Quantitative Interview questions and answers

[Puzzles]

does anyone have answers for #s 1,2, 4 and 5 in the original post? The OP didn't give them. I got:

1.sqrt(2),
2. 2e^(n2pi/6) where n is an integer from 0 to 5
4. 3/4
5. 1/8

 

[pruse]

does anyone have answers for #s 1,2, 4 and 5 in the original post? The OP didn't give them. I got:

1.sqrt(2),
2. 2e^(n2pi/6) where n is an integer from 0 to 5
4. 3/4
5. 1/8

1. correct.
2. you're missing an i (sqrt(-1)) in the exponent.
4. correct.
5. no. it's 1/4.

Note that 4 and 5 are effectively the same question.

 

[Puzzles]

5. no. it's 1/4.

Note that 4 and 5 are effectively the same question.

ok. i found the mistake in my work, I do get 1/4 for that one.

I actually didn't realize that 4 and 5 were the same thing. It's not obvious to me why that is, could you explain it?
edit: nvm figured it out
edit#2: Thanks for the answer

 

[pratikpoddar]

Generalized Semicircle Covering Points Problem

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

A nice generalization of this problem can be found here
If (n) points are drawn randomly on a circle, the probability of them being on the same semi-circle is (\frac{n}{2^{n-1}})

 

[Joy Pathak]

If person A flips 99 fair coins and obtains heads x times. Person B flips 100 fair coins and obtains heads y times. What is the probability that x < y?

 

[pruse]

If person A flips 99 fair coins and obtains heads x times. Person B flips 100 fair coins and obtains heads y times. What is the probability that x < y?

(\frac{1}{2})

For each sequence of tosses (S), let (S') be the sequence with heads replaced by tails and tails replaced by heads. Then there is a 1-1 correspondence

((A,B)\leftrightarrow(A',B'))

in which if one side has the property (x<y), then the other side has (x\geq y), showing that the two have equal probability.

 

[iniesta]

If person A flips 99 fair coins and obtains heads x times. Person B flips 100 fair coins and obtains heads y times. What is the probability that x < y?

not sure about this but isnt this just the difference of 2 binomial rvs?

 

[pruse]

not sure about this but isnt this just the difference of 2 binomial rvs?

sure, you can view it that way too, but the solution i gave above is more elegant

 

[ThinkDifferent]

This is from a phone interview with a tier 1 bank.

Let A be an n by n square matrix where n is odd. Each column and each row is a permutation of numbers from 1 to n. Prove that diagonal is a permutation of numbers from 1 to n.

 

[ThinkDifferent]

"Andy Nguyen likes this."

...you interviewed with them too, Andy? by the way, I cracked this one

 

[ace]

"Andy Nguyen likes this."

...you interviewed with them too, Andy? by the way, I cracked this one

n=3 -> 3x3 matrix

123 231 312
231 312 123
312 123 231

 

[WilliamKarasz]

This is from a phone interview with a tier 1 bank.

Let A be an n by n square matrix where n is odd. Each column and each row is a permutation of numbers from 1 to n. Prove that diagonal is a permutation of numbers from 1 to n.

New member here. Just wanted to throw in my 2 cents.
This is a misleading question, because the presumed true answer is actually false. Play sudoku lately?

[Edited to add a solved Sudoku puzzle]
6 9 1 5 4 7 2 3 8
3 8 4 2 6 1 9 7 5
5 7 2 3 8 9 4 1 6
8 3 6 9 7 5 1 4 2
7 1 9 6 2 4 5 8 3
2 4 5 8 1 3 6 9 7
9 6 7 1 3 2 8 5 4
4 5 8 7 9 6 3 2 1
1 2 3 4 5 8 7 6 9

 

[pruse]

New member here. Just wanted to throw in my 2 cents.
This is a misleading question, because the presumed true answer is actually false. Play sudoku lately?

[Edited to add a solved Sudoku puzzle]
6 9 1 5 4 7 2 3 8
3 8 4 2 6 1 9 7 5
5 7 2 3 8 9 4 1 6
8 3 6 9 7 5 1 4 2
7 1 9 6 2 4 5 8 3
2 4 5 8 1 3 6 9 7
9 6 7 1 3 2 8 5 4
4 5 8 7 9 6 3 2 1
1 2 3 4 5 8 7 6 9

I think what he's missing from the problem is that the matrix be symmetric (about the main diagonal). Then the main diagonal must indeed be a permutation of 1, ..., n.

 

[ThinkDifferent]

ehhh...matrix is symmetric by assumption! apologies, folks )

 

[pruse]

you sent 'em all on a wild goose chase

 

[ThinkDifferent]

well, I finally saw sudoku skills being useful in math Let's see how sudoku guy deals with the problem now

 

[iniesta]

basically you assume that there is some duplicate in the diagonal, then come up with an algorithm to fill in the matrix, eventually you'll run into a contradiction, ie. one of the columns/rows will have duplicates

 

[ThinkDifferent]

could you clarify the "an algorithm to fill in the matrix" part?

 

[iniesta]

could you clarify the "an algorithm to fill in the matrix" part?

well, just obey the constraints for as long as possible. ie fill the first column/row, second column/row, and so on

 

[ThinkDifferent]

Iniesta, could you be more specific? i.e. give a precise proof.
I have a different solution, so curious to see other way.