A pack of cards contains 4 aces, 4 kings, 4 queens and 4 jacks. -Maths 9th

A pack of cards contains 4 aces, 4 kings, 4 queens and 4 jacks. -Maths 9th
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1 Answer

Answer :

(d) \(rac{9}{20}\)Let S be the sample space for drawing 2 cards out of 4 aces, 4 kings, 4 queens and 4 jacks i.e, 16 cards. Then n(S) = 16C2 P(Drawing at least one ace) = 1 – P(Drawing no ace) Let E : Event of drawing no aces in the 2 drawn cards ⇒ n(E) = 12C2(Cards leaving aces = 16 – 4 – 12)∴ P(E) = \(rac{n(E)}{n(S)}\) = \(rac{^{12}C_2}{^{16}C_2}\) = \(rac{12 imes11}{16 imes15}\) = \(rac{11}{20}\)∴ P(drawing at least one ace) = 1 - \(rac{11}{20}\) = \(rac{9}{20}\) .
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Related questions

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 ... \(rac{4 imes3}{52 imes51}\) - \(rac{2}{52 imes51}\) = \(rac{660}{2652}\) = \(rac{55}{221}.\)

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

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

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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]

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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}\) = \(rac{26 imes25 imes24}{52 imes51 imes50}\) = \(rac{2}{17}.\)

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

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Last Answer : (a) \(rac{13}{25}\)Let E : Event of drawing a queen in a single draw the pack of 52 cards. As there are 4 queens in a pack of 52 cards,P(E) = \(rac{4}{52}\) = \(rac{1}{13}\)P(\(\bar{E}\)) = P(not ... {25}\). [Sum of a G.P with infinite terms = \(rac{a}{1-r}\) where a = 1st term, r = common ratio.]

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

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

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Description : Two decks of playing cards are well shuffled and 26 cards are randomly distributed to a player. -Maths 9th

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Two cards are drawn at random from a pack of 52 cards.what is the probability that either both are Red or both are king? A) 52/221 B) 55/190 C) 55/221 D) 19/221

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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

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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 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

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

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 \(rac{3}{7}\) = \(rac{4+6}{21}\) = \(rac{10}{21}.\)

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 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 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) = \(rac{n(A)}{n(S)}\) = \(rac{16}{36}\) = \(rac{4}{9}\).

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 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 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 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 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 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.

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

A cylindrical container of height 14 m and base 12 m contains oil. The oil is to be transferred to one cylindrical can, -Maths 9th

Description : A cylindrical container of height 14 m and base 12 m contains oil. The oil is to be transferred to one cylindrical can, -Maths 9th

Last Answer : answer:

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

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

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

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

Like New Pr Like New Products w oducts wants to i ants to improve its pack ve its packaging aft aging after readi er reading custo ng customer res mer responses t ponses to its customer its customer opinion poll. opinion poll. Which is not Which is not a function of a function of packaging? packaging? a. Its purpose is to c e is to contain and protect the product. b. It contains the brand mark. It contains the brand mark. c. It protects children. d. It determines product quality.

Description : Like New Pr Like New Products w oducts wants to i ants to improve its pack ve its packaging aft aging after readi er reading custo ng customer res mer responses t ponses to its customer its ... brand mark. It contains the brand mark. c. It protects children. d. It determines product quality.

Last Answer : d. It determines product quality.