}\) = \(rac{13}{32}\).
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Related questions

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.

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

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

A matchbox measures 4 cm×2.5cm×1.5cm. What will be the volume of a packet containing 12 such boxes? -Maths 9th

Description : A matchbox measures 4 cm×2.5cm×1.5cm. What will be the volume of a packet containing 12 such boxes? -Maths 9th

Last Answer : Dimensions of a matchbox (a cuboid) are l×b×h = 4 cm×2.5 cm×1.5 cm respectively Formula to find the volume of matchbox = l×b×h = (4×2.5×1.5) = 15 Volume of matchbox = 15 cm3 Now, volume of 12 such matchboxes = (15×12) cm3 = 180 cm3 Therefore, the volume of 12 matchboxes is 180cm3.

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

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⇒ ... \(rac{n(B)}{n(S)}\) = \(rac{21}{66}\) + \(rac{10}{66}\) = \(rac{31}{66}\).

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) ... \(rac{2}{6}\) = \(rac{2}{7}\) + \(rac{1}{6}\) = \(rac{19}{42} \).

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

Description : 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

Last 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

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

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.

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 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 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

Last Answer : hope its clear

Picture three boxes containing fruit. The first box is marked peaches, the second is marked oranges, and the third box is marked peaches and oranges. Each of the boxes is labeled incorrectly. How could you label each box correctly if you were allowed to select only one fruit from one of the boxes? -Riddles

Description : Picture three boxes containing fruit. The first box is marked peaches, the second is marked oranges, and the third box is marked peaches and oranges. Each of the boxes is labeled incorrectly. How could ... box correctly if you were allowed to select only one fruit from one of the boxes? -Riddles

Last Answer : First you select a fruit from the box marked peaches and oranges. If it was a orange you selected, you know that the box could only contain oranges. If it was a peach, you know that ... that box oranges and switch the other two incorrect labels around. Now all three would be correctly labeled.

Natalie had fourteen balls. Each one color. She alternated the colors each day, there was a blue, pink, purple, green, red, yellow, black, white, orange, turquoise, tan, neon pink, and brown. Each day of the week she also alternated fast food places. McDonald's, Chick-fil-A, In-and-out, Wendy's, Good eats, Burger King, and Arby's None in respective order. What day of the week will she go to Arby's? What ball will she have while she is at Arby's? -Riddles

Description : Natalie had fourteen balls. Each one color. She alternated the colors each day, there was a blue, pink, purple, green, red, yellow, black, white, orange, turquoise, tan, neon pink, and brown. Each day of the ... the week will she go to Arby's? What ball will she have while she is at Arby's? -Riddles

Last Answer : She will go to Arby's on Monday, for that is the first day of the week and A is the first letter of the alphabet. So on Tuesday she would go to Burger King. On Wednesday she would go to Chick- ... because red is the first color of the rainbow. So on Tuesday she would have a Orange ball. And so on.

Books are packed in piles each containing 20 books. Thirty five piles were examined for defective books and the results are given in the following table : -Maths 9th

Description : Books are packed in piles each containing 20 books. Thirty five piles were examined for defective books and the results are given in the following table : -Maths 9th

Last Answer : Total number of books = 700 (i) P(no defective books) = 400 / 700 = 4 / 7 (ii) P(more than 0 but less than 4 defective books) = 269 / 700 (iii) P(more than 4 defective books) = 13 / 700

Books are packed in piles each containing 20 books. Thirty five piles were examined for defective books and the results are given in the following table : -Maths 9th

Description : Books are packed in piles each containing 20 books. Thirty five piles were examined for defective books and the results are given in the following table : -Maths 9th

Last Answer : Total number of books = 700 (i) P(no defective books) = 400 / 700 = 4 / 7 (ii) P(more than 0 but less than 4 defective books) = 269 / 700 (iii) P(more than 4 defective books) = 13 / 700

Bulbs are packed in cartons each containing 40 bulbs. -Maths 9th

Description : Bulbs are packed in cartons each containing 40 bulbs. -Maths 9th

Last Answer : (i) P (a carton has no defective bulb) = 400/700 = 4/7 (ii) P (defective bulbs from 2 to 6) = P (2 defective bulbs) + P (3 defective bulbs) + P (4 defective bulbs) + P (5 defective bulbs) + P (6 ... + P (2 defective bulbs) + P (3 defective bulbs) = 400/700 + 180/700 + 48/700 + 41/700 = 669/700

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.

Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. -Maths 9th

Description : Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. -Maths 9th

Last Answer : Let l, b and h be the length, breadth and height of the box. Bigger Box: l = 25cm b = 20 cm h = 5 cm Total surface area of bigger box = 2(lb+lh+bh) = [2(25 20+25 5+20 5)] ... (546000 4)/1000 = Rs. 2184 Therefore, the cost of cardboard required for supplying 250 boxes of each kind will be Rs. 2184.

If a wooden box of dimensions 8 m x 7 m x 6 m is to carry boxes of dimensions 8 cm x 7 cm x 6 cm, then find the maximum number of boxes that can be carried in the wooden box. -Maths 9th

Description : If a wooden box of dimensions 8 m x 7 m x 6 m is to carry boxes of dimensions 8 cm x 7 cm x 6 cm, then find the maximum number of boxes that can be carried in the wooden box. -Maths 9th

Last Answer : Volume of wooden box = 800 cm × 700 cm × 600 cm Volume of box = 8 cm × 7 cm × 6 cm ∴ Number of boxes = volume of wooden box / volume of each box = 800 cm × 700 cm × 600 cm / 8 cm × 7 cm × 6 cm = 1000000

If a wooden box of dimensions 8 m x 7 m x 6 m is to carry boxes of dimensions 8 cm x 7 cm x 6 cm, then find the maximum number of boxes that can be carried in the wooden box. -Maths 9th

Description : If a wooden box of dimensions 8 m x 7 m x 6 m is to carry boxes of dimensions 8 cm x 7 cm x 6 cm, then find the maximum number of boxes that can be carried in the wooden box. -Maths 9th

Last Answer : Volume of wooden box = 800 cm × 700 cm × 600 cm Volume of box = 8 cm × 7 cm × 6 cm ∴ Number of boxes = volume of wooden box / volume of each box = 800 cm × 700 cm × 600 cm / 8 cm × 7 cm × 6 cm = 1000000

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

Description : 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

Last Answer : answer:

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

Last Answer : Solution :-

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.

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

Last Answer : Let radius of cylindrical rod =r ⇒ height =4r Volume of cylindrical rod =πr2h =πr2(4r) =4πr3 Volume of spherical balls of radius r=34​πr3 No. of balls =34​πr34πr3​=

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

If you have 4 green balls 2 red balls and 1 blue what is probability of getting a green?

Description : If you have 4 green balls 2 red balls and 1 blue what is probability of getting a green?

Last Answer : It is 1/7 if a ball is selected at random.

If A is the area of the right angled triangle and b is one of the sides containing the right angle, then what is the length of the -Maths 9th

Description : If A is the area of the right angled triangle and b is one of the sides containing the right angle, then what is the length of the -Maths 9th

Last Answer : answer:

A drawer contains 5 brown and 4 blue socks well mixed. A man reaches the drawer and pulls out 2 -Maths 9th

Description : A drawer contains 5 brown and 4 blue socks well mixed. A man reaches the drawer and pulls out 2 -Maths 9th

Last Answer : (c) \(rac{4}{9}\)n(S) = Total number of waysin which 2 socks can be drawn out of 9 socks (5 brown and 4 blue socks)= 9C2 = \(rac{|\underline9}{|\underline7|\underline2}\) = \(rac{9 imes8}{2}\) = 36Let A : Event of ... Required probability P(A) = \(rac{n(A)}{n(S)}\) = \(rac{16}{36}\) = \(rac{4}{9}\).

A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in figure. -Maths 9th

Description : A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in figure. -Maths 9th

Last Answer : Each shade of paper is divided into 3 triangles i.e., I, II, III 8 cm For triangle I: ABCD is a square [Given] ∵ Diagonals of a square are equal and bisect each other. ∴ AC = BD = 32 cm Height of AABD ... are: Area of shade I = 256 cm2 Area of shade II = 256 cm2 and area of shade III = 17.92 cm2

Rationalise the denominator in each of the following and hence evaluate by taking √2 = 1.414, √3 = 1.732 and √5 = 2.236 up to three places of decimal. -Maths 9th

Description : Rationalise the denominator in each of the following and hence evaluate by taking √2 = 1.414, √3 = 1.732 and √5 = 2.236 up to three places of decimal. -Maths 9th

