# Interview Questions

interview questions shared by candidates

## Interview Questions

I was asked about XSS, SQL Injection, Tools I have used for pen testing. 1 AnswerCross-site scripting (XSS) is a type of computer security vulnerability typically found in Web applications. XSS enables attackers to inject client-side script into Web pages viewed by other users. A cross-site scripting vulnerability may be used by attackers to bypass access controls such as the same origin policy. - SQL injection is a code injection technique, used to attack data-driven applications, in which malicious SQL statements are inserted into an entry field for execution (e.g. to dump the database contents to the attacker). |

### Business Analyst at Capital One was asked...

You have 3000 bananas at point A, which is 1000 feet removed from point B. You must move as many bananas to point B, but you can only carry 1000 bananas at any time, and traveling 1 feet requires you to eat 1 banana. You can drop off bananas at any point between A and B, and pick them up later. 29 Answers500 bananas 500? How? Could you pls explain? Don't worry it...got it Show More Responses Not very clear formulation of problem . I move V (in thousand) bananas on L feets., what I get in point L 1.V-L/V Or 2.(V-L) In first case I spend 1/V moving on 1 feet, but in second energy do not depend of load. Sorry my Error not 1/v but V and not (V-L/V) but (V-V*L) how do you mathematically formulate it? I mean I see how it's 500 from intuition. But, if I were to approach the problem from a purely mathematical way, how would I formulate it? @Shilpa, Could you share what you found? Thanks! It's not 500. Not sure if correct but I calc 666. It should be around 666 bananas. How did you arrive at 666? let us say you move x feet before you drop your bananas and then go back to point A, and let's say you pick 1000 bananas first time. You eat x bananas when you get to point x, and since you need go back to Point A, you will keep another x bananas, then you drop 1000-2x bananas at point X, next time suppose you pick y bananas, when you get to point X, you will eat y-x bananas, and pick former dropped 1000-2x bananas, all these bananas add up to 1000, so we got y-3x+1000=1000, that's y-3x=0. Since we can pick max 1000 bananas, so max y is 1000, x is 333, that is say, we can go as far as 333 feet and then we must go back. since after we get to point b and don't have any more bananas for us to go back to point A, seems 333 is the answer 1000-2x+1000-2x+1000-x=2000 x=200 (first stop point @ 200 feet) 1000-2y+1000-y=1000 y=333 (second stop point @ 533 feet) so you can move 1000-(1000-533)=533 to Point B. Hey ltss587ATgmailDOTcom - I understand how you got to 1000-2x+1000-2x+1000-x (3000-5x) which is effectively number of banas left at Point X. Why did you equate it to 2000? what is the rationale to it? Also how did you get the second equation? 1000-2y+1000-y=1000 Wei - In your equation, though it makes sense, but u didnot go back the third time to pick up the 1000 bananas. At this time, y-3x=0., you still have a 1000 bananas left? Could you please explan. I really appreciate it. Show More Responses Hey G It goes like this 1000-2x+1000-2x+1000-x=2000 ( equate it to 2000 because after this u will wont need to go back third time as the maximum u can carry will be over in two time) x=200 (first stop point @ 200 feet) 1000-2y+1000-y=1000 ( equate it to 1000 because after this u will wont need to go back second time as the maximum u can carry will be over once y=333 (second stop point @ 533 feet) so you can move the remaining 1000 distance in (1000-533)=533 to Point B. So 533 bananas left sorry guys its 833 bananas You have to do it a foot at at time to max out You'll have to carry 3 loads until you eat 1000 bananas so 3x = 1000 or x = 333 ft. next you have to carry 2 loads until you eat the next 1000 bananas so 2x = 1000 or x = 500 ft. Now you have exactly 1000 bananas left so only one load fo the remaining 167 feet this gives you 833 bananas left. It is 333 bananas!!! 833 is the maximum left and best explanation. Do it on a number line. Maximize each load and trip. The second step must be 2 trips closest to the goal ending with 1000 bananas because there be one final trip at maximum load the shortest distance. So stops at 333+500+167 gets the most product moved at least cost. Logistics. Technically this isn't really asking anything. Technically you don't have to eat any bananas. If you move them less than a foot each time, you don't have to eat any bananas. 750! 0 3000, pickup 1000 bananas and walk 1/2 feet, go back get another 1000 and put them @ 1/2 ft location and then get the rest bunch. 833 is correct. Starting at 1000 feet mark. Carry 1000 bananas drop 999 at 999 foot mark. Go back and bring 2*1000 similar way and drop at 999 ft. Now you have 3*999 bananas at 999 feet. Keep doing this until 3*x exceeds 1000. You are now at 666 foot mark with 1998 (3000-334*3) bananas. Now do two trips and drop bananas at each foot traveled. At 177 (833) foot mark you will have 1000 bananas. Now you will loose 177 bananas to end Show More Responses Incomplete question but this is making the presumption that the answer required is 'what is the largest number of bananas I get get to point B'? Te answer is 2 You must eat one banana per foot traveled. Whether forward or backwards, a foot is still a foot, so a banana must be eaten. If I take 1000 bananas and drop 334 of them off after 333 feet, I will have just enough to get back to point A. I grab another 1000. I walk 333 feet, arriving with 667 bananas. I pick up 333 of those I left the first time, leaving one, and walk another 333 feet. I reach the 666 feet mark with 667 in my possession. I leave two there. One my way back, I must pick up the remaining banana from the 333 foot mark in order to get back to point A. (665+1) I pick up the remaking thousand. At 666 feet, I have 334 left. I pick up the 2 that I left there, then eat the 334 necessary for me to reach point B. I have 2 left I grab the remaining 1000. I reach the 666 mark with 334 bananas. I take the one I left there. I walk 334 feet, arriving at point B with a single banana Sorry for the fat fingers in my answers! I ate too many bananas and was feeling ill, lol Second apology. Somehow seems as if my response sequence got messed up. I'll try this again! I must eat a banana whether traveling to or from point A. I take 1000 and walk 333 feet I arrive with 667 and leave 334 of them I get back to point A empty handed and grab another 1000. At 333 feet, I take 333 of those I left the first time, resuming my journey with 1000 bananas. I walk to the 666 feet mark arriving with 667. I leave 2 and head back with 665 in my possession. I pick up the one I left at the 333 point mark and make it back to point A empty handed. The last thousand is enough fuel to get me from A to B, so I arrive at point B with the two bananas I had left at 666 feet. The answer is 2 The answer can be googled, they offer a generalized form. For this particular example it is 533 533 |

