Lets hear some of them you guys have gotten in investment banking interviews...
Plus make sure to see my post <a href="http://www.wallstreetoasis.com/blog/how-i-got-a-770-on-the-gmat-with-minimal-effort-while-working-banking-hours%E2%80%A6">How I got a 770 on the GMAT with minimal effort while working banking... phantombanker's GMAT Study Guide |
A week or two ago there was an article written about how your weight might inversely effect you during the hiring process. During the conversation I dropped some personal information about me losing 130 (now 135 pounds). I was PM’d by somebody asking me to tell my story. Well I have some free... How finance saved and gave me a life |
Hey guys, it’s been some time since I posted regularly on this site. That’s what happens when you start working in finance; you don’t have any time. So why am I posting? It’s because I am seeing an injustice being done to a lot of military guys (and gals) who are discovering they want to... To all Military who want to work in IB/PE/Consulting |
<em>Mod (Andy) note: </em><a href="http://www.wallstreetoasis.com/blog/phantombankers-gmat-study-guide">Here is the link to his study guide</a>
Background: I recently took the GMAT and scored a 770 on my first and only try. Long story short, I studied and took... How I got a 770 on the GMAT with minimal effort while working banking hours… |
There was much wailing and gnashing of teeth at the close on Friday over the perceived lack of performance in the Facebook IPO. The stock priced at $38, opened at $42.50 (after a few hiccups at the NASDORK), and closed at exactly $38 after an effort by the syndicate members akin to the Spartans at... IPO Pricing 101 |
Hey Guys,
Been using this site for a while now and figured it would be beneficial for me to give back a bit. For starters, I thought I would detail my experiences using linkedin and cold emails to land a summer offer.
Coming from a semi-target undergrad b-school, I had my hands full.... How I used LinkedIn to get Interviews and Land Offers |
<a href="http://www.wallstreetoasis.com/blog/what-does-the-labor-participation-rate-say-about-our-society">Back on the topic of what the composition of the labor force says about our society</a>, the New York Times recently put out this <a... Grandpa And Grandma Are Taking Our Jobs |
The Clear Admit Shop currently has <em>free b-school "snapshot" guides</em> <a href="http://clearadmit.shop.studylink.com/index.cfm?onsale=1">available for download here.</a>
The guides aren't terribly comprehensive, but can serve as a... Free B-School Guides! |
There she is again: Diane Sawyer, the primary ABC evening news anchor, with another installment in their interminable series of reports on imports and employment in America. I could just strangle her.
One week their (alleged) reporters go to people’s houses and look at everything the occupants... Sawyer’s Syndrome: The ABCs of Trade and Employment |
Fellow Primates,
We are looking for 1-2 students on each campus to help WSO in its sales efforts to student clubs/career centers, and overall promotion at your school both online and on the ground. Below is a description of the position and benefits...thanks in advance for your help! <a... WSO Campus Representative Program |
| User | Silver Bananas |
|---|---|
| TNA | 848 |
| Edmundo Braverman | 760 |
| CompBanker | 747 |
| happypantsmcgee | 654 |
| IlliniProgrammer | 552 |
| UFOinsider | 469 |
| Nouveau Richie | 377 |
| Midas Mulligan Magoo | 369 |
| TheKing | 342 |
| cphbravo96 | 339 |
| User | Banana Points |
|---|---|
| TNA | 11678 |
| Edmundo Braverman | 11075 |
| WallStreetOasis.com | 10284 |
| happypantsmcgee | 9352 |
| UFOinsider | 8209 |
| CompBanker | 7797 |
| IlliniProgrammer | 7695 |
| Midas Mulligan Magoo | 5250 |
| monty09 | 5152 |
| cphbravo96 | 4434 |
What to have some fun? Try
What to have some fun? Try these two.
1. let's say A keep tossing a fair coin, until he get 2 consecutive heads, define X to be the number of tosses for this process; B keep tossing another fair coin, until he get 3 consecutive heads, define Y to be the number of the tosses for this process.
Calculate P{X>Y}
2. On an island there are snakes of 3 different colors. blue, red, yellow. Every time 2 snakes of different colors met, they both change their color to the 3rd color. eg, if a red snake meet a yellow snake, they both change to blue.
Now if we know at a certain time, there are 13 blue, 15 red, and 17 yellow snakes, question: could the snakes meet so that eventually all snakes become
same color?
Isn't the answer to number 2
Isn't the answer to number 2 no because there is an odd number of snakes? Unless there is a way to get it so there are 15 of each, which i can't figure out.
here's one
roll a die, and you get paid what the dice shows. if you want, but you don't have to, you can roll the die again and get paid what the second roll shows instead of the first. what is the expected value?
