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

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
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Answer: D) 
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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

Last Answer : Let A : Event of husband being selected B : Event of wife being selectedThen P(A) = \(rac{1}{7},\) P(B) = \(rac{1}{5}\)P(\(\bar{A}\)) = 1 - \(rac{1}{7}\) = \(rac{6}{7}\), P(\(\bar{B}\)) = 1 - ... {24}{35}\)(iv) P(at least one selected) = 1 - P(none selected) = 1 - \(rac{24}{35}\) = \(rac{11}{35}\)

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

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

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

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

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

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 

There are two groups, X and Y wrote an examination. The probability of X's pass is 3/5 and the probability of Y's pass is 5/7. What is the probability that only one of them is passed out? a) 15/16 b) 16/35 c) 12/43 d) 18/35

Description : There are two groups, X and Y wrote an examination. The probability of X's pass is 3/5 and the probability of Y's pass is 5/7. What is the probability that only one of them is passed out? a) 15/16 b) 16/35 c) 12/43 d) 18/35

Last Answer : Answer: B) Let X be the event of the group X pass Let Y be the event of the group Y pass Then, X'= Event of the group X's fail and Y'= event of the group Y's fail. Therefore, p(X) = 3/5 and p(Y) = 5/7, P(X') = 1 - P( ... )] + p[(Y And X')] = p(X) * p(Y') + P(Y) *P(X') = (6/35 + 10/35) = 16/35

A Package contains 12 pack of variety1 drink, 6 pack of variety2 drink and 8pack of variety3 drink. Three packsof them are drawn at random, what is the probability that the three are not of the same variety? a) 37/325 b) 288/325 c) 188/325 d) None of these

Description : A Package contains 12 pack of variety1 drink, 6 pack of variety2 drink and 8pack of variety3 drink. Three packsof them are drawn at random, what is the probability that the three are not of the same variety? a) 37/325 b) 288/325 c) 188/325 d) None of these

Last Answer : Answer: B) Total number of drink pack= 12+6+8= 26. Let S be the sample space. Then, n(S) = number of ways of taking 3 drink pack out of 26. Therefore, n(S) = 26C3 = 2600 Let Ebe the ... 296/2600=37/325 Then, the probability of taking 3 pack are not of the same variety = 1 - 37/325= 288/325

A box contains 3red, 8 blue and 5 green marker pens. If 2 marker pens are drawn at random from the pack, not replaced and then another pen is drawn. What is the probability of drawing 2 blue marker pens and 1 red marker pen? a) 3/20 b) 1/20 c) 7/20 d) 9/20

Description : A box contains 3red, 8 blue and 5 green marker pens. If 2 marker pens are drawn at random from the pack, not replaced and then another pen is drawn. What is the probability of drawing 2 blue marker pens and 1 red marker pen? a) 3/20 b) 1/20 c) 7/20 d) 9/20

Last Answer : Answer: B) Probability of drawing 1 blue marker pen =8/16 Probability of drawing another blue marker pen = 7/15 Probability of drawing 1 red marker pen = 3/14 Probability of drawing 2 blue marker pens and 1 red marker pen = 8/16*7/15*3/14=1/20

A Receptacle contains 3violet, 4purple and 5 black balls. Three balls are drawn at random from the receptacle. The probability that all of them are purple, is: A)3/55 B)7/55 C)1/55 D)9/55

Description : A Receptacle contains 3violet, 4purple and 5 black balls. Three balls are drawn at random from the receptacle. The probability that all of them are purple, is: A)3/55 B)7/55 C)1/55 D)9/55

Last Answer : Answer: C) Let S be the sample space. Then, n(S) = number of ways of drawing 3 balls out of 12 = 12C3 = 220 Let E = event of getting all the 3 purple balls. n(E) = 4C3= 4 P(E) = n(E)/n(S) = 4/220 = 1/55

