6 boys and 6 girls are sitting in a row randomly. The probability that boys and girls sit alternately is -Maths 9th

6 boys and 6 girls are sitting in a row randomly. The probability that boys and girls sit alternately is -Maths 9th
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(b) \(rac{1}{462}\)Let S be the sample space. Then, n(S) = Number of ways in which 6 boys and 6 girls can sit in a row = 12! Let E : Event of 6 girls and 6 boys sitting alternately. Then, the 6 girls and 6 boys can be arranged in alternate position in two ways. Ist way : We start with a boy. Then the arrangement is : B G    B G    B G    B G    B G    B G ∴ Number of ways of arranging 6 boys in 6 places = 6! Number of ways of arranging 6 girls in 6 places = 6! ∴ Number of ways of arranging 6 boys and 6 girls in alternate places = 6! × 6! Similarly, IInd way : Here we start with a girl. Then the arrangement is G B  G B    G B     G B     G B    G B ∴ Number of ways of arranging 6 boys and 6 girls alternately this way = 6! × 6!∴ n(E) = 6! × 6! + 6! × 6! = 2 × 6! × 6!∴ n(E) = \(rac{n(E)}{n(S)}\) = \(rac{2 imes6! imes6!}{12!}\)= \(rac{2 imes6 imes5 imes4 imes3 imes2 imes1}{12 imes11 imes10 imes9 imes8 imes7}\) = \(rac{1}{462}\).
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6 boys and 6 girls are seated in a row. Probability that all the boys sit together is -Maths 9th

Description : 6 boys and 6 girls are seated in a row. Probability that all the boys sit together is -Maths 9th

Last Answer : (d) \(rac{1}{132}\)Let S be the sample space for arranging 6 boys and 6 girls in a row. Then, n(S) = 12! If all 6 boys are to sit together, then consider the 6 boys as one entity. Now the ... ) = \(rac{7 imes6 imes5 imes4 imes3 imes2}{12 imes11 imes10 imes9 imes8 imes7}\) = \(rac{1}{132}\).

There are 6 numbered chairs placed around a circular table. 3 boys and 3 girls want to sit on them such that neither of two boys nor two girls -Maths 9th

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Last Answer : Since the chairs are numbered, so they are distinguishable. Therefore 3 boys can be arranged on 3 alternate chairs in 3! ways. 3 girls can be arrenged in 3! ways Also, the girls can be seated before the boys. Total number of required ways = 3! × 3! + 3! × 3! = 2 × (3!)2

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Last Answer : Have the 55 boys stand in a line. This can be done in 5!5! ways. For the moment, add a boy at each end, who will be removed when we're done. Now send the 33 girls, one at a time, to ... that each girl can either wedge herself between two boys or else stand next to a boy at one end or the other.

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750 families with 3 children were selected randomly and the following data recorded If a family member is chosen at random, compute the probability that it has : -Maths 9th

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Last Answer : (i) P(no boy child) =100 / 750 = 2 / 15 (ii) P (no girl child) = 120 /750 =4 / 25

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Last Answer : (c) \(rac{4}{1155}\)Let n(S) = Number of ways of selecting 3 numbers from 100 numbers = 100C3 Let E : Event of selecting three numbers divisible by both 2 and 3 from numbers 1 to 100 = Event of selecting three ... C_3}{^{100}C_3}\) = \(rac{16 imes15 imes14}{100 imes99 imes98}\) = \(rac{4}{1155}\).

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Last Answer : Total number of ways in which 10 person can sit around a circular table = 9! (∵ We shall keep one place fixed and the rest of the 9 places will be filled in (9 8 7 6 5 4 3 2 1) ways asthere is ... probability = \(rac{2 imes8!}{9!}\) = \(rac{2 imes8!}{9 imes8!}\) = \(rac{2}{9}.\)

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Description : If the probability of winning a race of an athlete is 1 / 6 less than the twice the probability of losing the race. -Maths 9th

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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

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Last Answer : As there are 200 integers, total number of exhaustive, mutually exclusive and equally likely cases, i.e, n(S) = 200 Let A : Event of integer chosen from 1 to 200 being divisible by 6⇒ n(A) = 33 \(\bigg(rac{200}{6}=33rac{1}{3}\ ... (rac{25}{200}\) - \(rac{8}{200}\) = \(rac{50}{200}\) = \(rac{1}{4}\).