### Operations Analyst at Capital One was asked...

Rate yourself on a scale of 1 to 10 how weird you are. 15 Answersjust know the company's culture and if they like uniqueness or not well, do they like uniqueness or not? fake Show More Responses about a 5.6719151431 "Extremely weird" since my scale shows 156 lbs. I am within two standard deviations from the norm... I am a 10. I am a software engineer. We are all very weird. please read this as, "hmmm, how will you answer this question?" My perception of weird and your perception are going to be different, so will the interviewer's. Always consider the audience that you are speaking to and make sure to add some, not a lot, of levity! One number lower than whoever came up with this question Maximus- love that answer- halarious. This is most likely fake, if it was asked it probably was just to amuse the interviewer, not so much of a deciding factor. This is to assess the candidate's perception, witt, and confidence, as well as ability to present a rational response to an "irrational" question. My anwer would be: Weirdness is a subjective topic, based on a scale of (1-10) one would first have to determine what the reference point is. I live life through my own eyes and judge people and events through my own moral compass. Therefore i would be a "1" and measure everything else with me as the reference point. There is no wrong or right answer, this is about confidence in presenting your argument. I would have a different answer of course if the question was about things other than weirdness, such as talent, uniqueness, difference, etc.. Show More Responses my answer would be "1" because normalcy is relative. 5 because im inbetween both a normal one....^_^ |

