A train leaves from Halifax, Nova Scotia heading towards Vancouver, British Columbia at 120 km/h. Three hours later, a train leaves Vancouver heading towards Halifax at 180 km/h. Assume there's exactly 6000 kilometers between Vancouver and Halifax. When they meet, which train is closer to Halifax?
Both trains would be at the same spot when they meet therefore they are both equally close to Halifax.
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.
A deliveryman comes to a house to drop off a package. He asks the woman who lives there how many children she has.
"Three," she says. "And I bet you can't guess their ages."
"Ok, give me a hint," the deliveryman says.
"Well, if you multiply their ages together, you get 36," she says. "And if you add their ages together, the sum is equal to our house number."
The deliveryman looks at the house number nailed to the front of her house. "I need another hint," he says.
The woman thinks for a moment. "My youngest son will have a lot to learn from his older brothers," she says.
The deliveryman's eyes light up and he tells her the ages of her three children. What are their ages?
Their ages are 1, 6, and 6. We can figure this out as follows:
Given that their ages multiply out to 36, the possible ages for the children are:
1, 1, 36 (sum = 38)
1, 2, 18 (sum = 21)
1, 3, 12 (sum = 16)
1, 4, 9 (sum = 14)
1, 6, 6 (sum = 13)
2, 2, 9 (sum = 13)
2, 3, 6 (sum = 11)
3, 3, 4 (sum = 10)
When the woman tells the deliveryman that the children's ages add up to her street number, he still doesn't know their ages. The only way this could happen is that there is more than one possible way for the children's ages to add up to the number on the house (or else he would have known their ages when he looked at the house number). Looking back at the possible values for the children's ages, you can see that there is only one situation in which there are multiple possible values for the children's ages that add up to the same sum, and that is if their ages are either 1, 6, and 6 (sums up to 13), or 2, 2, and 9 (also sums up to 13). So these are now the only possible values for their ages.
When the woman then tells him that her youngest son has two older brothers (who we can tell are clearly a number of years older), the only possible situation is that their ages are 1, 6, and 6.
In a supermarket, the first 25 customers of the day purchased an average of two items each.
After a further 15 customers, the average number of items purchased by each customer rose to eight.
What was the average number of items purchased by the last 15 customers only?