4.25 Another one: given a
4.25
Another one: given a circle, make n random cuts, what is expected length of longest piece?
I believe the answer to the
I believe the answer to the snake one is no. My explanation may not make a lot of sense though. Basically, in order for you to get all the snakes a single color you need to get the other 2 sets of snakes to be equal. Okay, so you have 13, 15, and 17 snakes. Every time you make a change, 2 numbers go down by 1 and one number goes up by 2. This means that you need 1 type of snake to be exactly 3 less than another in order to equal them out.
Right now the snakes are all 2 (or 4) apart in quantity. There is no change or set of changes you can do to make 2 colors seperated by 3. This is due to each mutation either closing the gap by 3 or keeping them equal. Hence, impossible.
Not the best formatted answer but that is my answer.
CompBanker
do they let you use paper or a calculator for these?
do they let you use paper or a calculator for these?
The Expected Value of Rolling Once and Twice ,,
I don't know why this called "Hardest Probability Questions" ..
I assume it is a kind of Joke ..
Lets Solve first Rolling Once ..\
The expected value of a random variable X is denoted by E(X). For a discrete random variable, E(X) is calculated as
Rolling a "1" has a probability of 1/6.
Rolling a "2" has a probability of 1/6.
Rolling a "3" has a probability of 1/6.
Rolling a "4" has a probability of 1/6.
Rolling a "5" has a probability of 1/6.
Rolling a "6" has a probability of 1/6.
Multiplying the values with their respective probability gives:
1 * 1/6 = 1/6
2 * 1/6 = 2/6
3 * 1/6 = 3/6
4 * 1/6 = 4/6
5 * 1/6 = 5/6
6 * 1/6 = 6/6
Adding them together gives:
1/6 + 2/6 + 3/6 + 4/6 + 5/6 + 6/6 = 3.5
Now solvin Rolling the Dice Twice ,,
EX. The random variable X has the following probability distribution: x pX(x)
2 1 / 36
3 2 / 36
4 3 / 36
5 4 / 36
6 5 / 36
7 6 / 36
8 5 / 36
9 4 / 36
10 3 / 36
11 2 / 36
12 1 / 36
The random variable X assumes a value equal to the sum of two dice rolls. Its expected value is calculated as
= 2(1/36) + 3(2/36) + 4(3/36) + 5(4/36) + 6(5/36) + 7(6/36) + 8(5/36) + 9(4/36) + 10(3/36) + 11(2/36) + 12(1/36)
= (1/36) (2 + 6 + 12 + 20 + 30 + 42 + 40 + 36 + 30 + 22 + 12)
= (252/36) = 7
In the long run, the average value of two dice rolls using regular dice is 7.
Probability of Two Dice Rolls
I think the previous poster forgot one fact that changes things a bit.
The key here is that the problem says "you can roll the die again and get paid what the second roll shows INSTEAD of the first" (caps added). This means that you if you decide to re-roll, you should disregard your first roll and just keep the second one.
The first time you roll the dice, you have two options:
- You keep you score
- You re-roll the die
Since the expected value (EV) of one dice roll is 3.5, it makes sense to keep your score if it is higher than 3.5, and re-roll if you get a score less than 3.5.
If you roll over 3.5, then the score you got must have been either 4, 5, or 6. Of these three numbers, the EV is 5.
The other possibility (which is equally likely) is that you get 1, 2, or 3. Then, it makes sense to re-roll. Again, on your second roll, your EV is 3.5.
Since you have two options, which are equally likely, the EV is:
1/2(the chance you don't reroll) + 1/2(the chance you re-roll and ignore your first score)
= 1/2(5) + 1/2(3.5)
= 1/2(8.5)
= 4.25
Note: If you had to keep the first roll and add it to your second score, it would always make sense to re-roll. Then, as the last poster mentioned, the EV would be 3.5+3.5 = 7.
- Claudia
First coin tossing problem
I think that P(X>Y) is 1/4. Here's why:
the odds that you get two straight heads on your 2nd roll is 1/4
the odds that you get two straight heads on your 3rd roll is (1/4)*(1/2)... 1/2 is the odds that first roll is tails
the odds that you get two straight heads on your 4rd roll is (1/4)*(1/2)*(1/2)
Getting the two straight heads by roll 3 or roll 4 has the same likelihood as getting 3 straight (1/8 and (1/8)*(1/2)). The only difference between X and Y is that Y can't happen in 2 rolls, whereas X has a 1/4 chance
RE: First Coin Tossing Problem
the odds that you get two straight heads on your 2nd roll is 1/4
the odds that you get two straight heads on your 3rd roll is (1/4)*(1/2)... 1/2 is the odds that first roll is tails
the odds that you get two straight heads on your 4rd roll is (1/4)*(1/2)*(1/2)
Downtown, you're overlooking the fact that there are two ways to create the event X=4
You can have the following for X=4 (where T is tails and H is heads)
T T H H
H T H H
The actual probabiliy function of X=x is:
f(x) = F_(x-1) * (1/2)^x, where F_(x-1) represents the Fibonnaci sequence and x=2,3,4,....
