Quantitative Interview questions and answers

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
 
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.
 
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
 
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

Swaptionz
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?
 
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.
 
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.
 
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
 
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.
 
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
 
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