Three letters are randomly selected from the 26 capital letters of the English alphabet. -Maths 9th

Three letters are randomly selected from the 26 capital letters of the English alphabet. -Maths 9th
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(d) \(rac{36}{1001}\)Required probability = \(rac{P( ext{2 blue balls}) imes{P}( ext{2 red balls})}{P( ext{4 balls out of 14 balls})}\) + \(rac{P( ext{2 green balls}) imes{P}( ext{2 black balls})}{P( ext{4 out of 14 balls})}\)
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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

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

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

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

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

Assume postal rates for 'light letters' are: $0.25 up to 10 grams; $0.35 up to 50 grams; $0.45 up to 75 grams; $0.55 up to 100 grams. Which test inputs (in grams) would be selected using boundary value analysis? A. 0, 9,19, 49, 50, 74, 75, 99,100 B. 10, 50, 75,100, 250, 1000 C. 0, 1,10,11, 50, 51, 75, 76,100,101 D. 25, 26, 35, 36, 45, 46, 55, 56

Description : Assume postal rates for 'light letters' are: $0.25 up to 10 grams; $0.35 up to 50 grams; $0.45 up to 75 grams; $0.55 up to 100 grams. Which test inputs (in grams) would be selected using boundary value analysis? A. 0, 9,19, ... C. 0, 1,10,11, 50, 51, 75, 76,100,101 D. 25, 26, 35, 36, 45, 46, 55, 56

Last Answer : C. 0, 1,10,11, 50, 51, 75, 76,100,101

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

Last Answer : answer:

In an examination there are 3 multiple choice questions and each question has 4 choices. If a student randomly selects answer for all -Maths 9th

Description : In an examination there are 3 multiple choice questions and each question has 4 choices. If a student randomly selects answer for all -Maths 9th

Last Answer : Probability of selecting a correct choice for a question = \(rac{1}{4}\)(∵ Out of 4 choices only one is correct)∴ Probability of answering all the three questions correctly = \(rac{1}{4}\)x \(rac{1}{4}\)x\ ... of not answering all the three questions correctly = 1 - \(rac{1}{64}\) = \(rac{63}{64}\).

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

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

Last Answer : (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 ... )= \(rac{2 imes6 imes5 imes4 imes3 imes2 imes1}{12 imes11 imes10 imes9 imes8 imes7}\) = \(rac{1}{462}\).

A management institute has six senior professors and four junior professors. Three professors are selected at random -Maths 9th

Description : A management institute has six senior professors and four junior professors. Three professors are selected at random -Maths 9th

Last Answer : (a) \(rac{5}{6}\)P(At least one junior professor is selected) = P(Selecting 1 Junior) P(Selecting 2 Seniors) + P(Selecting 2 Junior) P(Selecting 1 Senior) + P(Selecting all 3 Juniors)∴ Required probability = \(rac{^4C_1 imes^ ... }{30}\) = \(rac{15+9+1}{30}\) = \(rac{25}{30}\) = \(rac{5}{6}\).

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)

How many letters are there in the English alphabet? -Riddles

Description : How many letters are there in the English alphabet? -Riddles

Last Answer : answer:

Which of the letters of the English alphabet have only one line of symmetry ?

Description : Which of the letters of the English alphabet have only one line of symmetry ?

Last Answer : Which of the letters of the English alphabet have only one line of symmetry ?

To change selected text toall capital letters, click the change case button, then click– (A) UPPERCASE (B) UPPER ALL (C) CAPS LOCK (D) Lock Upper (E) Large Size

Description : To change selected text toall capital letters, click the change case button, then click– (A) UPPERCASE (B) UPPER ALL (C) CAPS LOCK (D) Lock Upper (E) Large Size

Last Answer : UPPERCASE

To change selected text to all capital letters, click the change case button and then click: a) UPPERCASE b) UPPER ALL c) CAPS LOCK d) Lock Upperbe) Large Size

Description : To change selected text to all capital letters, click the change case button and then click: a) UPPERCASE b) UPPER ALL c) CAPS LOCK d) Lock Upper e) Large Size

Last Answer : a) UPPERCASE

In a large population, 54 % of the people have been vaccinated. If 5 people are randomly selected, what is the probability that AT LEAST ONE of them has been vaccinated?