How to measure 9 minutes using only a 4 minute and 7 minute hourglass 14 Answershttp://www.abc.net.au/science/surfingscientist/pdf/conundrum9.pdf http://www.ronbarnette.com/Zeno/result7.html The key is understanding that you will have to use the two hourglasses together. Since this problem could be asked in many ways using different values for the hourglasses and the total amount of time, it's more important to understand how you use the tools rather than memorize a specific example. The question is used to determine those who can apply their knowledge to solve problems vs. those who memorize answers "from the book". Start both timers. After four minutes, the four-minute timer will have expired and the seven-minute timer will have three minutes remaining. Flip the four minute timer over. After seven minutes, the seven-minute timer will have expired and the four-minute timer will still have one minute left. Flip the seven-minute timer over. After eight minutes, the four-minute timer will have expired for the second time. The seven-minute timer will have accumulated one minute after it's last flip. Flip over the seven-minute timer and when it expires nine minutes will have elapsed. For extra measure, you can always throw in something like, "assuming the timers can be flipped over nearly instantly..." Show More Responses I think you made this much more complicated then it needed to be. Just flip the 7 minute hour glass, when it is done, flip the 4 minute, when you see the four minute is half way done you have 9 minutes, 7 + 2 = 9. Much easier yes?! Start both timers together. When the 4 minute timer is done, flip it. 7 minute timer will have 3 minutes left. When the 7 minute timer is done, the 4 minute timer will have 1 minute left. Now you can count to 9 minutes by simply leaving the 4 minute to expire (1 min), flip it and let it expire (4 min), flip it again and let it expire (4 min). 1 + 4 + 4 = 9 By trying unsuccessfully to show how smart he is, T$ flunked the interview question. The task is to measure nine minutes, not to guess. ultraman: all measurement systems and techniques are have an inherent accuracy and precision (or repeatability and reproduceability), so the important thing is to understand the level of accuracy and precision required, and know which technique will meet the level required. The most important thing in business is not running off to do the task you've been asked with a tool you have handy, but understanding the reason for completing the task, and knowing how the output from your work will be used by others, and how the accuracy and precision will impact of others down the value chain. Dodging the question never works, rd. Tim is correct. http://brainteaserbible.com/interview-brainteaser-hourglass-puzzle Actually, I agree with #T$. Ultraman and d-bag, the goal is to get the job done the most efficient and accurate way. The original question says it has to get done within 9 minutes as well. So you turn the 7 min timer over. While the seven minute timer is going, you take a measurement of the length of the 4 min timer, with the width of your thumb. Whatever number of thumb widths it is, divide it by 2 and that will (very accurately) give you the half way point on the hourglass which will equal to 2 minutes. Once the seven minute hourglass is done, flip the 4 minute timer. Once the 4 minute timer hits the halfway point you marked, you accurately measured 9 minutes. 1st timer 2nd timer time count 4 7...................start both timers 3 6................. 1min 2 5..................2mins 1 4..................3mins 0(flip) 3..................4mins completed 4 3..................4mins(assuming flip takes no time ideally) 3 2..................5mins 2 1..................6mins 1 0(flip)..........7mins 1 7..................7mins(again ideal flip) 0 6..................8mins(flip 2nd timer to count 1min) 0(as it is) 7..................9mins... The gist of this problem is that the only way to get an exact time is by having an hourglass empty out completely when you declare 9 minutes are reached. So let's take a holistic approach: the most time we can count using ONE of the hourglasses is 8 minutes: by starting the 4-minute hourglass, flipping it, then letting it go for another 4-minutes. Let's say we start both the 4-minute and the 7-minute hourglass at the same time. The 7-minute hourglass runs out before 8-minutes are up, so we flip it. When 8-minutes are up (measured by the 4-minute hourglass), we know 1 minute worth of sand has deposited since the 7-minute hourglass was flipped. So now flip that sand upside and down and let it drain completely, and we've reached 9 minutes. start 7 and 4 , flip 4 while 4 is gone but 7 is working , when 7 is gone , we have 1 min in 4 glass , start the timer , do 4 glass twice to count another two 4 mins : 1+ 4+ 4=9 4-4, 7-4 4 4-3 7-3 3 1-1 start timer 4 4 end timer Show More Responses Since this situation test our mental fortitude and problem solving skills, can we start by opening the lids and emptying the sand from the 4 min. Pour the sand from the 7 into the empty 4 so you are left with three. Poor out that three into a pile and fill the 7 min with the remains sand and repeat to get another three. So now you have two pile of three min sand for a total of 6. Fill up the 4 min and add the 2 pile of three min sand into the 7 min hour glass to have exactly 9 min. |

