The probability that in the random arrangement of the letters of the word ‘UNIVERSITY’the two I‘s do not come together is -Maths 9th

The probability that in the random arrangement of the letters of the word ‘UNIVERSITY’the two I‘s do not come together is -Maths 9th
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1 Answer

Answer :

(b) \(rac{4}{5}\)Let S be the sample space. Then, n(S) = Total number of waysin which the letters of the word ‘UNIVERSITY’ can be arranged = \(rac{10!}{2!}\)                        (∵ There are 2I‘s) Let E : Event of arranging the letters of the word ‘UNIVERSITY’such that 2 I’s remain together. Not considering the two I’s, the remaining 8 letters of ‘UNIVERSITY’ can be arranged in 8! ways. There seven places between these 8 letters and two on extreme ends which can be occupied by the two I’s, so that they are not together.—X—X—X—X—X—X—X—X—The places marked ‘–’ can be filled with I’s. ∴ Now 2I’s can be arranged in any of the 2 out of 9 places in 9C2 ways = \(rac{|\underline9}{|\underline7 imes|\underline2}\) = \(rac{9 imes8}{2}\) = 36 ways.∴ n(E) = 8! × 36∴ P(E) = \(rac{8! imes36}{rac{10!}{2!}}\) = \(rac{ ot8! imes36 imes2!}{10 imes9 imes ot8!}\) = \(rac{4}{5}\).
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Related questions

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Description : The letters of the word ‘SOCIETY’ are placed at random in a row. What is the probability that three vowels come together ? -Maths 9th

Last Answer : There are 7 letters in the word SOCIETY. ∴ Total number of ways of arranging all the 7 letters = n(S) = 7!. When the case of three vowels being together is taken, then the three vowels are considered as one unit, so the ... = 5! 3! ∴ Required probability = \(rac{5! imes3!}{7!}\) = \(rac{1}{7}\)

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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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Last Answer : The number of words formed from 'DAUGHTER' such that all vowels are together is 4320.

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Last Answer : Given: The word ‘SUNDAY’ Total number of letters in the word ‘SUNDAY’ is 6. So, number of arrangements of 6 things, taken all at a time is 6P6 = 6! = 6 ... of words using letters of ‘SUNDAY’ starting with ‘N’ and ending with ‘Y’ is 24

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Description : If all the L‘s occur together and also all I‘s occur together, when the letters of the word ‘HALLUCINATION’ are permuted, -Maths 9th

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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

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

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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

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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 : From a group of 3 man and 2 women, two person are selected at random. Find the probability that at least one women is selected. -Maths 9th

Last Answer : (b) \(rac{7}{10}\)Total number of ways of selecting 2 persons at random out of 5 persons = 5C2 ∴ n(S) = 5C2 = \(rac{|\underline5}{|\underline3|\underline2}\) = \(rac{5 imes4}{2 imes1}\) = 10Let A : Event of selecting ... = 2 3 + 1 = 7 ∴ Required probability = \(rac{n(A)}{n(S)}\) = \(rac{7}{10}\).

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Description : Five horses are in a race. Mr A. Selects two of the horses at random and bets on them. The probability that Mr A selected the winning horse is -Maths 9th

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

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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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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

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Description : 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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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

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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

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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

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A natural number is chosen at random from amongst the first 300. What is the probability that the number chosen is a multiple of 2 or 3 or 5 ? -Maths 9th

Description : A natural number is chosen at random from amongst the first 300. What is the probability that the number chosen is a multiple of 2 or 3 or 5 ? -Maths 9th

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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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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

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

Last Answer : (c) \(rac{1}{10}\)Let S be the sample space.Then n(S) = Number of triangles formed by selecting any three vertices of 6 vertices of a regular hexagon= 6C3 = \(rac{6 imes5 imes4}{3 imes2}\) = 20.Let A : Event that the ... Required probability = \(rac{n(A)}{n(S)}\) = \(rac{2}{20}\) = \(rac{1}{10}\).