Description : In a large population, 54 % of the people have been vaccinated. If 5 people are randomly selected, what is the probability that AT LEAST ONE of them has been vaccinated?

Last Answer : We will use Q and P to help solve this problem with Q representing the possibility that none of the randomly selected people are vaccinated and P representing the possibility that at least 1 randomly selected ... -0.0206=0.9794. The probability that at least 1 person has been vaccinated is 0.9794.

Management by exception implies focussing attention on– (A) All variations (B) Normal variations (C) Abnormal variations (D) Randomly selected variations

Description : Management by exception implies focussing attention on– (A) All variations (B) Normal variations (C) Abnormal variations (D) Randomly selected variations

Last Answer : Answer: Abnormal variations

A planning phase for an engineering component generated 80 engineering drawings. The QA team randomly selected 8 drawings for inspection. This exercise can BEST be described as example of: A. Inspection B. Statistical Sampling C. Flowcharting D. Control Charting

Description : A planning phase for an engineering component generated 80 engineering drawings. The QA team randomly selected 8 drawings for inspection. This exercise can BEST be described as example of: A. Inspection B. Statistical Sampling C. Flowcharting D. Control Charting

Last Answer : B. Statistical Sampling

.A planning phase for an engineering component generated 80 engineering drawings. The QA team randomly selected 8 drawings for inspection. This exercise can BEST be described as example of:A. Inspection B. Statistical Sampling C. Flowcharting D. Control Charting

Description : .A planning phase for an engineering component generated 80 engineering drawings. The QA  team randomly selected 8 drawings for inspection. This exercise can BEST be described as  example of: A. Inspection B. Statistical Sampling C. Flowcharting D. Control Charting

Last Answer : B. Statistical Sampling

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

Description : 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 ... sample B. Convenience sample C. Judgment sample D. Quota sample E. Area sample

Last Answer : A. Probability sample

Message can be sent more securely using DES by a. encrypting plain text by a different randomly selected key for each transmission b. encrypting plain text by a different random key for each message transmission and sending the key to the receiver using a public key system c. using an algorithm to implement DES instead of using hardware d. designing DES with high security and not publicizing algorithm used by it

Description : Message can be sent more securely using DES by a. encrypting plain text by a different randomly selected key for each transmission b. encrypting plain text by a different random key for each ... instead of using hardware d. designing DES with high security and not publicizing algorithm used by it  

Last Answer : b. encrypting plain text by a different random key for each message transmission and sending the key to the receiver using a public key system

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

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

Last Answer : (B) 1/32

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

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

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

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

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

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

80 bulbs are selected at random from a lot -Maths 9th

Description : 80 bulbs are selected at random from a lot -Maths 9th

Last Answer : Number of bulbs having life less than 900 hours = 10 + 12 + 23 = 45 P (a bulb has life less than 900 hours) = 45/80 = 9/16

1500 families with 2 children were selected -Maths 9th

Description : 1500 families with 2 children were selected -Maths 9th

Last Answer : (i) P (a family having 2 girls) = Number of families having 2 girls/Total number of families = 475/1500 = 19/60 (ii) P (a family having 1 girl) = Number of families having 1 girl/Total number of ... families = 211/1500 Sum of probabilities = 475/1500 + 814/1500 + 211/1500 = 1500/1500 = 1

An Insurance company selected 2000 drivers -Maths 9th

Description : An Insurance company selected 2000 drivers -Maths 9th

Last Answer : Total number of drivers = 2000 (i) Number of drivers who are 18-29 years old and have exactly 3 accidents in one year is 61 So, P (driver is 18-29 years old with exactly 3 accidents) = 61/2000 = 0.0305 ~ 0. ... = 440 + 505 + 360 = 1305 So, P (drivers with no accident) = 1305/2000 = 0.6525 = 0.653

In a group there are 3 women and 3 men. 4 people are selected at random from this group -Maths 9th

Description : In a group there are 3 women and 3 men. 4 people are selected at random from this group -Maths 9th