CASE: Cross-selling Credit Insurance to Cardholders direct mail: .50, 1% response rate, avg balance $1000, 5% claim insurance, etc. Profitable? How make more profitable? What if response rate doubled but claims doubled? Make chart of profit curve, what does it mean if..., etc. 12 Answerssee cross selling case on CO's website Hi Sir, Can you confirm if below is the correct answer. I believe its always unprofitable. Assume 100 direct mail offers were sent for card insurance, 1% response rate means, 1 person bought credit insurance Revenue = 1% of avg balance = (1/100)*1000 = 10$ Cost = Insurance claim Cost + Mailing Cost = (Response Count* claim rate * Avg Bal) + (No. of mails sent * Cost per mail) (1* (5/100)*1000) + (100 * 0.5) = 50 + 50 = 100 $ Selling credit insurance in this case is unprofitable even without the mkt cost. For other questions if we assume, response rate = x%, claim rate = y% with the assumption, 100 offers were sent for credit insurance Revenue = x*(1/100)*1000 = 10x Insurance Cost = x*(y/100)*1000 = 10xy Mkt Cost = 100 * 0.5 = 50 Profit = 10x-10xy-50 Assumption is that y% claim rate means, of x people who take the offer, y% of those x, file for claim, which means company has to cover their avg bal of 1000 for people who filed the claim, and hence is a loss for the company. At this point its a loss to sell credit insurance. Let me know if i am doing anything incorrectly. Great the way you laid it out, very thorough and clearly organizing your thoughts. Remember to think out loud/explain your thinking as you write during the case. Yes, something is incorrect, very small but key, changes the whole answer: you are assuming everything is in the same time unit. When you calculate revenue to be $10, you should realize right away that would be a MONTHLY amount, while claims et all would be annual. Helps to know that credit cards in the US regularly try to entice customers to add little $5-10/month services to their bills, whether insurance, credit report monitoring, etc so that $10 couldn't possibly be yearly, ie insurance less than 0.90/month. Show More Responses Thanks for your response and guidance. I knew the revenue is monthly, but i thought claim rate is also monthly and hence calculated profit for each credit insurance sold per month. In light of your clarification Profit per card insurance per year = (10*12)-50-50 = 20$ per year If we chose to calculate per month, we will need to consider monthly claim rate as (5/12) and also amortize the marketing expense over next 12 month. Profit per card insurance per month = 10-(50/12)-(50/12)= (20/12)$ per month The profitability equation (per card per year) = 120x-10xy-50 For calculating any of the break even rates (x or y assuming 1 is known), 120x-10xy-50 = 0. For graph, P = 120x-10xy-50 Let me know if my analysis/answer is accurate and up to the mark. Thanks a lot for all your guidance. Aspirant, why did you take the revenue as 1% of the average balance? I think the average balance is the revenue. Whereas the response rate is 1%. Assuming 100 people are sent the mail: Revenue : (Response rate * 1000) = 1 * 1000 = 1000 Cost : (100 * Mailing cost) + (Response rate * claim rate * 1000) = (100 * 0.5) + (1*(5/100)*1000) = 50 + 50 = 100 Profit = Revenue - Cost = 1000 - 100 = 900 (This is for 100 mails sent) So the profit per mail sent = 900/100 = $9 Now, assuming the response rate is x% (instead of 1% as given) Revenue = 1000x Cost = (100*0.5) + (x*(5/100)*1000) = 50 + 50x Profit = 1000x - (50+50x) = 950x - 50 Profit per person = (950x - 50)/100 = 9.5x - 0.5 To make it more profitable, try to increase response rate x. Now if response rate is doubled and claims doubled, Profit = (Revenue) - (Cost) =(2000) - (50 + 100) = 1850 So profit per person = 1850/100 = $18.5 Now to make a chart of profit curve, i.e profit vs response rate, we use Profit per person = 9.5x - 0.5 plot : y= 9.5x - 0.5 Meaning : To break even, we need 9.5x - 0.5 = 0 i.e x = 0.05 (approximately) So we need a response rate of atleast 0.05% to be profitable (or 1 in 2000 people to respond) Looks like I left out the price of the insurance: customers would pay 1% of monthly balance for insurance. @Aspirant 2: You are using response rate as a number and at the same time using claim rate correctly as 5/100. use Response rate as 1/100 wherever applicable. Revenue should be $10 Cost = 50.50 Profit(Loss) = 10-50.50 Profit (Loss) per mail sent = (10-50.50)/100 Why isn't churn rate mentioned in here? Isn't that an important factor? Or were those number given to you without probing? Thank you! =) @ ghachla: The response rate is 1% that is 1 out of 100. I assumed 100 people are sent the mail. Hence the response rate is 1. As for the claim rate, the people who don't respond can't make claims. So out of the people who respond, the claim rate is 5% (i.e 1 person responds in 100 and out of that 1, 0.05 make the claim); Or to make it more clearer, if we assume 10000 people are sent the mail, 100 respond (because of the 1% response rate) and 5 out of the 100 make claims (because of the 5% claim rate). @A2: I understand what you are saying. the words were just misleading. What about the insurance rate? That's revenue to the bank, right? "By intrvw candidate: Looks like I left out the price of the insurance: customers would pay 1% of monthly balance for insurance." = monthly avg balance x 10 x # of people who made claims? and then convert to annual numbers or monthly as the case may be? How do you suggest to answer the profit: annual numbers or monthly numbers? Hello, can someone tell me how to plot claim rate vs response rate. What units goes on the x and y axis. Also how to find response rate for maximum profit? This discussion was helpful, I have been thinking about this problem for a bit now. Only one thing, shouldnt equation for market cost be 50/x since if the response rate goes up, the cost per customer will go down as a function of the response rate. |

