What is the probability that a leap year selected at random will have 33 Sunday?

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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

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

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

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)

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

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) = \(rac{n(A)}{n(S)}\) = \(rac{4}{100}\) = \(rac{1}{25}\)

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

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 = \(rac{n(E)}{n(S)}\) = \(rac{2}{5}\).

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)

Description : Form, a cosmetic shop containing perfumes and does, a pair is selected at random. The probability that the selected pair will consist of one perfume a

Last Answer : Form, a cosmetic shop containing perfumes and does, a pair is selected at random. The probability ... of perfumes and does the shop can contain ?

Description : A mathematics book contains 250 pages. A page is selected at random. What is the probability that the number on the page selected is a perfect square?

Last Answer : A mathematics book contains 250 pages. A page is selected at random. What is the probability that the number on the page ... `(3)/(50)` D. `(7)/(125)`

Description : A box contains 50 tickets. Each ticket is numbered from 1 to 50. One ticket is selected at random, find the probability that the number on the ticket

Last Answer : A box contains 50 tickets. Each ticket is numbered from 1 to 50. One ticket is selected at random ... number on the ticket is not a perfect square.

Description : A Class IX English book contains 200 pages. A page is selected at random. What is the probability that the number on the page is divisible by 10?

Last Answer : A Class IX English book contains 200 pages. A page is selected at random. What is the probability that the number on the page is divisible by 10?

Description : Nine playing cards are numbered 2 to 10. A card is selected at random. What is the probability that the card will be an odd number? a. 1/9 b. 2/9 c. 4/9 d. 3/7

Last Answer : c. 4/9

Description : A number is selected at random from first 50 natural numbers. Find the probability that it is a multiple of 3 and 4. a. 3/25 b. 2/25 c. 1/25 d. 13/50

Last Answer : b. 2/25

Description : In a class there are 15 boys and 10 girls. Three students are selected at random. The difference between the probability that 2 boys and 1 girl are selected compared to 1 boy and 2 girls are selected is: a) 23/78 b) 19/88 c) 15/92 d) 4/23 e) 7/46

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)

Description : A box contains 15bolts of which 5 are defective. If 5 bolts are selected at random from the box, what is the probability that at least one of them is defective? a) 91/143 b) 101/143 c) 111/143 d) 121/143 e) 131/143

Last Answer : Probability that at least one is defective = 1 – probability that none is defective Probability that none is defective = 10C5 / 15C5 = 12/143 Required Probability = 1-12/143 = 131/143 Answer: e)

Description : What is the probability that a two digit number selected at random will be a multiple of '3' and not a multiple of '7'? e) 13/45 f) 14/45 g) 23/90 h) 19/45

Last Answer : Answer: a

Description : In a hostel, 40% of the students play cricket, 20% play chess and 10% both. If a student is selected at random, then the probability that he plays cricket or chess is: a) 1/2 b) 3/5 c) 1/4 d) 4/7

Last Answer : Answer: A) Given that, 40% play cricket; that is, P(C) = 40/100=4/10 20% play chess; that is, P(c) = 20/100 =2/10 And, 10% play both cricket and chess; that is, P(C And c) = 10/100 = 1/10 Now, we have ... P(C) + P(c) - P(C And c) = 4/20+2/10-1/10 =5/10=1/2 Hence, the required probability 1/2

Description : In a batch, 40% of the students offered Maths, 30% offered science and 15% offered both. If a student is selected at random, what is the probability that they has offered science or maths? A) 0.55 B) 0.65 C) 0.45 D) 0.75

Last Answer : Answer: A) P(M) = 0.40 P(S) =0.30 and P(M∩S) = 0.15 P(M∪S) = P(M) + P(S) - P(M∩S) = 0.55

Description : In a batch, there are 22 boys and 18 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is: a) 3754/8854 b) 4158/9880 c) 8514/9880 d) 2078/4920

Last Answer : Answer: B) Let , S - sample space E - event of selecting 1 girl and 2 boys. Then, n(S) = Number ways of selecting 3 students out of 40 = 40C3 = 9880 n(E) = 18C1 *22C2 = 18*231  = 4158 P(E) = n(E)/n(s) = 4158/9880 

Description : If the 10th of January is a Sunday in a leap year what day would the 11th of March be?