Derivation of f(x)
If you write it out, you will discover that there's
1 way to make X=3
2 ways to make X=4
3 ways to make X=5
5 ways to make X=6
8 ways to make X=7
The probability of two heads is (1/4) or equivalently, (1/2)^2
The probability of anything before that is (1/2)^(x-2).
Therefore f(x) = F_(x-1)*(1/2)^(x-2)*(1/2)^2
= F_(x-1)*(1/2)^x
If you check out the Wikipedia article on Fibonnaci Power Series, you can show that the sum of f(x) from x=2 to x=infinity does equal 1, and thus f(x) is a proper probability mass function.
Regarding event Y...
Y is a little bit more dfficult. Since X and Y are independent, the P(X>Y | Y=y) = P(X > y) = sum of f(x) from x=y+1 to x=infinity or equivalently, 1 - f(2) - f(3) ... f(y).
To show the P(X>Y) for all Y, we must calculate the infinite sum:
P(Y=3)*P(X>3)+P(Y=4)*P(X>4)+P(Y=5)*P(X>5)...
The PMF of Y=y is:
g(x)=G_(x-2)*(1/2)^x for x=3,4,5,6,...
where G_x= G_(x-1)+G_(x-2)+G_(x-3) with the seed G_0=0, G_1=1, and G_2=1
If you write it out, there's
1 way to make Y=4
2 ways to make Y=5
4 ways to make Y=6
7 ways to make Y=7
13 ways to make Y=8
.... and I don't know how to handle that sequence...
So.... that one problem probably takes the cake for hardest probability problem... perhaps there's some elegant solution to the problem
yes one or two of the
yes one or two of the previous explanations to the snake problem are correct. The snakes cannot all become the same colour simply because of the relative difference between sankes gaiven being divisible by two. By inspection , only a difference between any two snake numbers that is divisible by 3 will work . Hence the answer is no!
IdrisAlMalki wrote: I don't
I don't know why this called "Hardest Probability Questions" ..
I assume it is a kind of Joke ..
The random variable X assumes a value equal to the sum of two dice rolls. Its expected value is calculated as
= 2(1/36) + 3(2/36) + 4(3/36) + 5(4/36) + 6(5/36) + 7(6/36) + 8(5/36) + 9(4/36) + 10(3/36) + 11(2/36) + 12(1/36)
= (1/36) (2 + 6 + 12 + 20 + 30 + 42 + 40 + 36 + 30 + 22 + 12)
= (252/36) = 7
In the long run, the average value of two dice rolls using regular dice is 7.
dude, just add the expected value of two independent variables and you get 7. no need for all that calculations.
Very true. The expectation
Very true. The expectation operator is a linear one.
Prove that no three positive
Prove that no three positive integers a, b, and c can satisfy the equations (a^n)+(b^n) = (c^n) for any integer value of n greater than two.
balbasur wrote: Prove that no
Prove that no three positive integers a, b, and c can satisfy the equations (a^n)+(b^n) = (c^n) for any integer value of n greater than two.
sorry, i don't think andrew wiles posts here anymore.
Art Vandelay wrote: balbasur
Prove that no three positive integers a, b, and c can satisfy the equations (a^n)+(b^n) = (c^n) for any integer value of n greater than two.
sorry, i don't think andrew wiles posts here anymore.
never know <3
and if he did, no one here
and if he did, no one here would understand the proof anyway.
The answer to the snake
The answer to the snake problem is YES! They all can become one color, to proof this it is sufficient to show that there exist one solution where all snakes are one color, e.g.. one red and one yellow snake meet, this will lead to 14 blue, 14 red, and 16 yellow snakes. Now the 14 blue and 14 red snakes are meeting leading to all snakes being yellow. QED
the answer to the snake
the answer to the snake question is YES, agree with ralph! Coin question is 4.25, and dice quesiton is 3.5
balbasur wrote: Prove that no
Prove that no three positive integers a, b, and c can satisfy the equations (a^n)+(b^n) = (c^n) for any integer value of n greater than two.
Haha, we did that in IB HL math (Grade 12)
ralph240574 wrote: The answer
The answer to the snake problem is YES! They all can become one color, to proof this it is sufficient to show that there exist one solution where all snakes are one color, e.g.. one red and one yellow snake meet, this will lead to 14 blue, 14 red, and 16 yellow snakes. Now the 14 blue and 14 red snakes are meeting leading to all snakes being yellow. QED
Nope. One red and one yellow snake meeting will lead to 15 blue, not 14 blue. Compbanker was right earlier. You need a difference of 3 between 2 colours to make this possible.
HopefulIBGuy wrote: balbasur
Prove that no three positive integers a, b, and c can satisfy the equations (a^n)+(b^n) = (c^n) for any integer value of n greater than two.
Haha, we did that in IB HL math (Grade 12)
Sure, you solved this in grade 12! This is why it took more than 3 centuries to find a proof to this simple problem. It was only solved in 1994 by Andrew Wiles and Richard Taylor. (@ Art Vandelay: I'm not sure everyone got your post ;)
Read "Fermat's last theorem" by Simon Singh. Highly recommendable book!
Or have a quick look into wikipedia: http://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem
1. let's say A keep tossing a
1. let's say A keep tossing a fair coin, until he get 2 consecutive heads, define X to be the number of tosses for this process; B keep tossing another fair coin, until he get 3 consecutive heads, define Y to be the number of the tosses for this process.
Calculate P{X>Y}
Solution: 3/32
Problem can be thought of as "what is the probability that B gets 3 consecutive heads and A doesn't get 2 consecutive heads". Doesn't matter when A gets his heads as long as B gets it first.
This covers all cases because this defines when X>Y, and it becomes an easy problem.
P{A doesn't gets 2 consecutive heads} = 1 - P(A gets 2 consecutive heads} = 1 - (1/2)*(1/2) = 3/4.
P{B gets 3 consecutive heads} = (1/2)*(1/2)*(1/2) = (1/8).
P{both occuring} = (3/4) * (1/8) = 3/32.
*edit* nvm i think this is wrong
what does a 747 weigh but
what does a 747 weigh
but that was a recent consulting interview
Solution to problem 1 can be
Solution to problem 1 can be found here: http://pratikpoddarcse.blogspot.com/2009/10/lets-say-keep-tossing-fair-c...
What to have some fun? Try these two.
1. let's say A keep tossing a fair coin, until he get 2 consecutive heads, define X to be the number of tosses for this process; B keep tossing another fair coin, until he get 3 consecutive heads, define Y to be the number of the tosses for this process.
Calculate P{X>Y}
2. On an island there are snakes of 3 different colors. blue, red, yellow. Every time 2 snakes of different colors met, they both change their color to the 3rd color. eg, if a red snake meet a yellow snake, they both change to blue.
Now if we know at a certain time, there are 13 blue, 15 red, and 17 yellow snakes, question: could the snakes meet so that eventually all snakes become
same color?
Anyone figure out number 1
Anyone figure out number 1 yet? I don't get how to do it and it's driving me crazy.
What to have some fun? Try these two.
1. let's say A keep tossing a fair coin, until he get 2 consecutive heads, define X to be the number of tosses for this process; B keep tossing another fair coin, until he get 3 consecutive heads, define Y to be the number of the tosses for this process.
Calculate P{X>Y}
2. On an island there are snakes of 3 different colors. blue, red, yellow. Every time 2 snakes of different colors met, they both change their color to the 3rd color. eg, if a red snake meet a yellow snake, they both change to blue.
Now if we know at a certain time, there are 13 blue, 15 red, and 17 yellow snakes, question: could the snakes meet so that eventually all snakes become
same color?
Solution to Problem
Solution to Problem 1:
http://pratikpoddarcse.blogspot.com/2009/10/lets-say-keep-tossing-fair-c...
Best of Luck!
Anyone figure out number 1 yet? I don't get how to do it and it's driving me crazy.
What to have some fun? Try these two.
1. let's say A keep tossing a fair coin, until he get 2 consecutive heads, define X to be the number of tosses for this process; B keep tossing another fair coin, until he get 3 consecutive heads, define Y to be the number of the tosses for this process.
Calculate P{X>Y}
2. On an island there are snakes of 3 different colors. blue, red, yellow. Every time 2 snakes of different colors met, they both change their color to the 3rd color. eg, if a red snake meet a yellow snake, they both change to blue.
Now if we know at a certain time, there are 13 blue, 15 red, and 17 yellow snakes, question: could the snakes meet so that eventually all snakes become
same color?