Quantitative Interview questions and answers

Hi nsbzroro

I reconsider my answers.

1- The milk, water cup is the same, as you said.
2- the mean time to get back to A in the square problem, you are right as well.

Regards
 
You are correct, thanks Bob. Fixed the answer. :)

I get rid of ln on the left side but forgot about ln on the right side. If you can, please take a look at other questions as well.

I have a question regarding this. Sorry I don't know how to embed the Math properly.

If I have x^x^x^... = 4
Then by your logic, x = (4)^(1/4) = sqrt(2) also.

So x^x^x^... = 4 = 2?
How do I even know that the infinite power thing even converges?

Thanks!
 
I have a question regarding this. Sorry I don't know how to embed the Math properly.
Use ( tags
If I have x^x^x^... = 4
Then by your logic, x = (4)^(1/4) = sqrt(2) also.
So x^x^x^... = 4 = 2?
It's not the conclusion you should make. The (x^{x^{x^{...}}}=2) and (x^{x^{x^{...}}}=4) are two different series. What you did is to put one on top of another. If Y =2 then Y^Y=4.
How do I even know that the infinite power thing even converges?
What if they don't converge like in this case. If you take (sqrt{2}) and take power of itself forever, the series will only grow, not diverge.
 
new question

Hey guys,

so, if I have x% chance to die in an auto accident from NYC to Philly (say distance d1) what is the chance that I die in an auto accident from NYC to, say, Oregon (say distance d2)?

rio
 

Yuriy

MFE Alum
so, if I have x% chance to die in an auto accident from NYC to Philly (say distance d1) what is the chance that I die in an auto accident from NYC to, say, Oregon (say distance d2)?

We don't know anything about these routes :) maybe one is via a quiet road and the other via a busy highway :)
 
Two fun ones that are both pretty doable:

1. An airplane has N seats, and N passengers are waiting to board it, not in any particular order. Miraculously, everyone is assigned to a different seat on the airplane; however, the first passenger to board is a jerk and selects a seat at random. Thereafter, passengers board one at a time according to the following rule: If his or her assigned seat is vacant, the passenger sits there; otherwise, the passenger selects a vacant seat at random.

What's the probability that the last passenger to board gets his or her assigned seat?

2. We have two concentric circles. A chord of the larger circle is tangent to the smaller circle and has length 8. What's the area of the annulus--the region between the two circles?

1. The answer should be 0.5.
If the jerk picks his own seat, then the last passenger clearly gets his. If the jerk picks the last passenger's seat, the last passenger clearly can't.
If the jerk picks any other seat, say that of the Mth passenger, we can shrink the problem down to an airplane with (N-M) seats with (N-M) people, because everyone from the jerk to M-1 gets their own seat.
If there are only 2 seats, the probability is clearly 0.5, and we can shrink the N passenger problem down to this and shown above. There might be off-by-one errors in my argument though.
So the answer should be 0.5.

2. The answer is (16\pi). The inner circle's radius is perpindicular (sp?) to the tangent, so by the Pythagorean Theorem, (R_1^2-R_2^2 = 16).
R1 is the outer circle's radius, and R2 is the inner circles.
The area of the annulus is then 16*pi.

I love this thread! Thanks guys! Sorry about my English, not my native language.
 
Use ( tags

It's not the conclusion you should make. The (x^{x^{x^{...}}}=2) and (x^{x^{x^{...}}}=4) are two different series. What you did is to put one on top of another. If Y =2 then Y^Y=4.

What if they don't converge like in this case. If you take (sqrt{2}) and take power of itself forever, the series will only grow, not diverge.

I'm treating (sqrt{2}) to the power of itself infinitely many times as the limit of
(sqrt{2})
(sqrt{2}^sqrt{2})
(sqrt{2}^sqrt{2}^sqrt{2})
...

What does this sequence equal? I'm not convinced that it equals 2, we can show that it equals 4 depending on how we group them?
Anyone know if theres a way to bound that sequence?

It's a very clever problem, maybe too clever for me :-(
 
4. 3 points are randomly drawn on a circle. What is the probability of them being on the same semi-circle ?

Answer

The probability is 1 since three points always uniquely define a semi-circle. Two points define a diameter (and two semi-circles) and the third point tells us which semi-circle is defined.

Any thought on this ?
 
One more question looking for answer:

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?
--

I think the answer is 1/6 and it has come up only once in the forum. There are three different ways to get this answer.

1. The outcome of the dice and the man telling the truth or not are independant events. Therefore, the probability of the dice showing 6 is 1/6. (this has already been posted).

2. P[dice shows 6] = P[dice shows 6 | man tells the truth] + P[dice shows 6 | man tells a lie]

P[dice shows 6 | man tells the truth] = P[dice shows 6 AND man tells the truth] / P[man tells the truth] = (1/6)*(4/3) / (4/3) = 1/6

P[dice shows 6 | man tells a lie] = 0 because if the man lies and reports a 6, the dice outcome cannot be 6.

3. By listing all possible outcomes, we have the following couples (dice outcome, man report):

If the man tells the truth:
(1,1) (2,2) (3,3) (4,4) (5,5) (6,6)

If the man lies
(1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) (2,3) (2,4) (2,5) (2,6)
(3,2) (3,2) (3,4) (3,5) (3,6)
(4,1) (4,2) (4,3) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5)

There are 6 couples where the man reports 6 and only one of them where the dice actually shows 6, hence 1/6.
 
