Quantitative Interview questions and answers

[Karth]

A question posed previously was:

A man speaks the truth 3 out of 4 times. He throws a die and reports it to be a 6. What is the probability of it being a 6?

Here's my solution:
This is just a conditional probability problem. We are expected to compute P[really 6 | said 6]. Using Bayes's rule, this is just P[said 6 | really 6] / { P[said 6 | really 6] * P[really 6] + P[said 6 | not really 6] * P[not really 6] }. We know each of the probabilities here, so just substitute to get:
3/4 * 1/6 / (3/4 * 1/6 + 1/4 * 5/6) = 1/8 / (1/8 + 5/24) = 3/8

 

[Lyn Huang]

P[said 6 | really 6] / { P[said 6 | really 6] * P[really 6] + P[said 6 | not really 6] * P[not really 6] }.

The first term should be P[said 6, really 6], typo?

P[said 6 | not really 6] * P[not really 6]

This is not 1/4 * 5/6
should be (1/4 * 1/5) * 5/6 since the man has 5 choices not counting the real outcome.

 

[Karth]

The first term should be P[said 6, really 6], typo?


This is not 1/4 * 5/6
should be (1/4 * 1/5) * 5/6 since the man has 5 choices not counting the real outcome.

Oops, yes, typo in the numerator, but I think that I typed the substitution in correctly... this should be P[said 6 | really 6] * P[really 6] and not just P[said 6 | really 6].

I'm not sure if I follow your reasoning for the latter component. Could you explain this a bit more? Looking at P[said 6 | not really 6] * P[not really 6], I interpret P[said 6 | not really 6] as "the man lies," which yields a probability of 3/4. I think that we agree that P[not really 6] is 5/6 though.

 

[Lyn Huang]

A man speaks the truth 3 out of 4 times.

So P(lie) = 1/4
If the outcome is not 6, say it's 1, he lies by saying it's 2, 3, 4, 5, or 6. The probability he picks 6 should be 1/5 then

 

[Karth]

Sorry, I still don't understand Just to clarify, we're trying to compute P[said 6 | dice roll is not 6], right? Shouldn't this just be 1/4? One way to interpret this is to just say that the guy lies, and as such, the probability must be 1/4. Another way is to do enumeration. So...

P[said 6 | not really 6] =
P[said 6, not really 6] / P[not really 6] =
1 / P[not really 6] * { P[said 6, really 1] + P[said 6, really 2] + ... + P[said 6, really 5] } =
1 / P[not really 6] * { P[said 6|really 1] * P[really 1] + P[said 6|really 2] * P[really 2] + ... + P[said 6|really 5] * P[really 5] } =
1 / (5/6) * (1/4 * 1/6 * 5) = 6/5 * 1/4 * 1/6 * 5 = 1/4.

 

[Lyn Huang]

P[said 6|really 1] * P[really 1]

Following your logic this is 1/4 * 1/6, right?
What I meant is that it should be 1/4 * 1/5 * 1/6 since when the real outcome is 1 he can lie to you by telling it's any number from 2-6.

 

[Karth]

To clarify, I'm denoting "said 6" to be the event that the main claims that the number which appears is 6. I'm denoting "really 1" to be the event that the number that appears on the die is really in fact 1 (irrespective of what the man says). If that's the case, then I think that P[said 6 | really 1] is 1/4, since this amounts to the man telling a lie. For P[really 1], this is just 1/6 by simple probability. I'm still not sure where the 1/5 comes in.

 

[Lyn Huang]

He tells a lie in five possible ways. That's why I plug in 1/5.

 

[Karth]

There are five ways to tell a lie, but I don't think that this fact manifests itself in the problem.

 

[stephan acay]

No, the answer to this is \(x=\sqrt{2}\)

\(a=x^{x^{\ldots}}=2\)

\(ln(x^{x^{\ldots}}) = ln2\)

\(x^{x^{\ldots}}lnx = ln2\)

\(a lnx = ln2\)

\(2lnx = ln2\)

\(lnx = \frac{1}{2}ln2\)

\(lnx = ln\sqrt{2}\)

\(x = \sqrt{2}\)

There is alot easier way of doing this.

x^x^x^...=2

x^2=2
x=sqrt{2}

 

[stephan acay]

Q6

Answer is 2

 

[stephan acay]

I see that everybody is posting a question.

So this is from me for geometry lovers.

 

[Karth]

I see a way to do this problem with trig, but is there a pure geometric solution to doing this?

 

[stephan acay]

I see a way to do this problem with trig, but is there a pure geometric solution to doing this?

Yes. It can easly be solved with trig. But there is a pure geometric solution. And thats the challenging part.

 

[koupparis]

Got 30 using trig, stuck on the geometric solution...

 

[chaohan]

There is alot easier way of doing this.

x^x^x^...=2

x^2=2
x=sqrt{2}

I got a counterexample according to this method:

if x^x^x...^x=4, x=sqrt{2} as well.

Can anyone explain this?

 

[Michael M]

There is alot easier way of doing this.

x^x^x^...=2

x^2=2
x=sqrt{2}

This beautiful approach brings much ease to my mind.

 

[stephan acay]

I got a counterexample according to this method:

if x^x^x...^x=4, x=sqrt{2} as well.

Can anyone explain this?

This is a good point. I would be interested in learning this too.

dstefan

 

[Michael M]

And same for "... = 16 " too?

 

[chaohan]

This is a good point. I would be interested in learning this too.

I found the link about this: http://www.had2know.com/academics/power-tower-iterated-exponentiation-tetration.html shows that the result won't be larger than e. Otherwise it goes to infinity. Seems like my counter example is wrong...