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

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
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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) = P (C) = P (D). Since out of 16 balls, 11 are not white, therefore, P (A) = \(rac{11}{16}\).∴ Required probability = P (A) . P (B) . P (C) . P (D)= \(rac{11}{16}\) x \(rac{11}{16}\) x \(rac{11}{16}\) x \(rac{11}{16}\) = \(\big(rac{11}{16}\big)^4.\)
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Related questions

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

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 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 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 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

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

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)

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 )

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.

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)

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 Receptacle contains 3violet, 4purple and 5 black balls. Three balls are drawn at random from the receptacle. The probability that all of them are purple, is: A)3/55 B)7/55 C)1/55 D)9/55

Description : A Receptacle contains 3violet, 4purple and 5 black balls. Three balls are drawn at random from the receptacle. The probability that all of them are purple, is: A)3/55 B)7/55 C)1/55 D)9/55

Last Answer : Answer: C) Let S be the sample space. Then, n(S) = number of ways of drawing 3 balls out of 12 = 12C3 = 220 Let E = event of getting all the 3 purple balls. n(E) = 4C3= 4 P(E) = n(E)/n(S) = 4/220 = 1/55

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

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:

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

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

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 \(rac{5 imes4 imes3 imes2 imes1}{30 imes29 imes28 imes27 imes26}\) = \(rac{285}{5278}.\)

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

Find the probability that the three cards drawn from a pack of 52 cards are all black ? -Maths 9th

Description : Find the probability that the three cards drawn from a pack of 52 cards are all black ? -Maths 9th

Last Answer : Number of ways in which three cards can be drawn from a pack of 52 cards n(S) = 52C3. Let A : Event of drawing all the three cards as black Then, n(A) = 26C3 (∵There are 26 black cards)∴ P(A ... (rac{^{26}C_3}{^{52}C_3}\) = \(rac{26 imes25 imes24}{52 imes51 imes50}\) = \(rac{2}{17}.\)

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 ... ) + \(rac{6}{1326}\) - \(rac{1}{1326}\) = \(rac{330}{1326}\) = \(rac{55}{221}\).

Four cards are drawn from a full pack of cards. Find the probability that : -Maths 9th

Description : Four cards are drawn from a full pack of cards. Find the probability that : -Maths 9th

Last Answer : 4 cards can be drawn from a pack of cards in 52C4 ways ∴ Exhaustive number of cases = n(S) = 52C4 (a) There are 4 suits, each containing 13 cards. Let A : Event of drawing one card from each suit ⇒ Favourable number of ... = \(rac{15229}{54145}\) (∵ P(Event) + P(complement of event) = 1)

A pack contains 4 blue, 2 red and 3 black pens. If a pen is drawn at random from the pack, replaced and the process repeated 2 more times, what is the probability of drawing 2 blue pens and 1 black pen? a) 16/243 b) 16/283 c) 14/243 d) 23/729

Description : A pack contains 4 blue, 2 red and 3 black pens. If a pen is drawn at random from the pack, replaced and the process repeated 2 more times, what is the probability of drawing 2 blue pens and 1 black pen? a) 16/243 b) 16/283 c) 14/243 d) 23/729

Last Answer : a) 16/243

Consider a pack contains 2black, 9 white and 3 pink pencils. If a pencil is drawn at random from the pack, replaced and the process repeated 2 more times, What is the probability of drawing 2 black pencils and 1 pink pencil? a)3/ 49 b)3/686 c)3/14 d)3/545

Description : Consider a pack contains 2black, 9 white and 3 pink pencils. If a pencil is drawn at random from the pack, replaced and the process repeated 2 more times, What is the probability of drawing 2 black pencils and 1 pink pencil? a)3/ 49 b)3/686 c)3/14 d)3/545

Last Answer : Answer: B) Here, total number of pencils = 14 Probability of drawing 1 black pencil = 2/14 Probability of drawing another black pencil = 2/14 Probability of drawing 1 pink pencil = 3/14 Probability of drawing 2 black pencils and 1 pink pencil = 2/14 * 2/14 * 3/14 = 3/686

A box contains 3red, 8 blue and 5 green marker pens. If 2 marker pens are drawn at random from the pack, not replaced and then another pen is drawn. What is the probability of drawing 2 blue marker pens and 1 red marker pen? a) 3/20 b) 1/20 c) 7/20 d) 9/20

