A Puzzle

[All Seeing Eye]

I havn't read the thread so i'll just post my ramblings.

The first one is obviously 1 in 5 but the second question doesn't make sense, the minimum number of beeds you can take out to know it's the bag you are looking for would be 5 red beeds but I don't understand the 50% certainty part because if you only took 4 red beeds out there's only a 20% certainty it's the correct bag which is why IMO it's a trick question and they're probably looking out for someone who would point that out.

Take 1 red beed = 20% certainty
Take 2 red beeds = 20% certainty
Take 3 red beeds = 20% certainty
Take 4 red beeds = 20% certainty
Take 5 red beeds = 100% certainty

 

[mimic]

The 50% certainty would make sense if you returned the three beads that were removed to a random bag rather than the same bag that you took them from...but the question definitely says 'the bag' so I don't know what the deal is.

 

[Roger Wilco]

At least 50% certainty.

Draw 4 red beads, it's still a 1 in 5 chance of having a specific bag: 20%

Draw a 5th red bead and it instantly goes up to 100% certainty of having bag #5.

if drawing one red bead will give you a 30% probability of picking bag 5, I don't think you'll need 4 red beads to tell you you've got bag 5.

nah, the question asks how many to give you a 50% certainty that you got bag 5, not 100% certainty.

Yeah, that threw me off too. Although I have to admit, that I don't even understand your solution.

it's just the reverse of the first question.

doing the way you did:
- picking one red bead, the probability of getting bag 5 is 7/23;
- after you picked a red bead off bag 5, then there are 6 left in bag 5, and 22 in total, that give 6/22.
- and together they give ~57%
although I don't think it's right.

now, the way I did it: the probability of getting two red beads from bag 1-4 is p1 = 4/100 * 3/99; the probability of getting that from bag 5 is p2 = 7/100 * 6/99. So, the probability of picking bag 5 is p2/(p1*4+p2).

wait, we can use the Bayes theorem to do this using conditional probabilities. so, the probability of getting bag 5 given we have two red beads is written as P(B=5|X=2 red), where B is the bag number and X is two red beads. this is equals to: P(X|B) * P(B) / P(X). Now, P(X=2 red | B = 5) = p2 from above, P(B) = 1/5, P(X) is the tricky one. P(X) is the probability of picking 2 red beads regardless of the bag, I think it's 23/500 * 22/499. If that's the case, then P(B=5|X=2 red) = 41%. So that means you'll probable need 3 to give at least 50% certainty. But I can't be bothered with the maths anymore.

Ok, you get the job.

 

[Rosalind]

^ actually I made a small mistake, I thought my calculation for P(X) above is wrong.

The correct calculation:
P(X) = sum_B P(X|B)P(B)
= P(2 red | B = 1) P(B=1) + P(2 red | B = 2) P(B = 2) + ... + P(2 red | B=5) P(B=5)

So, P(B=5 | X = 2 red) = 46.7%, so you'll probably still need 3 to give at least 50% certainty. But, this time, I'm pretty sure it's correct.


edit: just saw this:

4) anyone who writes that the test is pointless and they're not going to waste their time working it out - fast-track to management

 

[Jadzia]

I havn't read the thread so i'll just post my ramblings.

The first one is obviously 1 in 5 but the second question doesn't make sense, the minimum number of beeds you can take out to know it's the bag you are looking for would be 5 red beeds but I don't understand the 50% certainty part because if you only took 4 red beeds out there's only a 20% certainty it's the correct bag which is why IMO it's a trick question and they're probably looking out for someone who would point that out.

Take 1 red beed = 20% certainty
Take 2 red beeds = 20% certainty
Take 3 red beeds = 20% certainty
Take 4 red beeds = 20% certainty
Take 5 red beeds = 100% certainty

Re-imagine the question:

Planets 1-5 each contain 6.6 billion inhabitants.

Planet 1 (Qo'Nos) has 1 human ambassador, the rest are Klingon.
Planet 2 (Romulus) has 1 human ambassador, the rest are Romulan.
Planet 3 (Cardassia) has 1 human ambassador, the rest are Cardassian.
Planet 4 (Vulcan) has 1 human ambassador, the rest are Vulcan.
Planet 5 (Earth) has 1 alien ambassador, the rest are Human.

You're star ship arrives in orbit about one of the planets but you don't know which one you're at. You lock your transporters randomly onto one of the inhabitants from the planet below, and as the person materialises on your bridge, you see he/she is human.

How sure are you that you're in orbit about Earth?

 

[Venardhi]

The ratio of alien to human in your problem is much more skewed than in the original. Really it is nothing like it.

 

[Roger Wilco]

^^^
It get's the point across though; the chance of getting a red bead from bag #5 clearly is not the same as from bags #1-4, so the answer to the first question cannot be 20%.

 

[All Seeing Eye]

I havn't read the thread so i'll just post my ramblings.

The first one is obviously 1 in 5 but the second question doesn't make sense, the minimum number of beeds you can take out to know it's the bag you are looking for would be 5 red beeds but I don't understand the 50% certainty part because if you only took 4 red beeds out there's only a 20% certainty it's the correct bag which is why IMO it's a trick question and they're probably looking out for someone who would point that out.