A carton contains 12 green and 8 blue bulbs .2 bulbs are drawn at random. Find the probability that they are of same colour. A) 91/47 B) 47/105 C) 47/95 D) 95/47

Description :  A carton contains 12 green and 8 blue bulbs .2 bulbs are drawn at random. Find the probability that they are of same colour. A) 91/47 B) 47/105 C) 47/95 D) 95/47

Last Answer : Answer: C) Let S be the sample space Then n(S) = no of ways of drawing 2 bulbs out of (12+8) = 20c2=20*19/2*1=190 Let E = event of getting both bulbs of same colour Then, n(E) = no of ways (2 bulbs out of 12) ... 12C2+ 8C2=(132/2)+(56/2) = 66+28 = 94 Therefore, P(E) = n(E)/n(S) = 94/190 = 47/95

Two dice are rolling simultaneously .What is the probability that the sum of the number on the two faces is divided by 5 Or 7. A) 13/36 B) 14/36 C) 11/36 D) 9/36

Description : Two dice are rolling simultaneously .What is the probability that the sum of the number on the two faces is divided by 5 Or 7. A) 13/36 B) 14/36 C) 11/36 D) 9/36 

Last Answer :  Answer: C) Clearly, n(S) = 6 x 6 = 36 Let E be the event that the sum of the numbers on the two faces is divided by 5or 7. Then,E = {(1,4),(1,6),(2,3),(2,5),(3,2),(3,4),(4,1),(4,3),(4,6),(5,2),(5,5),(6,1),(6,4)} n(E) = 11. Hence, P(E) = n(E)/n(S) = 11/36

Two dice are thrown together. The probability that the total score is a composite number is: A) 5/12 b) 12/7 c) 7/12 d) 12/5

Description : Two dice are thrown together. The probability that the total score is a composite number is: A) 5/12 b) 12/7 c) 7/12 d) 12/5

Last Answer :  Answer: C)  Clearly, n(S) = (6 x 6) = 36. Let E = Event that the sum is a composite number Then E= { (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 5), (3, 3), (3, 6), (4, 2), (4,4),(4, 5), (4, 6), ( ... 5,3),(5,4),(5,5),(6,2),(6,3),(6,4),(6,6) } n(E) = 21 P(E) = n(E)/n(S) = 21/36 = 7/12.

A question (sum) is given to three boys whose chances of solving it are 1/3,1/4 and 1/5 respectively. What is the probability that the question will be solved? A) 4/5 B) 3/5 C) 3/4 D) 7/5

Description : A question (sum) is given to three boys whose chances of solving it are 1/3,1/4 and 1/5 respectively. What is the probability that the question will be solved? A) 4/5 B) 3/5 C) 3/4 D) 7/5

Last Answer : Answer: B) 

A single coin is tossed 7 times. What is the probability of getting at least one tail? a) 127/128 b) 128/127 c) 2/128 d) 4/128

Description : A single coin is tossed 7 times. What is the probability of getting at least one tail? a) 127/128 b) 128/127 c) 2/128 d) 4/128

Last Answer : Answer: A) Consider solving this using complement. Probability of getting no tail = P(all heads) = 1/128 P(at least one tail) = 1 – P(all heads) = 1 – 1/128 = 127/128

Consider the example of finding the probability of selecting a red card or a 9 from a deck of 52 cards. A) 15/26 B) 26/15 C) 7/13 D) 13/7

Description : Consider the example of finding the probability of selecting a red card or a 9 from a deck of 52 cards. A) 15/26 B) 26/15 C) 7/13 D) 13/7

Last Answer : Answer: C) We need to find out P(R or 6) Probability of selecting a Red card = 26/52 Probability of selecting a 9 = 4/52 Probability of selecting both a red card and a 9 = 2/52  P(R or 9) = P(R) + P(9) – P(R and 9) = 26/52 + 4/52 – 2/52 = 28/52 = 7/13.