Description : A box contains 3red, 8 blue and 5 green marker pens. If 2 marker pens are drawn at random from the pack, not replaced and then another pen is drawn. What is the probability of drawing 2 blue marker pens and 1 red marker pen? a) 3/20 b) 1/20 c) 7/20 d) 9/20

Last Answer : Answer: B) Probability of drawing 1 blue marker pen =8/16 Probability of drawing another blue marker pen = 7/15 Probability of drawing 1 red marker pen = 3/14 Probability of drawing 2 blue marker pens and 1 red marker pen = 8/16*7/15*3/14=1/20

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

Two cards are drawn from a well shuffled pack of 52 cards one after another without replacement. -Maths 9th

Description : Two cards are drawn from a well shuffled pack of 52 cards one after another without replacement. -Maths 9th

Last Answer : Probability of drawing an ace in the first draw = \(rac{4}{52}.\)Probability of drawing a queen of opposite shade in the second draw = \(rac{2}{51}.\)Probability of drawing a queen in the first draw = \(rac{4}{52}.\) ... \(rac{2}{51}\) = \(rac{4}{663}.\) [ AND' and OR'Theorems]

A box contains 2 black, 3 orange and 4 pink ribbons. If two ribbons are drawn at random. What is the probability that both are orange? 1) 5/12 2) 1/13 3) 1/14 4) 1/12 5) 1/24

Description : A box contains 2 black, 3 orange and 4 pink ribbons. If two ribbons are drawn at random. What is the probability that both are orange? 1) 5/12 2) 1/13 3) 1/14 4) 1/12 5) 1/24

Last Answer : 4) 1/12

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

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 bag contains 2n + 1 coins. It is known that n of these coins have a head on both sides, whereas the remaining (n + 1) coins are fair. -Maths 9th

Description : A bag contains 2n + 1 coins. It is known that n of these coins have a head on both sides, whereas the remaining (n + 1) coins are fair. -Maths 9th

Last Answer : (a) 10As (n + 1) coins are fair P (Tossing a tail) = \(rac{rac{n+1}{2}}{2n+1}\) = \(rac{n+1}{2(2n+1)}\)∴ P (Tossing a head) = 1 - \(rac{n+1}{2(2n+1)}\) = \(rac{4n+2-n-1}{2(2n+1)}\) = \(rac{3n+1}{4n+2}\)Given, \(rac{3n+1}{4n+2}\) = \(rac{31}{42}\)⇒ 126n + 42 = 124n + 62 ⇒ 2n = 20 ⇒ n = 10.

All red face cards are removed from a pack of playing cards. The remaining cards are well shuffled and then a card is drawn at random from them. Find the probability that the drawn card is face card. a. 2/23 b. 7/44 c. 3/23 d. 4/25

Description : All red face cards are removed from a pack of playing cards. The remaining cards are well shuffled and then a card is drawn at random from them. Find the probability that the drawn card is face card. a. 2/23 b. 7/44 c. 3/23 d. 4/25

Last Answer : c. 3/23

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 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) 

Four boys and three girls stand in a queue for an interview. What is the probability that they will be in alternate positions ? -Maths 9th

Description : Four boys and three girls stand in a queue for an interview. What is the probability that they will be in alternate positions ? -Maths 9th

Last Answer : Total number of ways of arranging 4 boys and 3 girls, i.e., 7 people in a queue (row) = n(S) = 7! Let A : Event in which the 4 boys and 3 girls occupy alternate position. This is possible when the ... {4 imes3 imes2 imes1 imes3 imes2 imes1}{7 imes6 imes5 imes4 imes3 imes2 imes1}\) = \(rac{1}{35}.\)

A four digit number is formed by the digits 1, 2, 3, 4 with no repetition. The probability that the number is odd is -Maths 9th

Description : A four digit number is formed by the digits 1, 2, 3, 4 with no repetition. The probability that the number is odd is -Maths 9th

Last Answer : (a) \(rac{1}{2}\) Let S be the sample spaceThen n(S) = Number of four digit numbers that can be formed with out repetition1 of 4 digits1 of 3 digits1 of 2 digits1 of 1 digitThHTO= 4! = 4 3 2 1 ways = 24 ways. Let E : ... 1 = 12∴ P(E) = \(rac{n(E)}{n(S)}\) = \(rac{12}{24}\) = \(rac{1}{2}\).

