Four boys and three girls stand in a queue for an interview. What is the probability that they will be in alternate positions ? -Maths 9th

Four boys and three girls stand in a queue for an interview. What is the probability that they will be in alternate positions ? -Maths 9th
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1 Answer

Answer :

Total number of ways of arranging 4 boys and 3 girls, i.e., 7 people in a queue (row) = n(S) = 7! Let A : Event in which the 4 boys and 3 girls occupy alternate position. This is possible when the 4 boys occupy the 4 odd places standing form position 1 and the girls fill in the three even places between then as :∴ 4 boys can be arranged in a queue in 4! ways 3 girls can fill the gaps in the queue in 3! ways. ∴ n(A) = 4! × 3!∴ P(A) = \(rac{n(A)}{N(S)}\) = \(rac{4! imes3!}{7!}\) = \(rac{4 imes3 imes2 imes1 imes3 imes2 imes1}{7 imes6 imes5 imes4 imes3 imes2 imes1}\) = \(rac{1}{35}.\)
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Related questions

In a class, there are x girls and y boys, a student is selected at random, then find the probability of selecting a boy. -Maths 9th

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Last Answer : Number of boys = y Total students = (x + y) Thus,P(a boy)= y/(x+y)

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Last Answer : Number of boys = y Total students = (x + y) Thus,P(a boy)= y/(x+y)

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Description : 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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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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Description : A class consists of 80 students, 25 of them are girls and 55 boys. 10 of them are rich and 20 are fair complexioned. -Maths 9th

Last Answer : Let P (A) = Probability of selecting a fair complexioned person. ThenP(A) = \(rac{20}{80}\) = \(rac{1}{4}\)Let P(B) = Probability of selecting a rich person. Then P(B) = \(rac{10}{80}\) = \(rac{1}{8}\)Let P (C) = ... ) = \(rac{1}{4}\)x \(rac{1}{8}\)x \(rac{5}{16}\) = \(rac{5}{512}\) = 0.009.

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Description : A group of 2n boys and 2n girls is divided at random into two equal batches. -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 : 2 boys, 1 gi(Prl = (15c2×10c1) / 25c3 = 1050/2300 1 boy, 2 girls = (15c1×10c2) / 25c3 = 675/2300 Difference = (1050 - 675)/2300 = 375/2300 = 15/92 Answer: c)

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In the given figure, bisectors AP and BQ of the alternate interior angles are parallel, then show that l || m. -Maths 9th

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In the given figure, bisectors AP and BQ of the alternate interior angles are parallel, then show that l || m. -Maths 9th

Description : In the given figure, bisectors AP and BQ of the alternate interior angles are parallel, then show that l || m. -Maths 9th

Last Answer : Given, In the figure AP|| BQ, AP and BQ are the bisectors of alternate interior angles ∠CAB and ∠ABF. To show l || m Proof Since, AP|| BQ and t is transversal, therefore ∠PAB = ∠ABQ [alternate interior angles] ⇒ 2 ∠PAB = 2 ∠ABQ [multiplying both sides by 2]

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Description : In how many ways can the letters of the word “AFLATOON” be arranged if the consonants and vowels must occupy alternate places? -Maths 9th

Last Answer : 24 ways is the answer

Find the probability that the three cards drawn from a pack of 52 cards are all black ? -Maths 9th

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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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Three fair coins are tossed simultaneously. Find the probability of getting more heads than the number of tails. -Maths 9th

Description : Three fair coins are tossed simultaneously. Find the probability of getting more heads than the number of tails. -Maths 9th

Last Answer : (d) \(rac{1}{2}\)Let S be the sample space. Then, S = {HHH, HHT, HTH, HTT, THH, THT,TTH, TTT} ⇒ n(S) = 8 Let A : Event of getting more heads than number of tails. Then, A = {HHH, HHT, HTH, THH} ⇒ n(A) = 4∴ P(A) = \(rac{n(A)}{n(S)}\) = \(rac{4}{8}\) = \(rac{1}{2}.\)

Three identical dice are rolled. The probability that same number will appear on each of them is -Maths 9th

Description : Three identical dice are rolled. The probability that same number will appear on each of them is -Maths 9th

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Description : If three natural numbersfrom 1 to 100 are selected randomly, then the probability that all are divisible by both 2 and 3 is -Maths 9th

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}\) = \(rac{16 imes15 imes14}{100 imes99 imes98}\) = \(rac{4}{1155}\).

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

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Description : Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle with these vertices is equilateral equals : -Maths 9th

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A die is rolled three times. The probability of getting a larger number than the previous number each time is: -Maths 9th

Description : A die is rolled three times. The probability of getting a larger number than the previous number each time is: -Maths 9th

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If the three independent events E1, E2 and E3, the probability that only E1 occurs is α, E2 occurs -Maths 9th

Description : If the three independent events E1, E2 and E3, the probability that only E1 occurs is α, E2 occurs -Maths 9th

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Three girls Reshma, Salma and Mandeep are playing a game by standing on a circle of radius 5 m drawn in a park. -Maths 9th

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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

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A letter is taken out at random from ‘ASSISTANT’ and another is taken out from ‘STATISTICS’. The probability that they are same letters is -Maths 9th

Description : A letter is taken out at random from ‘ASSISTANT’ and another is taken out from ‘STATISTICS’. The probability that they are same letters is -Maths 9th

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Three students were made to stand on the points P, Q and S -Maths 9th

Description : Three students were made to stand on the points P, Q and S -Maths 9th

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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

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

Description : 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

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

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

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

Last Answer : Let probability of winning the race be p Probability of losing the race = 1 - p According to the statement of question, we have p = 2 (1 - p) - 1 / 6 ⇒ 6p = 12-12p -1 ⇒ 18p = 11 ⇒ p = 11 / 18 Hence, probability of winning the race is 11 / 18

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

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

Description : 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

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

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

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

Last Answer : Let probability of winning the race be p Probability of losing the race = 1 - p According to the statement of question, we have p = 2 (1 - p) - 1 / 6 ⇒ 6p = 12-12p -1 ⇒ 18p = 11 ⇒ p = 11 / 18 Hence, probability of winning the race is 11 / 18

Can the experimental probability of an event be a negative number? -Maths 9th

Description : Can the experimental probability of an event be a negative number? -Maths 9th

Last Answer : No, because the number of trials in which the event can happen cannot be negative and the total number of trials is always positive.

Can the experimental probability of an event be greater than 1? -Maths 9th

Description : Can the experimental probability of an event be greater than 1? -Maths 9th

Last Answer : No, as the number of trials can't be greater than the total number of trials.

In a throw of a die, find the probability of getting an even number. -Maths 9th

Description : In a throw of a die, find the probability of getting an even number. -Maths 9th

Last Answer : Total even number on a die = 3 P (getting an even numbers) = 3/6 = 1/2