In a Coupon, there are 30prizes and 75blanks. A Coupon is drawn at random. What is the probability of getting a prize? A) 2/7 B) 5/7 C) 1/5 D) 1/2

Description : In a Coupon, there are 30prizes and 75blanks. A Coupon is drawn at random. What is the probability of getting a prize? A) 2/7 B) 5/7 C) 1/5 D) 1/2

Last Answer : Answer: A) Total number of outcomes possible, n(S) = 30+75 = 105 Total number of prizes, n(E) = 30 P(E)=n(E)/n(S)=30/105=2/7

What is the probability that we could drink a toast to Schrödinger on his 126th birthday?

Description : What is the probability that we could drink a toast to Schrödinger on his 126th birthday?

Last Answer : See http://www.fluther.com/162587/speaking-of-cats-who-wants-to-have-a-google-doodle-giggle/

What is the probability of drawing a jack and a queen consecutively from a deck of 52 cards, without replacement? a) 4/664 b) 8/52 c) 4/663 d) 4/52

Description : What is the probability of drawing a jack and a queen consecutively from a deck of 52 cards, without replacement? a) 4/664 b) 8/52 c) 4/663 d) 4/52

Last Answer : Answer: C)  Probability of drawing a jack = 4/52 = 1/13 After drawing one card, the number of cards are 51. Probability of drawing a queen = 4/51. Now, the probability of drawing a jack and queen consecutively is 1/13 * 4/51 = 4/663

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

There are 2 vessels. 1st vessel contains 5white and 5 blue thread roll. 2nd vessel contains 4 white and 6 black thread roll. One roll is taken at random from first vessel and put to second vessel without noticing its color. Now a roll is chosen at random from 2nd vessel. What is the probability of the second roll being a white colored roll? A) 11/13 B) 9/11 C) 13/11 D) 5/12

Description : There are 2 vessels. 1st vessel contains 5white and 5 blue thread roll. 2nd vessel contains 4 white and 6 black thread roll. One roll is taken at random from first vessel and put to second vessel without noticing its color ... second roll being a white colored roll? A) 11/13 B) 9/11 C) 13/11 D) 5/12

Last Answer : Answer: B) Case 1: first was a white roll Now it is put in second vessel, so total white rolls in second vessel = 4+1 = 5, and total rolls in second vessel = 10+1 = 11 So probability of white roll ... = 5/11+4/11 = 9/11 (added the cases because we want one of these cases to happen and not both)

What is the probability of the occurrence of a number that is even or less than 3 when a fair die is rolled. a) 2/3 b)3/2 c)5/6 d)6/5

Description : What is the probability of the occurrence of a number that is even or less than 3 when a fair die is rolled. a) 2/3 b)3/2 c)5/6 d)6/5

Last Answer : Answer: A) Let the event of the occurrence of a number that is even be A' and the event of the occurrence of a number that is less than 3 be B'. We need to find P(A or B). P(A) = 3/6 (even numbers = 2,4,6) P(B) = 2/6 ( ... = P(A) + P(B) - P(A or B)  = 3/6 + 2/6 - 1/6 P(A or B) = 4/6=2/3

What is the probability of getting a 4 or a 6 when a die is thrown together? a) 2/3 b) 1/3 c) 3/6 d) 4/6

Description : What is the probability of getting a 4 or a 6 when a die is thrown together? a) 2/3 b) 1/3 c) 3/6 d) 4/6

Last Answer : Answer: B) Taking the individual probabilities of each number, getting a 4 is 1/6 and so is getting a 6. Applying the formula of compound probability, Probability of getting a 4 or a 6, P(4 or 6) = P(4) + P(6) – P(4 and 6) ==> 1/6 + 1/6 – 0 2/6 = 1/3

When a single die is rolled, the sample space is {1,2,3,4,5,6}.What is the probability of getting a 3 when a die is rolled? A) 1/2 B) 1/6 C) 6/1 D) 3/6

Description : When a single die is rolled, the sample space is {1,2,3,4,5,6}.What is the probability of getting a 3 when a die is rolled? A) 1/2 B) 1/6 C) 6/1 D) 3/6

Last Answer : Answer: B) No. of ways it can occur = 1 Total no. of possible outcomes = 6 So the probability of rolling a particular number (3) when a die is rolled = 1/6.