Last Answer : A : Selected 3 women and 1 man B : Selected 1 women and 3 men S : Selected 4 people from 6 people (3 + 3) Then n(A) = 3C3 3C1, n(B) = 3C1 3C3, n(S) = 6C4∴ Required probability = P(A) + P(B) = \(rac{ ... 3C_1}{^6C_4}\) + \(rac{^3C_1 imes^3C_3}{^6C_4}\)= \(rac{2 imes1 imes3}{15}\) = \(rac{2}{5}.\)

What is the probability that a number selected at random from the set of numbers {1, 2, 3, …, 100} is a perfect cube? -Maths 9th

Description : What is the probability that a number selected at random from the set of numbers {1, 2, 3, …, 100} is a perfect cube? -Maths 9th

Last Answer : (a) \(rac{1}{25}\) Let us assume S as the sample space in all questions. S means the set denoting the total number of outcomes possible. Let S = {1, 2, 3, , 100} be the sample space. Then, n(S) = 100 Let A : ... ∴Required probability P(A) = \(rac{n(A)}{n(S)}\) = \(rac{4}{100}\) = \(rac{1}{25}\)

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

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 = \(rac{n(A)}{n(S)}\) = \(rac{7}{10}\).

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

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

Last Answer : (b) \(rac{2}{5}\)As each horse has equal chance of winning the race, Number of ways in which one of the five horses wins the race = 5C1 ∴ n(S) =5C1 = \(rac{|\underline5}{|\underline4|\underline1}\) 5To find the chance ... n(E) = 2C1 = 2 ∴ Required probability = \(rac{n(E)}{n(S)}\) = \(rac{2}{5}\).

In how many ways can a team of 11 players be selected from 14 players when two of them play as goalkeepers only? -Maths 9th

Description : In how many ways can a team of 11 players be selected from 14 players when two of them play as goalkeepers only? -Maths 9th

Last Answer : As each team of 11 players has one goalkeeper and 10 team members, and out of 14 players there are 2 goalkeepers and 12 team members. = 12×112×2 = 132.

A box contains 5 different red and 6 different white balls. In how many ways can 6 balls be selected so that there are at least -Maths 9th

Description : A box contains 5 different red and 6 different white balls. In how many ways can 6 balls be selected so that there are at least -Maths 9th

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

A point is selected at random inside an equilateral triangle. From this point a perpendicular is dropped to each side. -Maths 9th

Description : A point is selected at random inside an equilateral triangle. From this point a perpendicular is dropped to each side. -Maths 9th

Last Answer : answer:

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

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 = \(rac{5! imes3!}{7!}\) = \(rac{1}{7}\)

A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, -Maths 9th

Description : A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, -Maths 9th

Last Answer : For the given triangle, we have a = 28 cm, b = 30 cm, c = 26 cm Area of the given parallelogram = Area of the given triangle ∴ Area of the parallelogram = 336 cm2 ⇒ base x height = 336 ⇒ ... be the height of the parallelogram. ⇒ h = 33628 = 12 Thus, the required height of the parallelogram = 12 cm

The median of the data 26,56,32,33,60,17,34,29,45 is 33. If 26 is replaced by 62, then find the new median. -Maths 9th

Description : The median of the data 26,56,32,33,60,17,34,29,45 is 33. If 26 is replaced by 62, then find the new median. -Maths 9th

Last Answer : Here, the given data in ascending order is 17, 29, 32, 33, 34, 45, 56, 60, 62 Now median is (9 + 1 / 2)th term i.e. , 5th term Hence, new median is 34.

If a + b + c = 9 and ab + bc + ca = 26, find a2 + b2 +c2. -Maths 9th

Description : If a + b + c = 9 and ab + bc + ca = 26, find a2 + b2 +c2. -Maths 9th

Last Answer : Find a2 + b2 +c2.

The median of the data 26,56,32,33,60,17,34,29,45 is 33. If 26 is replaced by 62, then find the new median. -Maths 9th

Description : The median of the data 26,56,32,33,60,17,34,29,45 is 33. If 26 is replaced by 62, then find the new median. -Maths 9th

Last Answer : Here, the given data in ascending order is 17, 29, 32, 33, 34, 45, 56, 60, 62 Now median is (9 + 1 / 2)th term i.e. , 5th term Hence, new median is 34.

