Logic riddles

logic

Find a short hidden message in the list of words below. carrot fiasco nephew spring rabbit sonata tailor bureau legacy corona travel bikini object happen soften picnic option waited effigy adverb report accuse animal shriek esteem oyster
Starting with the first two words, take the first and last letters, reading from left to right. Example: Carrot fiascO "from these pairs" the message is as follows: CONGRATULATIONS CODE BREAKER
81.84 %
60 votes
logicclean

Today is Admin's birthday. His five close friends Nell, Edna, Harish, Hsirah and Ellen surprised him with party. What is special with this list of these five names?
If you read the names from last to start, it reads the same.
81.65 %
45 votes
trickycleancrazylogic

A black dog stands in the middle of an intersection in a town painted black. None of the street lights are working due to a power failure caused by a storm. A car with two broken headlights drives towards the dog but turns in time to avoid hitting him. How could the driver have seen the dog in time?
It was daylight.
81.65 %
45 votes
logicmathcleanclever

Using only and all the numbers 3, 3, 7, 7, along with the arithmetic operations +,-,*, and /, can you come up with a calculation that gives the number 24? No decimal points allowed. [For example, to get the number 14, we could do 3 * (7 - (7 / 3))]
7 * ((3 / 7) + 3) = 24
81.65 %
45 votes
logicmathclean

What is the least number of people that need to be in a room such that there is greater than a 50% chance that at least two of the people have the same birthday?
Only 23 people need to be in the room. Our first observation in solving this problem is the following: (the probability that at least 2 people have the same birthday + the probability that nobody has the same birthday) = 1.0 What this means is that there is a 100% chance that EITHER everybody in the room has a different birthday, OR at least two people in the room have the same birthday (and these probabilities don't add up to more than 1.0 because they cover mutually exclusive situations). With some simple re-arranging of the formula, we get: the probability that at least 2 people have the same birthday = (1.0 - the probability that nobody has the same birthday) So now if we can find the probability that nobody in the room has the same birthday, we just subtract this value from 1.0 and we'll have our answer. The probability that nobody in the room has the same birthday is fairly straightforward to calculate. We can think of this as a "selection without replacement" problem, where each person "selects" a birthday at random, and we then have to figure out the probability that no two people select the same birthday. The first selection has a 365/365 chance of being different than the other birthdays (since none have been selected yet). The next selection has a 364/365 chance of being different than the 1 birthday that has been selected so far. The next selection has a 363/365 chance of being different than the 2 birthdays that have been selected so far. These probabilities are multiplied together since each is conditional on the previous. So for example, the probability that nobody in a room of 3 people have the same birthday is (365/365 * 364/365 * 363/365) =~ 0.9918 More generally, if there are n people in a room, then the probability that nobody has the same birthday is (365/365 * 364/365 * ... * (365-n+2)/365 * (365-n+1)/365) We can plug in values for n. For n=22, we get that the probability that nobody has the same birthday is 0.524, and thus the probabilty that at least two people have the same birthday is (1.0 - 0.524) = 0.476 = 47.6%. Then for n=23, we get that the probability that nobody has the same birthday is 0.493, and thus the probabilty that at least two people have the same birthday is 1.0 - 0.493) = 0.507 = 50.7%. Thus, once we get to 23 people we have reached the 50% threshold.
81.57 %
66 votes
logicsimpleclean

Suppose you want to send in the mail a valuable object to a friend. You have a box which is big enough to hold the object. The box has a locking ring which is large enough to have a lock attached and you have several locks with keys. However, your friend does not have the key to any lock that you have. You cannot send the key in an unlocked box since it may be stolen or copied. How do you send the valuable object, locked, to your friend - so it may be opened by your friend?
Send the box with valuable object and a lock attached and locked. Your friend attaches his or her own lock and sends the box back to you. You remove your lock and send it back to your friend. Your friend may then remove the lock she or he put on and open the box. Search: Man-in-the-middle attack
81.57 %
52 votes