A cartoon contains 15 torch lights out of which 3 are defective. Two torch light are chosen at random from this cartoon. The probability that at least one of these is defective is. A) 13/35 B) 14/35 C) 11/35 D) 17/35

Description : A cartoon contains 15 torch lights out of which 3 are defective. Two torch light are chosen at random from this cartoon. The probability that at least one of these is defective is. A) 13/35 B) 14/35 C) 11/35 D) 17/35

Last Answer : Answer: A) P (none is defective) = 12c2/15c2 = (12*11/2*1)/(15*14/2*1) = 66/105=22/35 P (at least one is defective) = (1 – 22/35) =13/35

one rupee coin is tossed twice. What is the probability of getting two consecutive heads ? A)1/2 B)1/4 C)3/4 D)4/3

Description : one rupee coin is tossed twice. What is the probability of getting two consecutive heads ? A)1/2 B)1/4 C)3/4 D)4/3

Last Answer : Answer: B) Probability of getting a head in one toss = 1/2 The coin is tossed twice. So 1/2 * 1/2 = 1/4 is the answer. Here's the verification of the above answer with the help of sample ... (H,H) whose occurrence is only once out of four possible outcomes and hence, our answer is 1/4.

Two cards are drawn at random from a pack of 52 cards.what is the probability that either both are Red or both are king? A) 52/221 B) 55/190 C) 55/221 D) 19/221

Description :  Two cards are drawn at random from a pack of 52 cards.what is the probability that either both are Red or both are king? A) 52/221 B) 55/190 C) 55/221 D) 19/221

Last Answer : Answer: C) We have n(s) = 52C2 = 1326. Let A = event of getting both red cards B = event of getting both king A∩B = event of getting king of red cards n(A) = 26C2 = 325, n(B)= 4C2= 6 and n(A∩B) = 2C2 = 1 P(A ... S) = 1/1326 P(A∪B) = P(A) + P(B) - P(A∩B) = (325+6-1) / 1326 = 330/1326 = 55/221

When two dice are thrown, find the probability of getting a greater number on the first die than the one on the second, given that the sum should equal 9. A) 1/2 B) 1/5 C) 2/5 D) 4/2

Description : When two dice are thrown, find the probability of getting a greater number on the first die than the one on the second, given that the sum should equal 9. A) 1/2 B) 1/5 C) 2/5 D) 4/2 

Last Answer : Answer: A) Let the event of getting a greater number on the first die be G. There are 4 ways to get a sum of 9 when two dice are rolled = {(3,6),(4,5),(5,4), (6,3)}. And there are two ways where the number on the ... Now, P(G) = P(G sum equals 9)/P(sum equals 9) = (2/36)/(4/36) = 2/4 =>1/2

If eight people participate in a Secret Santa gift exchange, what is the probability that the gift-giving will form a loop?

Description : If eight people participate in a Secret Santa gift exchange, what is the probability that the gift-giving will form a loop?

Last Answer : answer:I am not sure, please check it. As I see it, giving gifts can be seen as arrangements. We can then use permutations and combinations. To find the probability, we need sample space and an event. [] [] [] ... n(Event) / n(Sample Space) = 78 / 40320 = 0.001935 or 0.1935 % (approximately 0.2%).

How do I work backwards to find a probability curve?

Description : How do I work backwards to find a probability curve?

Last Answer : answer:Maybe this is the sort of thing you're looking for. I'm going to confess I didn't read it closely, just enough to gather it discusses regression analysis for bell curves. Based on ... of techniques are appropriate. Maybe sine regression? Sorry if you're looking for a more rigorous answer.