Solution ?

4. 3 points are randomly drawn on a circle. What is the probability of them being on the same semi-circle ?

Answer

The probability is 1 since three points always uniquely define a semi-circle. Two points define a diameter (and two semi-circles) and the third point tells us which semi-circle is defined.

Any thought on this ?

I think the question is about 3 random points on a GIVEN circle and not about any circle.

C++:
                       length of semi circle          (1/2)
I think the answer is = --------------------------- =   --------- = 3/4
                        length of largest arc           (2/3)
Now how did I arrive at 2/3 ?
You will get the longest arc when all 3 points are equidistant and it is 2/3 of circumference. In any other arrangement, you will be able to draw an arc that is less than 2/3s by picking the one of the other 2 points as a starting point.
 
since (x^{x^{\dots}} = 2), the exponent part can be replaced by 2 (i.e using the same equation)
(x^2 = 2)
so (x = \sqrt{2})


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})
 
Hey guys,

so, if I have x% chance to die in an auto accident from NYC to Philly (say distance d1) what is the chance that I die in an auto accident from NYC to, say, Oregon (say distance d2)?

rio

The answer should be x% since both numerator and denominator should increase in the same ratio
 
A STRATEGY PROBLEM

There are three urns A, B and U. Urn A is initially empty. Urn B contains two identical balls, one white, the other black. Urn U contains 2280 balls, all identical except for color. Of the balls in urn U, 1000 are white and 1280 are black. From urn U two balls are randomly drawn, away from our sight, and placed inside urn A. You cannot see the contents of urn A.

You are invited to play the following game that involves two stages. In the first stage you choose either urn A or urn B to randomly draw a single ball from. You may not choose both urns. If the ball drawn is white, then you wind $55,000 and if black, you win nothing and lose nothing. The drawn ball is returned to the urn from which it was picked. In the second stage, you are again to choose either urn A or urn B to randomly draw a single ball from. You many not choose both urns. If the ball drawn is white, then you win $55,000 and if black, you win nothing and lose nothing.

What strategy do you choose to play this game with if your goal is to maximize your winnings? In the first stage, which of the urns A or B do you choose to randomly draw a single ball from?

A strategy aims at a set of goals and actions and involves conscious choices of sequential decisions wherein later decisions are influenced by prior results.
 

Bastian Gross

German Mathquant
Without any utility function I would prefer a "B-B" strategy.
 
First step is to compute the probability of picking a white ball in A and B independently.
A: 1 - (1280 * 1279)/ (2280 * 2279) = 0.684 /* (1- P(Black,Black)) */
B : 1/2


So I will start with A
- if it turns out white, you win first turn
With one white ball in A, it eliminates the case of two black balls. So, for next turn, probability of picking a white ball is now more than 1/2. So A is obviously a better choice over B
- if it turns out black, you loose first turn
With one black ball in A, the probability that the second ball is white is less than 1/2. So, B is better choice as the probability is 1/2
 
Re: Strategy

Hi Bastian Gross,

Given that the problem has stated the goal of maximizing the winnings, do you need a utility function?

BTW, there are no costs or losses involved. So, why be concerned with utility?
 
Re: Strategy

Hi satyag,

I am not sure why you choose urn A in the first stage.

Isn't the probability of drawing a white ball from urn A in the first stage LESS than one-half? Even WITHOUT calculation, the probability must be less than .5 given that you start with more black balls in urn U than white balls. So, obviously, urn A would have a greater chance of producing black balls than white balls. Anyway, your number(s) are(is) incorrect.
 
when you pick 2 balls from U, it will be one of the following sequence

White, White
White, Black
Black, White
Black, Black

Probability of getting ATLEAST one white ball = 1 - probability of two black balls
= 1 - P(ball1=black) * P(ball2=black)
= 1 - (1280/2280) * (1279/2279)
= 0.684

So probability that it has ATLEAST one white ball is 0.684 as I mentioned before. If we are talking about a case where A endsup with EXACTLY one white ball, you are correct that the probability is less than 0.5.
Note that this is really the key difference between A and B. In B, we know that there is exactly one white ball. But in A, there is also a chance that both balls are white.
 
Re: Strategy

Hi satyag,

The issue is which urn you choose for the first stage. Isn't it obvious that the probability of drawing a white ball from urn A in the first stage is less than one-half? First address the question just asked.

Just as the probability for urn A to have at least one white ball is .684 (or ..685), the probability of urn A to have at least on black ball is .808, much greater than .684. Does this fact matter?

WHAT ABOUT THIS? ----> The fact that two balls were taken from urn U and placed in urn A does not affect the probability of drawing a white ball from urn A. The probability of drawing a white ball from urn A is precisely the same as the probability of drawing a white ball from urn U. Think about it: When you randomly select two balls from urn U and place them in urn A, urn A only acts as a temporary holding place. Now that you randomly choose a ball from urn A you really shouldn't care that this ball comes from urn A, it really comes from urn U. Don't you agree?

That is why the issue of FIRST STAGE - which I keep on emphasizing upon - is important. The probability of drawing a white ball from urn A in the first stage is 1000/2280=.439, less than one-half.

Thank you for your responses. Feel free to contribute. I do appreciate your interest.
 
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