If an integer P is chosen at random in the interval 0 ≤ p ≤ 5, the probability that the roots of the equation x^2 + px -Maths 9th

Description : If an integer P is chosen at random in the interval 0 ≤ p ≤ 5, the probability that the roots of the equation x^2 + px -Maths 9th

Last Answer : answer:

A basket contains 2 blue, 4 red, 3 green and 5 black balls. If 4 balls are picked at random, what is the probability that -Maths 9th

Description : A basket contains 2 blue, 4 red, 3 green and 5 black balls. If 4 balls are picked at random, what is the probability that -Maths 9th

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There are n letters and n addressed envelopes. If the letters are placed in the envelopes at random, -Maths 9th

Description : There are n letters and n addressed envelopes. If the letters are placed in the envelopes at random, -Maths 9th

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Find how many arrangements can be made with the letters of the word “MATHEMATICS” in which the vowels occur together? -Maths 9th

Description : Find how many arrangements can be made with the letters of the word “MATHEMATICS” in which the vowels occur together? -Maths 9th

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There are 10 persons who are to be seated around a circular table. Find the probability that two particular persons will always sit together. -Maths 9th

Description : There are 10 persons who are to be seated around a circular table. Find the probability that two particular persons will always sit together. -Maths 9th

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 = \(rac{2 imes8!}{9!}\) = \(rac{2 imes8!}{9 imes8!}\) = \(rac{2}{9}.\)

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

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A husband and a wife appear in an interview for two vacancies in the same post. The probability of husband‘s -Maths 9th

Description : A husband and a wife appear in an interview for two vacancies in the same post. The probability of husband‘s -Maths 9th

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The letters of the word ‘COCHIN’ are permuted and all the permutations are arranged in alphabetical order as in English dictionary. -Maths 9th

Description : The letters of the word ‘COCHIN’ are permuted and all the permutations are arranged in alphabetical order as in English dictionary. -Maths 9th

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

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

In a single throw of two dice, what is the probability of getting a sum of 9? -Maths 9th

Description : In a single throw of two dice, what is the probability of getting a sum of 9? -Maths 9th

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Two dice are thrown. Find the probability of getting an odd number on the first die and a multiple of 3 on the other. -Maths 9th

Description : Two dice are thrown. Find the probability of getting an odd number on the first die and a multiple of 3 on the other. -Maths 9th

Last Answer : Let A : Getting an odd number on first die; B : Getting a multiple of 3 on second die, thenA = {1, 3, 5}, B = {3, 6} ∴ P(A) = \(rac{3}{6}=rac{1}{2}\), P(B) = \(rac{2}{6}=rac{1}{3}\) ... B are independent∴ Required probability = P (A) . P (B) = \(rac{1}{2}\) x \(rac{1}{3}\) = \(rac{1}{6}\)

The probability of student A passing examination is 3/7 and of student B passing is 5/7 Assuming the two events “A passes”, -Maths 9th

Description : The probability of student A passing examination is 3/7 and of student B passing is 5/7 Assuming the two events “A passes”, -Maths 9th

Last Answer : p1 = P(A) = \(rac{3}{7}\), p2 = P(B) = \(rac{5}{7}\) ∴ q1 = P(\(\bar{A}\)) = 1 - P(A) = 1 - \(rac{3}{7}\) = \(rac{4}{7}\). q2 = P(\(\bar{B}\)) = 1 - P(B) = 1 - \(rac{5}{7}\) = \(rac{2}{7}\) ... passes) = p1 q2 + q1 p2 = \(rac{3}{7}\) x \(rac{2}{7}\) + \(rac{4}{7}\) x \(rac{5}{7}\) = \(rac{26}{49}.\)

If two coins are tossed once, what is the probability of getting at least one head ? -Maths 9th

Description : If two coins are tossed once, what is the probability of getting at least one head ? -Maths 9th