How are M&M's made? 11 AnswersI said that the chocolate was solid and then dipped into the candy shell They use hybrid dung beetles to roll the balls in candy coating! Going Green at M&M's with magic. that's why some of them talk. Show More Responses Reply back asking exactly how many M&Ms will you be expected to make each day. Say, "through teamwork!" Then turn to the nearest mirror, smile wildly and give an energetic thumbs-up! (It is best to assume it is a one-way mirror and that you are being watched). "Deliciously" One batch at a time. When a Mommy M and a Daddy M love each other very, very much.... Seriously, they make the chocolate first, cool it, separate them into their smaller spheres, then put into Panners where the sugar layers are added, dyed the different colors, then waxed (no kidding, very thin layer though). Then they go down a conveyor belt to get the M stamped on them. elves. i'm quite certain they are made by elves. how else would they talk, and refuse to get in the bowl. Two theories. 1.) They are coated with the shell and then while the shell is still liquid dropped from such a height that the shell cools on the way down therefore making the shell perfect with no seam or spot where it was set down. 2.) Put into molds, shell is pumped into molds then the candy is tumbled to remove the spot where the shell was injected, this will also make the candy shiney I have a feeling its probably number 2 but my mom posed this question to me when I was like 7 and I came up with the first answer, it always seemed much cooler Relocation |

How many numbers between 1 and 1000 contain a 3? 11 Answers300 numbers contain a 3, but you counted numbers of the form x33, 3x3, and 33x *twice* so you must subtract them 300-30=270, but you subtracted 333 once too many, so add it back 300-30+1=271. Answer is 271. 271 does not seem right . I think it is 111. Aren't there just 11 numbers containing 3 between 0 an 100, like 3,13,23 30,33 etc. So between 0 and 1000 there are 111; after adding in for 300. There are only 19 numbers between 1 and 100 containing '3': 3, 13, 23, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 43, 53, 63, 73, 83, 93. The same logic holds for 101-200, 301-400, 401-500, 501-600, 601-700, 701-800, 801-900, and 901-1000. That is 19 * 8 = 152. There are 20 numbers between 201-300 that contain a 3: 203, 213, 223, 230, 231, 232, 233, 234, 235, 236, 237, 238, 239, 243, 253, 263, 273, 283, 293, 300. That brings the sum up to 152 + 20 = 172. Finally, 99 numbers from 301-400 contain a 3. The final sum is 172+99=271. Of course, there are much more intelligent ways to count. For example, instead of counting how many numbers contain a 3, count how many do NOT contain a 3. That means that there are 9 possibilities for the 1st number (0-9 except 3), 9 possibilities for the 2nd number, and 9 possibilities for the 3rd number. Thus, there are 9 * 9 * 9 = 729 that do not contain a 3 which means that there are 1000 - 729 = 271 numbers that do contain a 3. Show More Responses Thanks Candidate, or more simply For 300 to 399 = 100 numbers For other x00 to x99 = 19 numbers each = 19 x 9 = 171 numbers Total of 271 numbers I would go with elimination of one character from a base 10 numbering system gives you a base 9 numbering system. 9^3 = 729 permutations of a base 9 numbering system (a system with no number 3) with 3 digits, since 10^3 = 1000; 10^3 - 9^3 = 271 thus 271 numbers have a 3 in them. http://brainteaserbible.com/interview-brainteaser-numbers-between-0-and-1000-at-least-one-5 basically the same thing 1000 does not contain a 3. So, count the number of 3 digit numbers without a three. There are 9 choices for the first entry, 9 for the second, and 9 for the third. So, there are 729 numbers without a 3, and 1000-729 = 271 with a 3. I have a much simpler and faster method: let A be the cardinality of numbers between 1 and 1000 that contain a 3 and let A' be the cardinality of numbers between 1 and 1000 that do not contain a 3. There are 3 digits that can take the form of (0,1,2,3,4,5,6,7,8,9), so 10 possibilities.. To obtain A' cardinality we have 9 possibilities because 3 is excluded so A' = 9^3 = 729. Hence, the amount of numbers that don't have a 3 from 1 to 1000 is 729 so to obtain the amount that does contain at least one 3 is : 1000 - 729 = 271 lol im in 7th grade and this question is easy to me. first you count the numbers which DON't have a three. there are 9 choices for the first digit, 9 for the second digit, and 9 for the third. You probably noticed that this counts 0 to 999 instead of 1 to 1000 but its okay because, we count the same amount of numbers. 9^3=729 -1000=|271| (i like using absolute value cause it makes me look cool, its just a way to show "difference" in math) 3xx 100 a3x a is not 3 here to reduce duplication, 10*9 ab3 a , b are neither not 3 , 9*9 =271 You can use this formula to work it out for any power of 10: Tn+1=9Tn+10^n. Tn being the number of threes in the numbers between 1 and the previous power of 10. Tn+1 is simply saying the number of threes in the next power (the one you are working out). The 10^n is the power of 10 that you add on. This is the previous power of ten. You must start off knowing that there is one 3 between 1 and 10 For 100: Tn+1=9*1+10^1=19 For 1000: Tn+1=9*19+10^2=271 For 10 000: Tn+1=9*271+10^3=3439 |