Take 1 red beed = 20% certainty
Take 2 red beeds = 20% certainty
Take 3 red beeds = 20% certainty
Take 4 red beeds = 20% certainty
Take 5 red beeds = 100% certainty

Re-imagine the question:

Planets 1-5 each contain 6.6 billion inhabitants.

Planet 1 (Qo'Nos) has 1 human ambassador, the rest are Klingon.
Planet 2 (Romulus) has 1 human ambassador, the rest are Romulan.
Planet 3 (Cardassia) has 1 human ambassador, the rest are Cardassian.
Planet 4 (Vulcan) has 1 human ambassador, the rest are Vulcan.
Planet 5 (Earth) has 1 alien ambassador, the rest are Human.

You're star ship arrives in orbit about one of the planets but you don't know which one you're at. You lock your transporters randomly onto one of the inhabitants from the planet below, and as the person materialises on your bridge, you see he/she is human.

How sure are you that you're in orbit about Earth?

There's about a 1 in 1 chance you beam a Human up at Earth (not exactly 1 in 1 cos of the alien ambassador) but a 1 in 6.6 Billion chance of beaming up a Human from the other worlds.
It's more likely you're at Earth because the odds of you beaming up a human on all the other worlds is far less.

But that's not really what the second question was about. How can you have 50% certainty from the minimum amount of beeds picked? you pick out 4 red beeds there's no way to be certain you've picked from bag 5 because all 5 bags have 4 red beeds in, the only way to have any certainty is to pick out 5 red beeds but once you've done that you're at 100% certainty.
It's not about probability it's about certainty.

 

[Rosalind]

I realised that the two questions are exactly the same if you think about the questions as conditional probabilities: p = P(B=5|X). The first question give you X and you need to find p, while the second question gives you p and you need to find X.

which means my answer to the first questions wasn't exactly right. Again, back to Bayes Thereom:

P(B = 5 | X) = P(X | B=5) P(B=5) / P(X)
where
P(X) = sum_B P(X | B=1)P(B=1) + ... + P(X | B=5) P(B=5)
and X = 1 red and 2 black beans.

So
P(X|B=1) = P(X|B=2) = P(X|B=3) = P(X|B=4) ~= 4/100 * 96/100 * 95/100 and
P(X|B=5) ~= 7/100 * 93/100 * 92/100
and
P(B=1) = ... = P(B=5) = 1/5

So after drawing 1 red and 2 black beads from a bag, the probability that you draw from bag 5 is 29.10%, so I was right with the answer but wrong in getting there, but since the numbers were small, it didn't matter.

 

[trampledamage]

What job was this for?

Hope it's not anything I would ever want to apply for - I'd be screwed!

 

[the 4th hanson bro]

Getting a job as his assistant isn't a long term position usually.

How true.

You know, if I set this sort of math question on a job application, I would divide the completed applications as follows:


3) anyone who gets the right answer and can't prove they did it themselves - sales

What's wrong with being in sales I ask?

4) anyone who writes that the test is pointless and they're not going to waste their time working it out - fast-track to management

My first thought was "who cares?" Is that close enough?

 

[Holdfast]

edit: just saw this:

4) anyone who writes that the test is pointless and they're not going to waste their time working it out - fast-track to management

I was serious.

3) anyone who gets the right answer and can't prove they did it themselves - sales

What's wrong with being in sales I ask?

Hey it was a compliment! It meant that a) you were persuasive enough to get someone to do it for you; and b) you were smart enough to get the right person to do it for you. You picked the right target and you convinced them. THAT'S sales material.

4) anyone who writes that the test is pointless and they're not going to waste their time working it out - fast-track to management

My first thought was "who cares?" Is that close enough?

Yes. But your inability to phrase those sentiments delicately enough not to offend the underlings that believe such frippery to be worth doing means you'd better make CEO quickly, before the revolution.

 

[RJDiogenes]

I realised that the two questions are exactly the same if you think about the questions as conditional probabilities: p = P(B=5|X). The first question give you X and you need to find p, while the second question gives you p and you need to find X.

which means my answer to the first questions wasn't exactly right. Again, back to Bayes Thereom:

P(B = 5 | X) = P(X | B=5) P(B=5) / P(X)
where
P(X) = sum_B P(X | B=1)P(B=1) + ... + P(X | B=5) P(B=5)
and X = 1 red and 2 black beans.

So
P(X|B=1) = P(X|B=2) = P(X|B=3) = P(X|B=4) ~= 4/100 * 96/100 * 95/100 and
P(X|B=5) ~= 7/100 * 93/100 * 92/100
and
P(B=1) = ... = P(B=5) = 1/5

So after drawing 1 red and 2 black beads from a bag, the probability that you draw from bag 5 is 29.10%, so I was right with the answer but wrong in getting there, but since the numbers were small, it didn't matter.

 

[Roger Wilco]

^^
*saved*

 

[Jadzia]

Probability is easy...

 

[RJDiogenes]

^^
*saved*

Probability is easy...

Ah, another Sidney Harris fan. This is my all-time favorite: