Quantitative Interview questions and answers

kean

Mathematics Student
Incorrect

Unfortunately your answer is incorrect. For one thing, I would like to clarify that I do not mean to test anyone in this forum. I just pick some look simple but tricky questions to share with members.

My intention is to help developing probability knowledge. I urge everyone not to use this intention as solving assignment questions. No good.:-ss

I will post the answer three day later. By the way, please show your calculation. Thanks.


I say 11.

11 has .1944 chance whereas 12 has .1666 chance.

am I right?
 
I found a simple way.
For x+y+z=11, the probability is sum of 1/6*P(y+z=a), a is from 5 to 10
For x+y+z=12, the probability is sum of 1/6*P(y+z=a), a is from 6 to 11
The difference: 1/6*[P(y+z=5)-P(y+z=11)]>0
so P(x+y+z=11) > P(x+y+z=12)
 

kean

Mathematics Student
This is the answer

This was a famous question Pascal asked Femat in the 17th century.


A sum of 11 is obtained with the following 6 combinations:
(6, 4, 1) (6, 3, 2) (5, 5, 1) (5, 4, 2) (5, 3, 3) (4, 4, 3).
A sum of 12 is obtained with the following 6 combinations:
(6, 5, 1) (6, 4, 2) (6, 3, 3) (5, 5, 2) (5, 4, 3) (4, 4, 4).

Each combination of 3 distinct numbers corresponds to 6 permutations, while each
combination of 3 numbers, two of which are equal, corresponds to 3 permutations.
Counting the number of permutations in the 6 combinations corresponding to a sum
of 11, we obtain 6 + 6 + 3 + 6 + 3 + 3 = 27 permutations. Counting the number of
permutations in the 6 combinations corresponding to a sum of 12, we obtain 6 + 6 +
3+3+6+1 = 25 permutations. Since all permutations are equally likely, a sum of 11
is more likely than a sum of 12.
Note also that the sample space has 63 = 216 elements, so we have P(11) =
27/216, P(12) = 25/216.:D
 

kean

Mathematics Student
New Quiz

Call the throw of a pair of dice lucky if the sum is 7 or 11.

Two players each toss a pair of dice (independently of one another) until each makes a lucky throw.

Find the probability that they take the same number of throw.
 

kean

Mathematics Student
correction

The sample space is 6^3.


This was a famous question Pascal asked Femat in the 17th century.


A sum of 11 is obtained with the following 6 combinations:
(6, 4, 1) (6, 3, 2) (5, 5, 1) (5, 4, 2) (5, 3, 3) (4, 4, 3).
A sum of 12 is obtained with the following 6 combinations:
(6, 5, 1) (6, 4, 2) (6, 3, 3) (5, 5, 2) (5, 4, 3) (4, 4, 4).

Each combination of 3 distinct numbers corresponds to 6 permutations, while each
combination of 3 numbers, two of which are equal, corresponds to 3 permutations.
Counting the number of permutations in the 6 combinations corresponding to a sum
of 11, we obtain 6 + 6 + 3 + 6 + 3 + 3 = 27 permutations. Counting the number of
permutations in the 6 combinations corresponding to a sum of 12, we obtain 6 + 6 +
3+3+6+1 = 25 permutations. Since all permutations are equally likely, a sum of 11
is more likely than a sum of 12.
Note also that the sample space has 63 = 216 elements, so we have P(11) =
27/216, P(12) = 25/216.:D
 
I was just reading the first post and got different answers for these two questions. Feel free to comment :)

3. The hour and minute hands of a clock meet at 12'oclock. When will be the first time they meet again ?
Answer
: After 1 hour, 5 minutes 27.2727 seconds

I got 1hour, 5 minutes, 5.0847 seconds...

