Brainteaser/Puzzle Corner (Part 5)

5 or 6?

 

wrong

 

the limit

 

wrong!

 

they got the wrong answer and walked back?

 

@EMR

As many as you damn well please, you just can't fit them all in it at the same time

@BoP the problem with your math suggesting that ( 1/10 ) ^ infinity is zero therefor its zero both ways is that you are finding the probabllity that all the digits are 3 when you want to find the probabllity that at least 1 is three. Actually calculating the probability that at least one digit is three is difficult. However, you don't need to do that since it is equal to 1 - P( no digits are three ). And again, P( no digits 3 ) is zero.

 

What kind of mathematician are you? I have no idea what you are talking about, but that may be because the highest math course I've had so far is Calc1.

 

its just probability, I got that brainteaser from the professor who teaches probability. Every Friday she gives a brainteaser, most of them are probability related, this is the one she gave last friday. The person who is impressive is LordKerwyn for getting the answer so quickly.

 

Well quanta I actualy typer out the way you would find the answer by using adding up to find the probability.

If there is any holes in that equation point them out i am actually curious if it would work.

 

Well, you made me look into it and I think a gemoetric distribution fits this problem.

(1 - p)x - 1*p where x is an element of the natural numbers. Since p is .1 we would get
sum x = 1..infinity (1 - 0.1)x - 1*.1 = 1 This would be the probablility that the first digit is a 3 plus the probability that the second digit a three and the first isn't plus ... plus the probability that the xth digit is a three and all the previous digits aren't.

What you have is a binomeal distribution and I think it may also work. If we want to find the probability that at least 1 is a three the only thing missing that I can see is that I think you need an extra summation

{sumation as A = 1 to infinity} {summation as X = 1 to A} (.9)^(X-A)*(.1)^(A)*(X ncr A) = 1

But again, the simplest solution is to use a uniform distribution and find 1 - P( no digit is 3 ) which is more formally

limit as A -> infinity (1 - (.9)A) = 1

 

Your geometric series comes out to 1 just for the record. But I would be hesitant to use a geometric series because if i rmember correctly it doesnt take into account the various combinations by having more than 1 3 at once. But I dont remember enough of stats toa argue that point atm. Also I think only one summation would be needed in my equation since we are just looking at one thing that is infinitely long and the combinations that can come out of it. So i think a lim would work and I am pretty sure it would make it nicer when solving it later.

 

EatMeReturns
The answer is one, because after that one the dropship isn't vacant anymore.

Edit: Yeehaa I won!

 

needler wins!

 

Ha I knew this riddle had a stupid answer..

Nowhere in the problem did it say that the ship had to remain vacant in order to fill more Marines.

The ship without Marines is vacant, and you can fill a vacant ship with as many people you like. The problem did not state that the ship had to stay vacant. Review your gramatic next time.

Anyways, I have a riddle I'm cooking up that I will post here soon.

 

Okay:

Carlos, a Terran, was trying to figure out how much space does a Zealot, a Stalker, and an Immortal occupy inside a Phase Prism. They can only tell the percentage when the Prism is full, which is when a blue light shines. So they did the following experiment:

Firstly, Carlos placed 3 Immortals inside the Phase Prism, no light. He added a Probe. Blue Light.

Then, he removed 2 Immortals, the Probe, and added 3 Stalkers, blue light.

Lastly, he removed a Stalker and an Immortal and added 3 Zealots. No light. He added a Probe, and then Blue light.

Then the Phase Prism powered down and he couldn't recharge it nor do any more experiments with it.

After this, Carlos still could not figure it out. He spent countless hours thinking and still could not solve this problem. When he was about to give up, a scientist came up to him and said: "Carlos, we have discovered that the Probe fills up half the space of a Zealot.

Carlos then after a few scribbling figured out the exact amount of space a Zealot, a Stalker and an Immortal fills up.

What did he find?

*Advice* - You can give the answer in a fraction. (i.e. the Zealot the Stalker and the Immortal occupy exactly 89/112 of the total space inside a Phase Prism), if needed.

Prize: 1 Power-up + 200 minerals.

 

an immortal takes up 9/28 of the space
a stalker takes up 6/28 of the space
a zealot takes up 2/28 of the space
and a probe takes up 1/28 of the space

 

I had a go at this but the third experiment ruins everything >.<
My guess is:
probe: 12.3% -> zealot: 24.6%
immortal: ~29.23%
stalker: ~23.59%

I'll revise my formulae again...

 

You are not supposed to give the answer for each of the units, but 1 Zealot, 1 Immortal and 1 Stalker together. Both the answers are wrong, anyways.

 

Haha maybe you should review your grammar as well ;D

 

a zealot a stalker and an immortal take up 17/28 of the space