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

A card is drawn at random from a well shuffled pack of 52 cards -Maths 9th
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(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) = \(rac{13}{52}\) + \(rac{13}{52}\) + \(rac{4}{52}\) - \(rac{1}{52}\) - \(rac{1}{52}\) = \(rac{28}{52}\)(∵ There are 13 clubs, 13 hearts and 4 jacks, 1 jack of clubs and 1 jack of hearts)P(Z) = \(rac{4}{52}\) + \(rac{13}{52}\) + \(rac{4}{52}\) - \(rac{1}{52}\) - \(rac{1}{52}\) = \(rac{19}{52}\)(∵ 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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Related questions

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

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

What is the probability a well shuffled pack of 52 cards a card is drawn at random find the probability that it is either a heart or a queen?

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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]

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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

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Description : Two players A and B play a game by alternately drawing a card from a well-shuffled pack of playing cards, replacing the card each time after draw. -Maths 9th

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.]

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Description : One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn. -Maths 9th

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

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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 pack of 52 cards. What is the probability that both of them are either black or queen cards? a) 55/442 b) 54/221 c) 55/221 d) 51/221

Description : Two cards are drawn at random from a pack of 52 cards. What is the probability that both of them are either black or queen cards? a) 55/442 b) 54/221 c) 55/221 d) 51/221

Last Answer : c) 55/221

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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Last Answer : Answer: C) We have n(s) = 52C2 = 1326. Let A = event of getting both red cards B = event of getting both king A∩B = event of getting king of red cards n(A) = 26C2 = 325, n(B)= 4C2= 6 and n(A∩B) = 2C2 = 1 P(A ... S) = 1/1326 P(A∪B) = P(A) + P(B) - P(A∩B) = (325+6-1) / 1326 = 330/1326 = 55/221

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

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

Last Answer : (b) \(rac{23}{26}\)Total number of ways in which 3 letters can be selected from 26 letters = 26C3. If A is not to be included in the choice, there are 25 letters left, so number of ways in which 3 letters can be ... 25}C_3}{^{26}C_3}\) = \(rac{25 imes24 imes23}{26 imes25 imes24}\) = \(rac{23}{26}\).

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Description : In how many ways can a pack of 52 cards be divided equally among four players in order? -Maths 9th

Last Answer : Distribution of 52 cards can be equally divided among four players. Hence, number of ways is (13!)4! 52! ​ 4!= (13!) 52! ​

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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!

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)

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Description : Three cards are drawn at random from an ordinary pack of cards. Find out the probability that they will consist of a king, aqueen and an ace?

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

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

Last 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 ... \(rac{11}{20}\)∴ P(drawing at least one ace) = 1 - \(rac{11}{20}\) = \(rac{9}{20}\) .

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 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

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

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

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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

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

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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

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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

Do playing card manufacturers print all 52 (or 54) cards at once?

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Last Answer : I do not know, but I would think that they print all cards onto a single sheet of cardboard, then cut them out. Otherwise, you would need a separate printer for each card, which seems ludicrous to me.

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There are 100 cards numbered from 1 to 100 in a box. If a card is drawn from the box and the probability of an event is 1/2, then the number of favour

Description : There are 100 cards numbered from 1 to 100 in a box. If a card is drawn from the box and the probability of an event is 1/2, then the number of favour

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There are 10 cards numbered from 1 to 10 in a box. If a card is drawn randomly, then find the probability of getting an even numbered card.

Description : There are 10 cards numbered from 1 to 10 in a box. If a card is drawn randomly, then find the probability of getting an even numbered card.

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If a red suit is drawn from an ordinary deck of cards what is the probability that the card is a diamond?

Description : If a red suit is drawn from an ordinary deck of cards what is the probability that the card is a diamond?

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Description : 4. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.30). Show that F is the mid-point of BC. -Maths 9th

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Description : A student wrote the equations of the lines a and b drawn in the following graph as y =1 and 2x + 3y =6. Is he right? -Maths 9th

Last Answer : Clearly, line a is parallel to X-axis at a distance of 1 unit in positive direction of Y-axis, therefore its equation is y = 1. Also, if we draw the graph of line 2x + 3y = 6, then its graph should intersect X - axis at (3,0 ... Base Height = 1/2 BC AC = 1/2 1 3 / 2 = 3 / 4 sq unit.

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Description : X and y are points on the side LN of the triangle LMN , such that LX = XY = YN . Through X, a line is drawn parallel to LM to meet MN at Z. -Maths 9th

Last Answer : Here, △XZM and △XZL are on the same base (XZ) and lie between the same parallels (XZ || LM). ∴ ar(△XZL) = ar( △XZM) Adding ar(△XZY) on both sides , we have ar(△XZL) + ar(△XZY) = ar(△XZM) + ar(△XZY) ⇒ ar(△LZY) = ar(quad.MZYX)

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Description : In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. -Maths 9th

Last Answer : According to question prove that ar (ABCD) = ar (APQD).

