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

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

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 drawing 2 socks of same colour ⇒ A = Drawing 2 brown socks out of 5 brown socks or Drawing 2 blue sock out of 4 blue socks⇒ n(A) = 5C2 + 4C2 = \(rac{|\underline5}{|\underline3|\underline2}\) + \(rac{|\underline4}{|\underline2|\underline2}\) = \(rac{5 imes4}{2}\) + \(rac{4 imes3}{2}\) = 10 + 6 = 16∴ Required probability P(A) = \(rac{n(A)}{n(S)}\) = \(rac{16}{36}\) = \(rac{4}{9}\).
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Related questions

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

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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 \(rac{11}{16}\) = \(\big(rac{11}{16}\big)^4.\)

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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

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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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 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

Last Answer : Total number of nuts and bolts in the box = 150 + 50 = 200 Number of nuts and bolts rusted = 1 / 2 × 200 = 100 P(a rusted nut or bolt) = 100 / 200 = 1 / 2

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

Last Answer : Total number of nuts and bolts in the box = 150 + 50 = 200 Number of nuts and bolts rusted = 1 / 2 × 200 = 100 P(a rusted nut or bolt) = 100 / 200 = 1 / 2

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

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

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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 - \(rac{11}{20}\) = \(rac{9}{20}\) .

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 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

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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.

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 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

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Why can't I buy a pack of brown dress socks that are all the same color?

Description : Why can't I buy a pack of brown dress socks that are all the same color?

Last Answer : Because Mr. Monk isn’t enough of a consumer to support this manufacturing decision. :-)

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

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Last Answer : Answer: C) 

The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find -Maths 9th

Description : The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find -Maths 9th

Last Answer : Radius of well = (r) = 3.5/2 m Depth of well = (h) = 10 m (i) Inner curved surface area of well = 2 πrh = 2 x 22/7 x 3.5/2 x 10 = 110 m2 (ii) Cost of plastering 1 m2 = ₹ 40 ∴ Cost of plastering 110 m2 = ₹110 X 40 = ₹4400

The inner diameter of a circular well is 3.5m. It is 10m deep. Find (i) its inner curved surface area, (ii) the cost of plastering this curved surface at the rate of Rs. 40 per m2. -Maths 9th

Description : The inner diameter of a circular well is 3.5m. It is 10m deep. Find (i) its inner curved surface area, (ii) the cost of plastering this curved surface at the rate of Rs. 40 per m2. -Maths 9th

Last Answer : Inner radius of circular well, r = 3.5/2m = 1.75m Depth of circular well, say h = 10m (i) Inner curved surface area = 2πrh = (2 (22/7 ) 1.75 10) = 110 Therefore, the inner curved surface ... area = Rs (110 40) = Rs.4400 Therefore, the cost of plastering the curved surface of the well is Rs. 4400.

ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that: (i) ABCD is a square (ii) Diagonal BD bisects ∠B as well as ∠D. -Maths 9th

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Last Answer : Solution: (i) ∠DAC = ∠DCA (AC bisects ∠A as well as ∠C) ⇒ AD = CD (Sides opposite to equal angles of a triangle are equal) also, CD = AB (Opposite sides of a rectangle) ,AB = BC = CD = AD Thus ... interior angles) ⇒ ∠CBD = ∠ABD Thus, BD bisects ∠B Now, ∠CBD = ∠ADB ⇒ ∠CDB = ∠ADB Thus, BD bisects ∠D

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]

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

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 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

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.]

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

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).

If the sum as well as the product of roots of a quadratic equation is 9, then the equation is: -Maths 9th

Description : If the sum as well as the product of roots of a quadratic equation is 9, then the equation is: -Maths 9th

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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}\) = \(rac{25 imes24 imes23}{26 imes25 imes24}\) = \(rac{23}{26}\).

The weight of a man is four times the weight of a child. Write an equation in two variables for this situation. -Maths 9th

Description : The weight of a man is four times the weight of a child. Write an equation in two variables for this situation. -Maths 9th

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A Police man fires six bullets on a decoit. The probability that the decoit will be killed by one bullet is 0.6. What is the probability that the -Maths 9th

Description : A Police man fires six bullets on a decoit. The probability that the decoit will be killed by one bullet is 0.6. What is the probability that the -Maths 9th

Last Answer : Let Ai be the event that the decoit is killed by the ith bullet (1 ≤ i ≤ 6). Then \(\bar{A}_i\) is the event that the decoit is not killed, ∴ P (Ai) = 0.6 and P(\(\bar{A}_i\)) = 1 − 0.6 = 0.4∴ Probability ... decoit p = 0.6 ⇒ q = 1 − p = 1 − 0.6 = 0.4Required probability = qqqqqq = (q)6 = (0.4)6.