ABC is an isosceles triangle in which altitude BE and CF are drawn to equal sides AC and AB respectively (Fig. 7.15). Show that these altitudes are equal. -Maths 9th

Description : ABC is an isosceles triangle in which altitude BE and CF are drawn to equal sides AC and AB respectively (Fig. 7.15). Show that these altitudes are equal. -Maths 9th

Last Answer : In △ABE and △ACF, we have ∠BEA=∠CFA (Each 90 0 ) ∠A=∠A (Common angle) AB=AC (Given) ∴△ABE≅△ACF (By SAS congruence criteria) ∴BF=CF [C.P.C.T]

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 \(rac{5}{7}\) = \(rac{26}{49}.\)

Among 15 players, 8 are batsman and 7 are bowlers. Find the probability that a team is chosen of 6 batsman and 5 bowlers ? -Maths 9th

Description : Among 15 players, 8 are batsman and 7 are bowlers. Find the probability that a team is chosen of 6 batsman and 5 bowlers ? -Maths 9th

Last Answer : The chosen consists of players (6 + 5). ∴ Number of ways of selecting 11 players out of 15 players = n(S) = 15C11 Let A : Event of choosing 6 batsmen of 8 batsmen and 5 bowlers of 7 bowlers Then, n(A) = 8C6 ... 7 imes6}{2}\) x \(rac{4 imes3 imes2 imes1}{15 imes14 imes13 imes12}\) = \(rac{28}{65}.\)

The probability of A, B, C solving a problem are 1/3, 2/7 and 3/8 respectively. -Maths 9th

Description : The probability of A, B, C solving a problem are 1/3, 2/7 and 3/8 respectively. -Maths 9th

Last Answer : Let E1, E2, E3 be the eventsthat the problem issolved by A, B, C respectively and let p1, p2, p3 be corresponding probabilities. Then,p1 = P(E1) = \(rac{1}{3}\), p2 = P(E2) = \(rac{2}{7}\), p3 = P(E3) = \(rac{3}{8}\) ... }{8}\) = \(rac{25}{168}\) + \(rac{5}{42}\) + \(rac{5}{28}\) = \(rac{25}{56}.\)

Two dice are rolled once. Find the probability of getting an even number on the first die, or a total of 7. -Maths 9th

Description : Two dice are rolled once. Find the probability of getting an even number on the first die, or a total of 7. -Maths 9th

Last Answer : (c) \(rac{7}{12}\)Total number of ways in which 2 dice are rolled = 6 6 = 36 ⇒ n(S) = 36 Let A : Event of rolling an even number of 1st dice B : Event of rolling a total of 7 ⇒ A = {(2, 1), (2, 2) , (2, 6), (4 ... (rac{18}{36}\) + \(rac{6}{36}\) - \(rac{3}{36}\) = \(rac{21}{36}\) = \(rac{7}{12}\).

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 Package contains 12 pack of variety1 drink, 6 pack of variety2 drink and 8pack of variety3 drink. Three packsof them are drawn at random, what is the probability that the three are not of the same variety? a) 37/325 b) 288/325 c) 188/325 d) None of these

Description : A Package contains 12 pack of variety1 drink, 6 pack of variety2 drink and 8pack of variety3 drink. Three packsof them are drawn at random, what is the probability that the three are not of the same variety? a) 37/325 b) 288/325 c) 188/325 d) None of these

Last Answer : Answer: B) Total number of drink pack= 12+6+8= 26. Let S be the sample space. Then, n(S) = number of ways of taking 3 drink pack out of 26. Therefore, n(S) = 26C3 = 2600 Let Ebe the ... 296/2600=37/325 Then, the probability of taking 3 pack are not of the same variety = 1 - 37/325= 288/325

A carton contains 12 green and 8 blue bulbs .2 bulbs are drawn at random. Find the probability that they are of same colour. A) 91/47 B) 47/105 C) 47/95 D) 95/47

Description :  A carton contains 12 green and 8 blue bulbs .2 bulbs are drawn at random. Find the probability that they are of same colour. A) 91/47 B) 47/105 C) 47/95 D) 95/47

Last Answer : Answer: C) Let S be the sample space Then n(S) = no of ways of drawing 2 bulbs out of (12+8) = 20c2=20*19/2*1=190 Let E = event of getting both bulbs of same colour Then, n(E) = no of ways (2 bulbs out of 12) ... 12C2+ 8C2=(132/2)+(56/2) = 66+28 = 94 Therefore, P(E) = n(E)/n(S) = 94/190 = 47/95

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.