Last Answer : A Friday.

Description : If the 10th of January is a Sunday in a leap year what day would the 11th of March be?

Last Answer : A Friday.

Description : What is the probability of getting 53 Sundays or 53 Tuesdays or 53 Thursdays in a non–leap year ? -Maths 9th

Last Answer : A non-leap year consists of 365 days. Therefore in a non-leap year there are 52 complete weeks and 1 day over which can be one of the seven days of the week. Possible outcomes n(S) = 7 = {Sunday, Monday, Tuesday, Wednesday, Thursday, ... ⇒ n(A) = 3∴ P(A) = \(rac{n(A)}{n(S)}\) = \(rac{3}{7}.\)

Description : If the date of the last Saturday of last month and the first Sunday of this month do not add up to 33, what month are you in? -Riddles

Last Answer : The same month you are reading this.

Description : Is it odd that I'm 33 and have never been selected for jury duty?

Last Answer : Seems a little odd to me. I was selected for jury duty pretty much as soon as I registered to vote. That was almost seven years ago though, and I haven’t been chosen once since then.

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

Description : `A={1, 2, 3, ..., 9}`. If three numbers are selected from set A and are arranged, the probability that three numbers are either in increasing order or

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Description : A day is selected in a week, find the probability that the day is Monday.

Last Answer : A day is selected in a week, find the probability that the day is Monday.

Description : In a company there are 11 executives: six women five men. Four are selected at attend a management seminar. Find the probability that all four selected are men.?

Last Answer : i dont know

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.

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

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

Description : Murali and his wife appear in an interview for two vacancies in the same post. The probability of murali's selection is (1/6) and the probability of wife's selection is (1/4). What is the probability that only one of them is selected ? A) 8/25 B) 1/7 C) 3/4 D) 1/3

Last Answer : Answer: D) 

Description : In a certain city, 5% of all the persons in town have unlisted phone numbers. If you select 100 names at random from that city's phone directory, how many people selected will have unlisted phone numbers? -Riddles

Last Answer : None. If their names are in the phone directory, they do not have unlisted phone numbers!

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

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) = \(rac{ ... 3C_1}{^6C_4}\) + \(rac{^3C_1 imes^3C_3}{^6C_4}\)= \(rac{2 imes1 imes3}{15}\) = \(rac{2}{5}.\)

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 = \(rac{^4C_1 imes^ ... }{30}\) = \(rac{15+9+1}{30}\) = \(rac{25}{30}\) = \(rac{5}{6}\).

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:

Description : The type of sampling in which each member of the population selected for the sample is returned to the population before the next member is selected is called _________. a. Sampling without replacement b. Sampling with replacement c. Simple random sampling d. Systematic sampling

Last Answer : b. Sampling with replacement

Description : When each member of a population has an equally likely chance of being selected, this is called: a. A nonrandom sampling method b. A quota sample c. A snowball sample d. A Random Sample

Last Answer : d. A Random Sample

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

Description : What is the probability of a random string of numbers having a consistent pattern?

Last Answer : I think if it’s truely random then it is stochastic, but true randomness may not even be possible, so who knows?

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

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

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

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

Description : An integer is chosen at random from the first two hundred positive integers. What is the probability that the integer chosen is divisible by 6 or 8 ? -Maths 9th

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 \(\bigg(rac{200}{6}=33rac{1}{3}\ ... (rac{25}{200}\) - \(rac{8}{200}\) = \(rac{50}{200}\) = \(rac{1}{4}\).

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

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

Last Answer : (b) \(rac{7}{99}\)There are (7 + 5) = 12 balls in the bag. 4 balls can be drawn at random from 12 balls in 12C4 ways. ∴ n(S) = 12C4 = \(rac{|\underline{7}}{|\underline3|\underline4}\) = \(rac{7 imes6 imes5}{3 ... ) = 35∴ Required probability = \(rac{n(A)}{n(S)}\) = \(rac{35}{495}\) = \(rac{7}{99}\).

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