Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

2) 9:00 pm

4) 1992

10 minutes.

Let's say you have five dots in a line and another five dots in a line some distance away.

The total number of ways of drawing five lines, connecting the dots on one side with the dots on the other and each line connecting two dots that aren't connected by any other lines, will give the total number of ways of selecting the two questions, but we can't have the lines going straight across because that would be equivalent to selecting the same question for the same student.

Here's an equation using the inclusion-exclusion principle:

48,120 represents the total number of ways of drawing the five lines with at least one line going straight across.

That isn't what we want. We want the total number of ways of drawing five lines with none going straight across. So...

Since each two questions for each student can be selected in two different ways we divide by 2^5.

I'm not sure exactly why my earlier answer was wrong.

Anyways, good job anonimnystefy! This is a hard problem.

**Fruityloop**- Replies: 5

I remember reading somewhere that if the probability of an event occurring is too small, say 10^-50, then it is basically equal to zero.

In other words, it will never happen given even an infinite amount of time and trials. That seems odd because you would think that

with an infinite amount of trials every event that could happen, will happen, no matter how small the probability. But, let's say I ask you to keep flipping a coin until you get 1,000,000,000,000 heads in a row. It seems that this will never happen no matter how long you try. Maybe this explains part of the St. Petersberg paradox (which is discussed elsewhere on this site). Thoughts anyone?

OK. The 44 permutations for the second question needs to be run through for all possible permutations of the first question which is 120 or 5!. We are counting things twice so we divide by 2.

Is this right?

The chances of at least one match contain other situations where there are more than just pairs of people matching birthdays. Three people sharing the same birthday for example. Maybe that accounts for the discrepancy?

I'm not sure what you did but you put 1.0299682 instead of 0.00010299682. Always remember that the probability of an event occurring is always a number between 0 and 1. Just keep trying. You can do it.

I have the same answer as you but I used a 20 x 20 Markov chain. The way you got the answer involves some math that is way beyond me.

I'm glad I got the right answer.

I think the general formula, for n people and exactly x pairs is...

So for exactly 3 pairs of people out of 23 having the same birthday the probability is 0.01832728.

I'm not absolutely sure if this is correct, but I think this is right.

A derangement problem with none of the items in its own slot.

**Fruityloop**- Replies: 20

You have two boxes. We will call them box #1 and box #2. Box #1 has 20 wooden disks inside that are numbered 1 to 20. Box #2 is empty.

You have a 20 sided die. You roll the die and whichever number comes up on the die, the disk with that number gets moved to the other box.

So, if it is in box #1 it gets moved to box #2, if it is in box #2 it gets moved to box #1.

Starting with all 20 disks in box #1, on average how many rolls are required for box #2 to contain all of the disks and box #1 is empty?

People buying powerball tickets...

9y - 17 = 11x

y-(17/9) = x + (2x/9)

y-x-1 = (8+2x)/9

Because x and y are integers so is (8+2x)/9.

(8+2x)/9=z

8+2x = 9z

4+x=4z+(z/2)

4+x-4z=(z/2)

z=2w

we have

8+2x=18w

so we have

x=9w - 4

y=11w - 3

So we have 2 solutions x=-4 and y=-3 also x=5 and y=8

To get the other number we subtract y from 11.

so we have 83 as one solution and -38 as another solution also -83 is another solution.

The other problem is this, again 3 people A, B, and C.

A and B working together can prepare a shipment in 8 days.

A and C working together can prepare a shipment in 9 days.

B and C working together can prepare a shipment in 10 days.How long would C take working alone?

The book says 22 days. I am getting 23 and 7/31 days.That is what I am getting also.

Yeah, I don't understand the solution in the book. They have the following equations...

(A + B)/4 = 8

(A + C)/4 = 9

(B + C)/4 = 10

So, A = 14, B = 18, and C =22.

I honestly have no idea what they're doing.

This is back a few pages..

Now supposing 2 die had faces of 2,4,6,8,10,12 and 2 other die had faces of 1,3,5,7,9,11.

All four die are thrown once. What is the probability of them summing to 20 or more?

Is it 545/648?

**Fruityloop**- Replies: 6

I have a book which I think has the wrong answers to 2 problems.

In the first problem there are 3 people, we'll call them A, B, and C.

A can make 5 scarves while C makes 2.

B makes 4 scarves while A makes 3.

A's scarf takes 5 times as much cloth as B's scarf.

Three of B's scarves take as much cloth as 5 of C's scarves.

C's scarves are 4 times as warm as B's scarves.

A's scarves are 3 times the C's scarves.

Who's the best overall?

Interestingly, I have 2 books which have this exact same problem, one book is from 1885 and the other book

is from 2013. I believe that the book from 1885 has the correct answer, it claims that the best overall is C.

The book from 2013 claims the answer is A. The book from 2013 makes the mistake (which is pointed out

by the older book) of adding the numbers once the proportions are established, when they should be multiplied instead.

The other problem is this, again 3 people A, B, and C.

A and B working together can prepare a shipment in 8 days.

A and C working together can prepare a shipment in 9 days.

B and C working together can prepare a shipment in 10 days.

How long would C take working alone?

The book says 22 days. I am getting 23 and 7/31 days.

Try moving the 1.5^2 to the right side of the equation and see what you can do.

I agree that with the ever-increasing temperatures, it is becoming more and more likely to set new higher temperature records.

My idea was to figure out what would be an average number of years since a new temperature record was set and then

take an average for all 365 days from a weather station and do a comparison. When I have time maybe I will do that.

It seems you get two equations with two variables.

One equation for the lateral area has two variables, say x and y.

One equation for the volume is another equation with x and y.

5. the base is an isosceles triangle with a height of 8 and a base of 3 (sides of 6)

I'm a little unclear with this.

I feel like such a ding-dong. I think I know the answer now.

Let's say that there are records going back N years. The probability of each year being the one that holds the record is 1/N.

So the average number of years since the latest record (which is the record overall of course) is

I generated 65 random numbers between 1 and 1300 inclusive and did this 50 times.

The average number of numbers since the all-time high number was 30.28 which is close to 33 which is what we would expect.

Let's say you have temperature records going back 130 years, the average number of years since the all-time high

record would be 65.5. To me this seems counter-intuitive because it seems like later years would be more likely to hold

the all-time high records.

Thank you for your response Bobbym.

This is kind of a strange problem because we are dealing with something varying around an average that is unknown.

True, the setting of 50% is somewhat arbitrary, but the first year that there is a record we assume that the rainfall (or temperature)

record is just as likely to be above as below the average, since we don't know what the average is.

because each of the following 4 years has 50% chance of breaking it or not.

I'm comparing the following 4 years only to the first year.

There are 2 possible sequences for the first 2 years, with only one sequence having the 2nd year setting a new record. So we have a probability of 1/2 for the 2nd year setting a new record. Similar reasoning shows that the probability of the 3rd year setting a new record is 1/3 and so on...