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

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
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1 Answer

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 drawing a queen) = 1 - P(E) = 1 - \(rac{1}{13}\) = \(rac{12}{13}\)∴ P(A wins) = P(A draws a queen in the 1st or 3rd or 5th or ...draws with B not drawing the queen in the 2nd or 4th or 6th or ...draws)= P(E or \(\bar{E}\) \(\bar{E}\) E or \(\bar{E}\) \(\bar{E}\) \(\bar{E}\) \(\bar{E}\) E or ........) = P(E) + P(\(\bar{E}\)) . P(\(\bar{E}\)) . P(E) + P(\(\bar{E}\)) P(\(\bar{E}\)) P(\(\bar{E}\)) . P(\(\bar{E}\)) . P(E) + ... ∞= \(rac{1}{13}\) + \(\bigg(\)\(rac{12}{13}\) x \(rac{12}{13}\) x \(rac{1}{13}\)\(\bigg)\)+ \(\bigg(\)\(rac{12}{13}\) x \(rac{12}{13}\) x \(rac{12}{13}\) x \(rac{12}{13}\) x \(rac{1}{13}\)\(\bigg)\)+ ... ∞= \(rac{1}{13}\) \(\bigg[\)1 + \(\bigg(rac{12}{13}\bigg)^2\)+\(\bigg(rac{12}{13}\bigg)^4\)+ ... ∞\(\bigg]\)= \(rac{1}{13}\) \(\bigg[rac{1}{1-\big(rac{12}{13}\big)^2}\bigg]\) = \(rac{1}{13}\) \(\bigg[rac{1}{rac{(169-144)}{169}}\bigg]\)= \(rac{1}{13}\) x \(rac{169}{25}\) = \(rac{13}{25}\). [Sum of a G.P with infinite terms = \(rac{a}{1-r}\)  where a = 1st term, r = common ratio.]
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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}\)= \(rac{44}{52}\) + \(rac{16}{52}\) = \(rac{28}{52}\) = \(rac{7}{13}\) .

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

Description : A card is drawn at random from a well shuffled pack of 52 cards -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 : (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.

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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!

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

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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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Eight persons P, Q, R, S, W, X, Y and Z in which 4 females are facing towards and 4 males facing away from the center. Males and females are seated alternately. I. X is male, W position is 3rd to the right of X and exactly opposite to Q. II. P is 2nd to the left of W. III. S and R are facing each other as well R & Y are neighbors. Who are all females in the given arrangement? a) S, Q, R, W b) P, Q, Y, W c) P, Q, R, W d) P, Q, R, S

Description : Eight persons P, Q, R, S, W, X, Y and Z in which 4 females are facing towards and 4 males facing away from the center. Males and females are seated alternately. I. X is male, W position is 3rd to the right of X and exactly opposite to Q. ... a) S, Q, R, W b) P, Q, Y, W c) P, Q, R, W d) P, Q, R, S

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Description : For drawing a frequency polygon of a continuous frequency distribution, we plot the points whose ordinates are the frequencies of the respective classes and abscissae are, respectively -Maths 9th

Last Answer : NEED ANSWER

For drawing a frequency polygon of a continuous frequency distribution, we plot the points whose ordinates are the frequencies of the respective classes and abscissae are, respectively -Maths 9th

Description : For drawing a frequency polygon of a continuous frequency distribution, we plot the points whose ordinates are the frequencies of the respective classes and abscissae are, respectively -Maths 9th

Last Answer : (c) Class marks i.e., the mid-point of the classes are abscissa of the points, which we plot for frequency polygon.

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

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

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

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1. Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points? -Maths 9th

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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

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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

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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

Cards with numbers 1, 2, 3, ........... 100 are -Maths 9th

Description : Cards with numbers 1, 2, 3, ........... 100 are -Maths 9th

Last Answer : (i) Favourable cards are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, i.e., 10. P (prime number less than 30) = 10/100 or 1/10 (ii) Favourable cards are 35, 70. P (card is a multiple of 5 and 7) = 2/100 or 1/50 ( ... 42, 49, 56, 63, 77, 84, 91, 98 i.e., 32 P(card is a multiple of 5 or 7) = 32/100 or 8/25.

Operating method in which the transmission is made alternately in each direction of a telecommunication channel A. Semi-duplex operation B. Duplex operation C. Half-duplex operation D. Simplex operation

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Description : How many players can play any given game on the same Game Boy device?

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A common assumption about the players in a game is that a. Neither player knows the payoff matrix b. The players have different information about the payoff matrix c. Only one of the players pursues a rational strategy d. The specific identify of the players is irrelevant to the play of the game

Description : A common assumption about the players in a game is that a. Neither player knows the payoff matrix b. The players have different information about the payoff matrix c. Only one of the players pursues a rational strategy d. The specific identify of the players is irrelevant to the play of the game

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

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

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The work done by a body on application of a constant force is the product of the constant force and distance travelled by the body in the direction of force. Express this in the form of a linear equation in two variables and draw its graph by taking the constant force as 3 units. -Maths 9th

Description : The work done by a body on application of a constant force is the product of the constant force and distance travelled by the body in the direction of force. Express this in the form of a linear equation in two variables and draw its graph by taking the constant force as 3 units. -Maths 9th

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Draw a graph of the equation x + Y = 5 & 3x - 2y =0 on the same graph paper. Find the coordinates of the point whose two lines intersect. -Maths 9th

Description : Draw a graph of the equation x + Y = 5 & 3x - 2y =0 on the same graph paper. Find the coordinates of the point whose two lines intersect. -Maths 9th

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Draw a graph of the equation x - Y = 4 & 2x+ 2y =4 on the same graph paper find the coordinates of the point whose two lines intersect. -Maths 9th

Description : Draw a graph of the equation x - Y = 4 & 2x+ 2y =4 on the same graph paper find the coordinates of the point whose two lines intersect. -Maths 9th

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