Last Answer : Rationalisation of denominator

Rationalise the denominator in each of the following and hence evaluate by taking √2 = 1.414, √3 = 1.732 and √5 = 2.236 up to three places of decimal. -Maths 9th

Description : Rationalise the denominator in each of the following and hence evaluate by taking √2 = 1.414, √3 = 1.732 and √5 = 2.236 up to three places of decimal. -Maths 9th

Last Answer : Rationalisation of denominator

Three circles of radius a, b, c touch each other externally. The area of the triangle formed by joining their centres is: -Maths 9th

Description : Three circles of radius a, b, c touch each other externally. The area of the triangle formed by joining their centres is: -Maths 9th

Last Answer : (a) \(\sqrt{(a+b+c).a.b.c}\)As shown in the figure, AB = a + b, BC = b + c, CA = a + c∴ Area of ΔABC = \(\sqrt{s(s-AB)(s-BC)(s-CA)}\)where, s = \(rac{1}{2}\) (AB + BC + CA)= \(rac{a+b+b+c+c+a}{2}\) = a + b + ... (\sqrt{(a+b+c)[(a+b+c)-(a+b)][(a+b+c)-(b+c)][(a+b+c)-(c+a)]}\)= \(\sqrt{(a+b+c).a.b.c}\)

Three identical dice are rolled. The probability that same number will appear on each of them is -Maths 9th

Description : Three identical dice are rolled. The probability that same number will appear on each of them is -Maths 9th

Last Answer : (d) \(rac{1}{36}\)Total number of outcomes when three identical dice are rolled, n(S) = 6 6 6 = 216 Let A : Event of rolling same number or each dice ⇒ A = {(1, 1, 1), (2, 2, 2), (3, 3, 3), (4, 4, 4), (5, 5, 5), ... 6)} ⇒ n(A) = 6 ∴ P(A) = \(rac{n(A)}{n(S)}\) = \(rac{6}{216}\) = \(rac{1}{36}\).

A die is rolled three times. The probability of getting a larger number than the previous number each time is: -Maths 9th

Description : A die is rolled three times. The probability of getting a larger number than the previous number each time is: -Maths 9th

Last Answer : (b) \(rac{5}{24}\)Total number of ways three die can be rolled = 6 6 6 = 216 A larger number than the previous number can be got in the three throws as (1, 2, 3), (1, 2, 4), (1, 2, 5) ( ... , 5, 6). ∴ Total number of favourable cases = 20∴ Required probability =\(rac{20}{216}\) = \(rac{5}{24}\).

A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. -Maths 9th

Description : A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. -Maths 9th

Last Answer : (c) \(rac{11}{3^5}\)Probability of guessing a correct answer = \(rac{1}{3}\)Probability of guessing an incorrect answer = \(rac{2}{3}\)∴ Probability of guessing 4 or more correct answers = 5C4 \(\big(rac{1}{3}\big)^4\)\(\big(rac{2 ... )^5\) = 5 x \(rac{2}{3^5}\) + \(rac{1}{3^5}\) = \(rac{11}{3^5}\).

In how many ways can a pack of 52 cards be divided into 4 sets, three of them having 16 cards each and the fourth just 4 cards? -Maths 9th

Description : In how many ways can a pack of 52 cards be divided into 4 sets, three of them having 16 cards each and the fourth just 4 cards? -Maths 9th

Last Answer : First we divide 52 cards into two sets which contains 1 and 51 cards respectively is 1! 51! 52! Now 51 cards can be divided equally in three sets each contains 17 cards (Here order of sets is not important) in 3!(17!) ... ways Hence, the required number of ways = 1! 51! 52! 3! (17!) 3 51!

If three cubes of copper, each with an edge 6 cm, 8 cm and 10 cm respectively are melted to form a single cube, -Maths 9th

Description : If three cubes of copper, each with an edge 6 cm, 8 cm and 10 cm respectively are melted to form a single cube, -Maths 9th

Last Answer : 20.8 cm Let the edge of the single cube be ‘a’ cm. Then, total volume melted = Volume of cube formed ⇒ (6)3 + (8)3 + (10)3 = a3 ⇒ a3 = 216 + 512 + 1000 = 1728 ... Diagonal of the new cube = 3–√a=(3–√×12)3a=(3×12) cm = 20.8 cm (approx.)

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 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)