reinout-g

A random person:

:| How to get the general formula of the set of numbers if there is a difference in two consecutive differences

blizzard:

googol has 3 special pots.Each pot contains 484 spiders.How many spiders does he have altogether

pancakes:

what is a algebraic expression

yara:

the area that lies either to the left of 1.56 or to the right of 2.30

Alan:

Here's my offering:

dfdx.PNG

Nobody Special:

reinout-g:

Here are the new probability puzzles!
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Easier one:
A special machine contains 100 red, 100 green and 100 blue marbles.
The machine performs the following four operations.

First, the machine randomly picks 50 marbles from the entire batch.

Secondly the machine divides the marbles into one group of 20 and one group of 30

Furthermore, the machine picks and shuffles 5 marbles of the group of 20 and 10 marbles of the group of 30.

Finally, the machine randomly chooses one of the final 15 marbles it picked and shows it to you.

Assuming you cannot see anything which happens inside the machine.

What are the odds the machine will show you a red marble?

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Alright, I'm gonna \try this one. ^_^
Since there are all 100 marbles per color, and the machine chooses randomly, then there would be an equal chance that you would get any color.
So there would be a 33.33...% chance of red, blue, or green. Right? ^_^

And I'd agree with @stuff on the other one.

Bertie:

reinout-g:

Here are the new probability puzzles!
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Easier one:
A special machine contains 100 red, 100 green and 100 blue marbles.
The machine performs the following four operations.

First, the machine randomly picks 50 marbles from the entire batch.

Secondly the machine divides the marbles into one group of 20 and one group of 30

Furthermore, the machine picks and shuffles 5 marbles of the group of 20 and 10 marbles of the group of 30.

Finally, the machine randomly chooses one of the final 15 marbles it picked and shows it to you.

Assuming you cannot see anything which happens inside the machine.

What are the odds the machine will show you a red marble?

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Hard one :

Steve has two routes (A and B) to travel to his girlfriend which he is indifferent about (see my amazing illustrational skills )

knipsel11.png

Steve has decided to play a game which makes him decide which route to pick.
He always waits where the road splits between route A and B until he either sees a train going from A to B or from B to A.
If the first train Steve sees goes from A to B he always takes route A and if the first train Steve sees is a train going from B to A he always takes route B.
There are equally as many trains going from A to B as there are going from B to A.
Nevertheless Steve has noticed that for every five times he visits his girlfriend he takes route A four times and route B only once.

How can this be?

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Reinout-g

edit: I updated a flaw in the illustration, 'work' was meant to be 'girlfriend'

Easier one.
Ignore the flim-flam, via a circuitous route a single marble has been taken at random from the original 300. It's 2 to 1 against it being a red marble.

Harder one.
In principle there can be any number of trains travelling between A and B, (and B and A), but start by thinking in terms of one train in either direction.

Suppose that the train from A to B passes on the hour and the train from B to A at twelve minutes past the hour. If 'our man' arrives at the split in the road at some random time, there will be a 48 minute window within which the next train will be the one from A to B, (12 min past the hour through to 60min past the hour, or 12.01 and 59.99 min past the hour if you're going to be pedantic) and a 12 minute window when the next train will be the one from B to A, (0 min to 12 min). The time interval for A is four times as long as the interval for B so it's four times more likely that the next train will be the A to B train.

Now suppose that there are two trains in either direction, the A to B ones on the hour and half hour, the B to A ones at 6 minutes and 36 minutes past the hour. Combining the windows, there will be a total of 48 minutes, (24+24), during which the next train will be from A to B, and 12 minutes, (6+6), when the next train will be from B to A. Again, that's 4 to 1 in favour of A.

We can do this for higher numbers of trains, though it does become somewhat impracticable after a while !
For three trains in either direction the timings for the A to B trains could be** on the hour, 20 minutes past and 40 minutes past, while the timings for the B to A trains could be at 4 minutes, 24 minutes and 44 minutes past the hour.

** The timings need not be exactly as stated, and the intervals need not be equal. So long as the 'windows' for the A to B trains add to 48 minutes it will always be 4 to 1 in favour of A to B.

stuff:

This is so because the majority of the B trains must pass while he is at work, and then the A trains start to roll in when he leaves work for his girlfriend.