12 1 2
+----------+----------+
min hour
hand hand

can be interpreted as


0 5 10
+----------+----------+
min hour
hand hand

at 1pm the minute hand is at 12 position, the hour hand is at 1 position.
The minute hand travels at 1 per minute
The hour hand travels at 1/60 per minute

at intersection, 5 + t/60 = t
Hence t=300/59= 5 min 5.0847... sec


4. 3 points are randomly drawn on a circle. What is the probability of them being on the same semi-circle ?
Answer
: 1/2 or 50%

There are 2 possible outcomes to this set up. But I found some are more likely than others:

Say the circle is divided into A and B and the points P1, P2, P3. Area A = Area B hence prob(each point in each semi circle) = 0.5

Hence if you draw a probability tree, you'd find the probability is 1/4.

6. Calculate \(\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\ldots}}}} \)
Answer: 2

Let a= \(\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\ldots}}}} \)
ln(a)
= 1/2 x ln2 + 1/4 x ln 2 + 1/8 x ln2 + ...
=ln2 (1/(1-0.5))
=> a=4
 
I was just reading the first post and got different answers for these two questions.
The correct answers are discussed in later posts. I haven't updated my first posted for a while.
Let a= (\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\ldots}}}} )
ln(a)
= 1/2 x ln2 + 1/4 x ln 2 + 1/8 x ln2 + ...
=ln2 (1/(1-0.5))
=> a=4
Check your calculation. It should be
(\ln a = \frac{1}{2} \ln (2 + a))
(a^2=2+a \Rightarrow a =2)
There are easier way to do this without using log but the idea is the same.
 
Consider first the circumstance where the first point--(x_1)--is on the interval ((0,\frac{1}{2})). Where can we place the second point in order to make the desired result happen?

It's fairly clear that the interval ((\frac{1}{2}, x_1 + \frac{1}{2})) is the answer to that question. This interval has length (x_1).

I don't quite agree.

If (x_1) is the length of the first portion broken off, and (x_2), the length of the second portion, to form a triangle, we need (x_1<0.5)and (0.5<x_1+x_2><1)
i.e.</x_1+x_2>(x_2) between (0.5-x_1) and (1-x_1)<x_1+x_2><x_2><x_2><x_2>

(\int_0^{\frac{1}{2}}{\int_{\frac{1}{2}-x_1}^{1-x_1}\frac{x_1}{1}*\frac{x_2}{1-x_1}dx_2 dx_1)

Integrating, I got 0.2017
</x_2></x_2></x_2></x_1+x_2>
 
The correct answers are discussed in later posts. I haven't updated my first posted for a while.

Check your calculation. It should be
(\ln a = \frac{1}{2} \ln (2 + a))
(a^2=2+a \Rightarrow a =2)
There are easier way to do this without using log but the idea is the same.

*Bangs head*


Thanks :)
 

DominiConnor

Quant Headhunter
There is a large pile of Brainteaser questions at Wilmott Forums - Brainteaser Forum
They're set (and often answered) by finance types.
One very large bank has told me they've given up trying to avoid asking the ones to be found there, since their views is that anyone smart enough to be able to do that range of BTs is someone they want to hire.
 
JP Morgan interview question for a summer analyst position.

"Imagine you are in a room. There is a door, but in order to open it, you have to insert a table-tennis ball into the hole in the door. There is a table-tennis ball in the room, but it’s in a very deep hole in the ground so that you cannot reach it. Given a feather and an empty box of Kellogg’s corn flakes, how would you get out of the room?"
 

Yuriy

MFE Alum
Put the feather in the box of corn flakes, say magic words, take it out and touch the ball, the ball will attract to the feather :)
 
King has 1000 bottles of wine. One of the bottles is poisoned. He is using subjects to find out which bottle is poisoned. Objective is to minimize the number of people you use. The rule is you can't let some people drink, wait and let others drink. Everyone will drink at the same time...What is the optimal strategy to minimize the number of people used while finding the poisoned bottle?
 
"Imagine you are in a room. There is a door, but in order to open it, you have to insert a table-tennis ball into the hole in the door. There is a table-tennis ball in the room, but it's in a very deep hole in the ground so that you cannot reach it. Given a feather and an empty box of Kellogg's corn flakes, how would you get out of the room?"


well I am not sure, but where does the question say that I was in the room with a ping pong ball -- i am infact in my own room fighting with Ito ;)
 
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