Would anyone have a suggestion for this math/probability problem, please?

Description : Would anyone have a suggestion for this math/probability problem, please?

Last Answer : Homework, right?

When you find someone who drowned in a puddle, what is the probability that he drowned in it by accident versus being murdered, simply based on how difficult it is to drown in a puddle by accident?

Description : When you find someone who drowned in a puddle, what is the probability that he drowned in it by accident versus being murdered, simply based on how difficult it is to drown in a puddle by accident?

Last Answer : Drowning in a puddle?? Oh ok I get it, well I’d guess it would be high that it was murder.

Does a Bernoulli trial have to have a .5 probability?

Description : Does a Bernoulli trial have to have a .5 probability?

Last Answer : No. A Bernoulli trial just has to have only two possible outcomes. Heads or Tails? Boy or Girl? etc. Although these examples might appear to have a pr of .5, it doesn't have to. For example, if ... female But suppose there are 15 males in the room, the probability of picking a male is .75, not .5

In MSM college, 35% of the students study Tamil and English. 40% of the students study English. What is the probability of a student studying Tamil given he/she is already studying english? A) 0.675 B) 0.580 C) 0.875 D) 0.725

Description : In MSM college, 35% of the students study Tamil and English. 40% of the students study English. What is the probability of a student studying Tamil given he/she is already studying english? A) 0.675 B) 0.580 C) 0.875 D) 0.725

Last Answer : Answer: C) P(T and E) = 0.35 P(E) = 0.40 P(T/E) = P(T and E)/P(E) = 0.35/0.40 = 0.875

Consider a pack contains 2black, 9 white and 3 pink pencils. If a pencil is drawn at random from the pack, replaced and the process repeated 2 more times, What is the probability of drawing 2 black pencils and 1 pink pencil? a)3/ 49 b)3/686 c)3/14 d)3/545

Description : Consider a pack contains 2black, 9 white and 3 pink pencils. If a pencil is drawn at random from the pack, replaced and the process repeated 2 more times, What is the probability of drawing 2 black pencils and 1 pink pencil? a)3/ 49 b)3/686 c)3/14 d)3/545

Last Answer : Answer: B) Here, total number of pencils = 14 Probability of drawing 1 black pencil = 2/14 Probability of drawing another black pencil = 2/14 Probability of drawing 1 pink pencil = 3/14 Probability of drawing 2 black pencils and 1 pink pencil = 2/14 * 2/14 * 3/14 = 3/686

Tickets numbered 1 to 37 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 4 or 10? A) 11/37 B) 37/11 C) 12/37 D) 37/12

Description : Tickets numbered 1 to 37 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 4 or 10? A) 11/37 B) 37/11 C) 12/37 D) 37/12

Last Answer : Answer: A) Here, S = {1, 2, 3, 4, ...., 36,37}. Let E = event of getting a multiple of 4 or 10= {4,8,12,16,20,24,28,32,36,10, 30}. P(E) = n(E)/n(S) = 11/37

What is the channel capacity of a noisy channel with conditional probability of error p = 1/2?

Description : What is the channel capacity of a noisy channel with conditional probability of error p = 1/2? A) 0 B) 1 C) Infinity D) 2  

Last Answer : What is the channel capacity of a noisy channel with conditional probability of error p = 1/2?  A) 0 B) 1 C) Infinity D) 2  

M speaks truth in 45% of cases and N in 65% of cases. In what percentage of cases are they likely to contradict each other, narrating the same incident? A) 57.5 B) 55.5 C) 53.8 D) 51.5

Description : M speaks truth in 45% of cases and N in 65% of cases. In what percentage of cases are they likely to contradict each other, narrating the same incident? A) 57.5 B) 55.5 C) 53.8 D) 51.5