In figure, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. -Maths 9th

Description : In figure, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. -Maths 9th

Last Answer : Given ABCDE is a pentagon. BP || AC and EQ|| AD. To prove ar (ABCDE) = ar (APQ) Proof We know that, triangles on the same base and between the same parallels are equal in area. Here, ΔADQ and ΔADE lie on the ... ar (ΔACD) = ar (ΔADE) + ar (ΔACB) + ar (ΔACD) ⇒ ar (ΔAPQ) = ar (ABCDE) Hence proved.

If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral, so formed is cyclic. -Maths 9th

Description : If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral, so formed is cyclic. -Maths 9th

Last Answer : Given ΔABC is an isosceles triangle such that AB = AC and also DE || SC. To prove Quadrilateral BCDE is a cyclic quadrilateral. Construction Draw a circle passes through the points B, C, D and E.

ABCD is a parallelogram. A circle through A, B is so drawn that it intersects AD at P and BC at Q. -Maths 9th

Description : ABCD is a parallelogram. A circle through A, B is so drawn that it intersects AD at P and BC at Q. -Maths 9th

Last Answer : Given ABCD is a parallelogram. A circle whose centre O passes through A, B is so drawn that it intersect AD at P and BC at Q To prove Points P, Q, C and D are con-cyclic. Construction Join PQ ... Thus, the quadrilateral QCDP is cyclic. So, the points P, Q, C and D are con-cyclic. Hence proved.

A student wrote the equations of the lines a and b drawn in the following graph as y =1 and 2x + 3y =6. Is he right? -Maths 9th

Description : A student wrote the equations of the lines a and b drawn in the following graph as y =1 and 2x + 3y =6. Is he right? -Maths 9th

Last Answer : Clearly, line a is parallel to X-axis at a distance of 1 unit in positive direction of Y-axis, therefore its equation is y = 1. Also, if we draw the graph of line 2x + 3y = 6, then its graph should intersect X - axis at (3,0 ... Base Height = 1/2 BC AC = 1/2 1 3 / 2 = 3 / 4 sq unit.

X and y are points on the side LN of the triangle LMN , such that LX = XY = YN . Through X, a line is drawn parallel to LM to meet MN at Z. -Maths 9th

Description : X and y are points on the side LN of the triangle LMN , such that LX = XY = YN . Through X, a line is drawn parallel to LM to meet MN at Z. -Maths 9th

Last Answer : Here, △XZM and △XZL are on the same base (XZ) and lie between the same parallels (XZ || LM). ∴ ar(△XZL) = ar( △XZM) Adding ar(△XZY) on both sides , we have ar(△XZL) + ar(△XZY) = ar(△XZM) + ar(△XZY) ⇒ ar(△LZY) = ar(quad.MZYX)

In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. -Maths 9th

Description : In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. -Maths 9th

Last Answer : According to question prove that ar (ABCD) = ar (APQD).

In figure, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. -Maths 9th

Description : In figure, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. -Maths 9th

Last Answer : Given ABCDE is a pentagon. BP || AC and EQ|| AD. To prove ar (ABCDE) = ar (APQ) Proof We know that, triangles on the same base and between the same parallels are equal in area. Here, ΔADQ and ΔADE lie on the ... ar (ΔACD) = ar (ΔADE) + ar (ΔACB) + ar (ΔACD) ⇒ ar (ΔAPQ) = ar (ABCDE) Hence proved.

If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral, so formed is cyclic. -Maths 9th

Description : If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral, so formed is cyclic. -Maths 9th

Last Answer : Given ΔABC is an isosceles triangle such that AB = AC and also DE || SC. To prove Quadrilateral BCDE is a cyclic quadrilateral. Construction Draw a circle passes through the points B, C, D and E.

ABCD is a parallelogram. A circle through A, B is so drawn that it intersects AD at P and BC at Q. -Maths 9th

Description : ABCD is a parallelogram. A circle through A, B is so drawn that it intersects AD at P and BC at Q. -Maths 9th

Last Answer : Given ABCD is a parallelogram. A circle whose centre O passes through A, B is so drawn that it intersect AD at P and BC at Q To prove Points P, Q, C and D are con-cyclic. Construction Join PQ ... Thus, the quadrilateral QCDP is cyclic. So, the points P, Q, C and D are con-cyclic. Hence proved.