If a + b + c = 9 and ab + bc + ca = 26, find a2 + b2 +c2. -Maths 9th

Description : If a + b + c = 9 and ab + bc + ca = 26, find a2 + b2 +c2. -Maths 9th

Last Answer : Find a2 + b2 +c2.

What is the supplement of complement of 26 degrees -Maths 9th

Description : What is the supplement of complement of 26 degrees -Maths 9th

Last Answer : The complement of 26 degrees is (90-26). The supplement of (90-26) is 180-(90-26)=116 degrees.

What is the supplement of complement of 26 degrees -Maths 9th

Description : What is the supplement of complement of 26 degrees -Maths 9th

Last Answer : The complement of 26 degrees is (90-26). The supplement of (90-26) is 180-(90-26)=116 degrees.

Find a letter from the Bengali alphabet , which is divided into two numbers by addition , subtraction , multiplication , 12 , 8 , 26 and 3, respectively, what is the letter ?

Description : Find a letter from the Bengali alphabet , which is divided into two numbers by addition , subtraction , multiplication , 12 , 8 , 26 and 3, respectively, what is the letter ?

Last Answer : The letter ' ঔ ' ... If you divide it side by side, 9 above and 3 below. Which is 9 + 3 = 12 , 9-3 = 8 , 9 * 3 = 26 , 9/3 = 3.

Use capital letters, full stops, question marks, commas and inverted commas wherever necessary in the following para-graph. -English 9th

Description : Use capital letters, full stops, question marks, commas and inverted commas wherever necessary in the following para-graph. -English 9th

Last Answer : An arrogant lion was wandering through the jungle one day. He asked the tiger, Who is stronger than you ? You, O! lion, replied the tiger. Who is more fierce than a leopard? asked the lion. You, ... Look , said the lion, there is no need to get mad just because you don't know the answer.

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

Last Answer : Total number of ways of placing n letters in n envelopes = n! All the letters can be placed correctly in only 1 way ∴ Probability of placing all the letters in the right envelopes = \(rac{1}{n!}\) ∴ Probability that all the letters are not placed in the right envelope = 1 – \(rac{1}{n!}\) .

The letters ofthe word ‘NATIONAL’are arranged atrandom. What is the probability that the last letter will be T ? -Maths 9th

Description : The letters ofthe word ‘NATIONAL’are arranged atrandom. What is the probability that the last letter will be T ? -Maths 9th

Last Answer : (c) \(rac{1}{8}\)Let S be the sample space.Then n(S) = Total number of ways in which the letters of word NATIONAL can be arranged= \(rac{8!}{2!\,2!}\) (∵There are 2A s and 2N's in 8 letters)Let E : Event of ... !}{2!\,2!}}\) = \(rac{7!}{8!}\) = \(rac{7!}{8 imes7!}\) = \(rac{1}{8}\).

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

Description : 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

Last 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) ... ! imes36}{rac{10!}{2!}}\) = \(rac{ ot8! imes36 imes2!}{10 imes9 imes ot8!}\) = \(rac{4}{5}\).

How many words can be formed from the letters of the word “DAUGHTER” so that the vowels always come together? -Maths 9th

Description : How many words can be formed from the letters of the word “DAUGHTER” so that the vowels always come together? -Maths 9th

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

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

Last Answer : (i) There are 11 letters in the word 'MATHEMATICS' . Out of these letters M occurs twice, A occurs twice, T occurs twice and the rest are all different. Hence, the total number of arrangements of ... 4!2!=12. Hence, the number of arrangement in which 4 vowels are together =(10080×12)=120960.

How many words can be formed from the letters of the word “SUNDAY” so that the vowels never come together? -Maths 9th

Description : How many words can be formed from the letters of the word “SUNDAY” so that the vowels never come together? -Maths 9th

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

All the words that can be formed using the letters A, H, L, U, R are written as in a dictionary -Maths 9th

Description : All the words that can be formed using the letters A, H, L, U, R are written as in a dictionary -Maths 9th

Last Answer : No. of words starting with A are 4!=24 No. of words starting with H are 4!=24 No. of words starting with L are 4!=24 These account for 72 words Next word is RAHLU and the 74th word RAHUL.