Last Answer : Answer: D) Let M = Event that M speaks the truth N = Event that N speaks the truth Then P(M) = 45/100 = 9/20 P(N) = 65/100 = 13/20 P(M-lie) = 1-9/20 = 11/20 P(N-lie) = 1-13/20 = 7/20 Now, M and N ... = P(M).P(N-lie) + P(M-lie).P(N) = 9/20*7/20 + 11/20*13/20 =206/400 = 206/400*100= 51.5%

Pradeesh gets a chance of 40% to win 1st round of a game and a Priya gets a chance of 55% to win 2nd round of the game. In what % of cases are they likely to contradict each other, narrating the same incident? a) 49% b) 54% c) 51% d) 38%

Description : Pradeesh gets a chance of 40% to win 1st round of a game and a Priya gets a chance of 55% to win 2nd round of the game. In what % of cases are they likely to contradict each other, narrating the same incident? a) 49% b) 54% c) 51% d) 38%

Last Answer : Answer: C) Let A be the event that a pradeesh wins 1st round Let B be the event that a priya wins 2nd round. Then, A' = Event that the pradeesh losses 1st round and B' = event that the priya losses 2nd ... =72/400+132/400 = 204/400 We have to find the %. Required % = (204/400)x100 = 51%.

You expect me to believe it is a mere coincidence that two digit products of multiples of 9 add up to 9?

Description : You expect me to believe it is a mere coincidence that two digit products of multiples of 9 add up to 9?

Last Answer : Believe what you want. it’s a fact. Think about it – 9 is one less than 10, so anything multiplied by 9 will be (10 times that number) minus (1 times that number). it’s really pretty simple logic.

If you sat 1000 monkeys down at 1000 3D-printer Makerbots, would you eventually end up with an A-bomb?

Description : If you sat 1000 monkeys down at 1000 3D-printer Makerbots, would you eventually end up with an A-bomb?

Last Answer : No, because they do not have access to weapons grade uranium, nor the eplosives to compress the urianium beyond critical mass.

Why doesn't this work (combinations and permutations)?

Description : Why doesn't this work (combinations and permutations)?

Last Answer : answer:It seems to me that in the multiplication you're attempting that you're double-counting some of the groups, aren't you? I'm having a hard time visualizing this, but I went to Wolfram ... first two women's choices are counted) The sum of these permutations is the same 285 that you expected.

A group of people all randomly pick another member in the group. How many people are not picked by anybody? {see details}?

Description : A group of people all randomly pick another member in the group. How many people are not picked by anybody? {see details}?

Last Answer : answer:The birthday problem is easier to solve than this because it's not asking for the expected number of matching birthdays, it just asks whether there are matching birthdays. To do the same for your ... it! My method wouldn't be hard to do with a computer program for X of reasonable sizes.

Can you find the optimum strategy for guessing the highest number?

Description : Can you find the optimum strategy for guessing the highest number?

Last Answer : answer:There seems to be very few rules here. As it is a thought experiment, I assume that there is no limit on the numbers and that a number could be repeated. I would also assume that the ... the probability of being correct is 1/n. Did you miss out any important rules that would alter this?

A little Monday morning math.

Description : A little Monday morning math.

Last Answer : I agree with you. Initially, three doors leads to a probability of 1 in 3 for each door. Person chooses Door 1 , with a probability of 33.3% that the car is behind it. When Door 3 is opened ... , in essence, the contestant has two choices: Door 1 and Door 2. The probability is 50% for each door.