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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

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Description : Three equal cubes are placed adjacently in a row. Find the ratio of total surface area of the new cuboid to that of the sum of the surface areas of the three cubes. -Maths 9th

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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

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In a recent research study initiated by Tel-E-Fone Telecommunications, survey calls were made to a randomly selected group in which every member has an equal chance of selection. This type of sample selection is called A. Probability sample B. Convenience sample C. Judgment sample D. Quota sample E. Area sample

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Description : What is the probability that a randomly selected bit string of length 10 is a palindrome? (A) 1/64 (B) 1/32 (C) 1/8 (D) ¼

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A box contains 5 cone and 4 chocobar ice-creams. Preethi eats 3 of them, by randomly choosing. What is the probability of choosing 1 chocobar and 2 cone ice-creams? a) 63/10 b) 20/63 c) 10/63 d) 63/20

Description :  A box contains 5 cone and 4 chocobar ice-creams. Preethi eats 3 of them, by randomly choosing. What is the probability of choosing 1 chocobar and 2 cone ice-creams? a) 63/10 b) 20/63 c) 10/63 d) 63/20

Last Answer :  Answer: C) Probability of choosing 1 cone= 5/9 After taking out 1 cone, the total number is 8 . Probability of choosing 2nd cone = 4/8 Probability of choosing 1chocobaricecream out of a total of 7 = 4/7 So the final probability of choosing 2 cone and 1chocobar ice cream = 5/9*1/2*4/7 =10/63

A coin is tossed 500 times and we get Heads : 285 and tails : 215 times. When a coin is tossed at random, what is the probability of getting a. head? b. tail? -Maths 9th

Description : A coin is tossed 500 times and we get Heads : 285 and tails : 215 times. When a coin is tossed at random, what is the probability of getting a. head? b. tail? -Maths 9th

Last Answer : Given, Total number of events = 500 No. of times heads occur = 285 Probability of getting head when coin is tossed at random = 285/500 = 57/100 No. of times tails occur = 215 Probability of getting tails when coin is tossed at random = 215/500 = 43/100

Two coin are tossed 400 times and we get a. Two Heads : 112 times b. One Head : 160 times c. No Head : 128 times. When two coins are tossed at random, what is the probability of getting a. Two Heads b. One Head c. No Head -Maths 9th

Description : Two coin are tossed 400 times and we get a. Two Heads : 112 times b. One Head : 160 times c. No Head : 128 times. When two coins are tossed at random, what is the probability of getting a. Two Heads b. One Head c. No Head -Maths 9th

Last Answer : Given, Total number of events = 400 (a) No. of times two heads occur = 112 Probability of getting two heads = 112/400 = 7/25 (b) No. of times one heads occur = 160 Probability of getting one heads = 160/400 = 2/5 (c) No. of times no heads occur = 128 Probability of getting no heads = 128/400 = 8/25

The probability of guessing the correct answer to a certain question is x/ 2. -Maths 9th

Description : The probability of guessing the correct answer to a certain question is x/ 2. -Maths 9th

Last Answer : Here, probability of guessing the correct answer = x / 2 And probability of not guessing the correct answer = 2 / 3 Now, x / 2 + 2 / 3 = 1 ⇒ 3x + 4 = 6 ⇒ 3x = 2 ⇒ x = 2 / 3

In a throw of a die, find the probability of not getting 4 or 5. -Maths 9th

Description : In a throw of a die, find the probability of not getting 4 or 5. -Maths 9th

Last Answer : Required probability = 1 – P(4) – P(5) =1- 1 / 6 - 1 / 6 = 4 / 6 = 2 / 3

A coin is tossed 500 times and we get Heads : 285 and tails : 215 times. When a coin is tossed at random, what is the probability of getting a. head? b. tail? -Maths 9th

Description : A coin is tossed 500 times and we get Heads : 285 and tails : 215 times. When a coin is tossed at random, what is the probability of getting a. head? b. tail? -Maths 9th

Last Answer : Given, Total number of events = 500 No. of times heads occur = 285 Probability of getting head when coin is tossed at random = 285/500 = 57/100 No. of times tails occur = 215 Probability of getting tails when coin is tossed at random = 215/500 = 43/100

Two coin are tossed 400 times and we get a. Two Heads : 112 times b. One Head : 160 times c. No Head : 128 times. When two coins are tossed at random, what is the probability of getting a. Two Heads b. One Head c. No Head -Maths 9th

Description : Two coin are tossed 400 times and we get a. Two Heads : 112 times b. One Head : 160 times c. No Head : 128 times. When two coins are tossed at random, what is the probability of getting a. Two Heads b. One Head c. No Head -Maths 9th

Last Answer : Given, Total number of events = 400 (a) No. of times two heads occur = 112 Probability of getting two heads = 112/400 = 7/25 (b) No. of times one heads occur = 160 Probability of getting one heads = 160/400 = 2/5 (c) No. of times no heads occur = 128 Probability of getting no heads = 128/400 = 8/25