Last Answer : When two coins are tossed once, there are four possible outcomes, i.e., S = {HH, HT, TH, TT} ∴ Total number of outcomes = n(S) = 4 Let A : Event of getting at least one head ⇒ A = {HH, HT, TH} ⇒ n(A) = 3∴ P(A) = \(rac{n(A)}{n(S)}\) = \(rac{3}{4}.\)

Two unbiased dice are rolled. Find the probability of getting a multiple of 2 on one die and a multiple of 3 on the other die ? -Maths 9th

Description : Two unbiased dice are rolled. Find the probability of getting a multiple of 2 on one die and a multiple of 3 on the other die ? -Maths 9th

Last Answer : When two unbiased dice are rolled, the possible out comes are∴ n(S) = 36 Let A : getting a multiple of 2 on one die and a multiple of 3 on the other die. ⇒ A = {(2, 3), (2, 6), (4, 3), (4, 6), (6, 3), (6, 6), (3, 2), ( ... (3, 6), (6, 2), (6, 4)} ⇒ n(A) = 11∴ P(A) = \(rac{n(A)}{n(S)} =rac{11}{36}.\)

Find the probability that a two digit number formed by the digit 1, 2, 3, 4 and 5 is divisible by 4. -Maths 9th

Description : Find the probability that a two digit number formed by the digit 1, 2, 3, 4 and 5 is divisible by 4. -Maths 9th

Last Answer : The two digit numbers can be formed by putting any of 5 digits at the one 's place and also one of the 5 digits at ten's place. So, Total number of 2-digit numbers that can be formed using these 5-digits = 5 5 = ... 52}, i.e, 5 in number. ∴ Required probability = \(rac{5}{25}\) = \(rac{1}{5}.\)

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

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 dice are rolled simultaneously. The probability of getting a multiple of 2 on one dice and a multiple of 3 on the other is -Maths 9th

Description : Two dice are rolled simultaneously. The probability of getting a multiple of 2 on one dice and a multiple of 3 on the other is -Maths 9th

Last Answer : (c) \(rac{11}{36}\)Total number of outcomes when two identical dice are rolled, n(S) = 6 6 = 36 Let A : Event of rolling a multiple of 2 on one die and a multiple of 3 on the other die ⇒ A = {(2, 3), (2, 6), (4, 3), (4, ... , 4), (3, 6)} ⇒ n(A) = 11 ∴ P(A) = \(rac{n(A)}{n(S)}\) = \(rac{11}{36}\).

Two dice are rolled once. Find the probability of getting an even number on the first die, or a total of 7. -Maths 9th

Description : Two dice are rolled once. Find the probability of getting an even number on the first die, or a total of 7. -Maths 9th

Last Answer : (c) \(rac{7}{12}\)Total number of ways in which 2 dice are rolled = 6 6 = 36 ⇒ n(S) = 36 Let A : Event of rolling an even number of 1st dice B : Event of rolling a total of 7 ⇒ A = {(2, 1), (2, 2) , (2, 6), (4 ... (rac{18}{36}\) + \(rac{6}{36}\) - \(rac{3}{36}\) = \(rac{21}{36}\) = \(rac{7}{12}\).

A die is thrown twice. What is the probability that at least one of the two throws comes up with the number 3 ? -Maths 9th

Description : A die is thrown twice. What is the probability that at least one of the two throws comes up with the number 3 ? -Maths 9th

Last Answer : (b) \(rac{11}{36}\)Let S = total ways in which two dice can be rolled ⇒ n(S) = 6 6 = 36 Let A : Event of throwing 3 with 1st dice, B : Event of throwing 3 with 2nd dice. Then, A = {(3, 1), (3, 2), (3, 3), (3, 4), ... ) - P(A ∩ B)= \(rac{6}{36}\) + \(rac{6}{36}\) - \(rac{1}{36}\) = \(rac{11}{36}\).

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

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

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