At the time of marriage, a man was 6 years older to his wife. 12 years after their marriage, his age is 6/5 times the age of his wife. What was wife’s age at the time of marriage? a) 18 years b) 24 years c) 30 years d) 36 years e) None of these

Description : At the time of marriage, a man was 6 years older to his wife. 12 years after their marriage, his age is 6/5 times the age of his wife. What was wife’s age at the time of marriage? a) 18 years b) 24 years c) 30 years d) 36 years e) None of these

Last Answer : ratio of their ages, 12 years hence = 6 : 5  Difference in ages = 1 ratio = 6 years wife’s age (12 years hence) = 5 ratio = 5 × 6 years = 30 years wife’s age at the time of marriage = 30 – 12 = 18 years Answer: a)

A man has 3 wife's, each one has 3 houses, in one of houses for every wife his mother in law lives. He goes to first house with money, mother doubles it, he leaves half, he goes to second wife, mother doubles it again, he leaves the same amount as to first wife, he goes to third house and mother doubles it again, he leaves the same amount as to first one,but goes home with no money. How much money did he have? 2 wife are sister and they have a same mother? -Riddles

Description : A man has 3 wife's, each one has 3 houses, in one of houses for every wife his mother in law lives. He goes to first house with money, mother doubles it, he leaves half, he goes to second wife, ... with no money. How much money did he have? 2 wife are sister and they have a same mother? -Riddles

Last Answer : This answer was deleted by moderators

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

Rohit married 21 years ago at the age of 90 years. His wife's present age 99years. If 24 years later the average age of rohit, his wife and their son was 90 years, then what is son's present age? A) 15yr B) 14yr C) 13yr D) 12yr

Description : Rohit married 21 years ago at the age of 90 years. His wife's present age 99years. If 24 years later the average age of rohit, his wife and their son was 90 years, then what is son's present age? A) 15yr B) 14yr C) 13yr D) 12yr

Last Answer : Answer: D) Rohit present age = 111years total age of family after 24 yrs = 90*3= 270 present age of rohit & his wife = 111+99 = 210yrs present age of rohit, his wife & his son= 270- 24*3 =198 present age of son= 210-198=12yrs

Your an FBI agent and you are assigned to the case of a murder. A man named Julious Jill was murderered by someone in the house. At least that's what people say. You are to interview everyone that was in the house on Sunday at 6:30pm. You interview the kids first you ask them 'what were you doing on Sunday at 6:30' the kids said we were playing with our toys. You interview the wife and you ask the same thing. The wife says she was cleaning the kids room because the maid said she was going out with friends. So Mts. Jill said she will clean. Next you ask the maid what she was doing the maid says she was getting the mail. Who murderered Mr. Jill? -Riddles

Description : Your an FBI agent and you are assigned to the case of a murder. A man named Julious Jill was murderered by someone in the house. At least that's what people say. You are to interview everyone that ... maid what she was doing the maid says she was getting the mail. Who murderered Mr. Jill? -Riddles

Last Answer : The maid: the maid said she was getting the mail on a Sunday the mail doesn't come on Sundays and she lied to Mrs. Jill about going out with friends.

What to do if wife has intercourse during menstruation ? I have committed a sin. I had sex during my wife's menstruation. God will forgive me for what I do now.

Description : What to do if wife has intercourse during menstruation ? I have committed a sin. I had sex during my wife's menstruation. God will forgive me for what I do now.

Last Answer : It is haraam to have sexual intercourse knowingly while menstruating. The Qur'an says Allah has majide, ويسألونك عن المحيض قل هو أذى فاعتزلوا النساء في المحيض ولا تقربوهن حتى يطهرن فإذا تطهرن فأتوهن من حيث ... Rayek 1/196 ; Fathul Qadir 1/146 ; Fatawal Walwalizia 1/56 ; Addurrul Mukhtar 1/296)

D's mother's father is C's father's father-in-law. How are C and D related if D is male and C female? Options: 1) D is C's wife's brother 2) C is sister of D's wife 3) D is C's brother 4) C is mother of D's wife

Description : D's mother's father is C's father's father-in-law. How are C and D related if D is male and C female?  Options: 1) D is C's wife's brother 2) C is sister of D's wife 3) D is C's brother 4) C is mother of D's wife

Last Answer : Correct Answer: D is C's brother