### Business Analyst at Capital One was asked...

Let's say we're playing Russian roulette. The revolver has SIX barrels, THREE of which contain bullets. I give you two options: A. Spin, shoot. Spin, shoot. Spin, shoot. (For a total of three times). B. Spin once and pull the trigger twice. Which option would you pick and why? 13 AnswersA. Your chances of surviving are (3/6)*(3/6)*(3/6) = 1/8. B. Your chances of surviving are 1/3. Therefore, you want option B. Correct me if I'm wrong but aren't the chances of surviving scenario B based on conditional probability...the probability you survive the second trigger pull given that you have already survived the first... (3/6)*(2/5)/(3/6)= 40% chance of survival I forgot to add a crucial part of the question in my original post, which is that you know that the three bullets are right next to each other. Thus, the first spin determines whether you survive, and there are two out of the six chambers that are safe to "land on," hence 1/3. In my original wording of the question, I think your chances of survival would just be (3/6)*(2/5) = 1/5. Show More Responses A: 3/6*3/6*3/6 = 1/8 B: 3/6*2/3 = 1/3 OP, I have a question to you. Can you explain about the cases. Do you think prior knowledge of Credit card industries and other banking concepts are important. I am from engineering background and i am that familiar with banking concepts. You should have a basic knowledge of credit card concepts, e.g. credit limits, balances, interest rates. The interviewer will be happy to explain any uncommon terms that they use during the case (e.g. I wasn't familiar with the term "utilization ratio" and my interviewer was happy to tell me what it meant). You don't need to know about the crazy derivatives or anything like that. Put it this way: The interviewers won't penalize you if you ask them what a particular term means during a case. However, if they ask you what consumers might care about in choosing a credit card or bank, you should be able to name basic concepts such as the ones above. Thanks for your response. this problem is conditional probability because previous events effect probability of current. If the chambers are loaded with 3 bullets at random: % of survival = 3/6 * 2/5 *1/4 = 5% chance of shooting 3 blanks consecutively. If the bullets are in 3 consecutive chambers, then your % drastically changes. 50% chance on first shot. If blank, then you know that 3 permutations are no longer available: chamber 123,612, 561. Therefore chamber 5 has a 66% chance to contain a bullet chamber 6 has a 33% chance to contain a bullet chamber 2 has a 33% chance of containig the bullet chamber 3 has a 66% chance to contain a bullet chamber 4 must contain a bullet Chance of dying with 3 consecutive shots if the bullets have to be consecutive is 0.5*0.666*0.333 = 11% --Spin-shoot-spin-shoot-spin-shoot: We are selecting a random barrel each time (50% chance of containing a bullet) Probability of surviving = (1/2)*(1/2)*(1/2) = 1/8 = .125 --Spin-shoot-shoot First chamber has 50% chance of containing bullet. If we survive the first shot, there are still 3 bullets (2 blanks) left, but only five remaining chambers Probability of surviving = (1/2)*(2/5) = 2/10 = .2 .2 > .125, so the second option is safer. Martin is right except for the last part. It would be Probability of surviving = (1/2)*(2/3) = .33 This is because if you survived the first shot you are in 3 of the possible blank chambers that are all next to each other. If you first shot blank #1, you will move and shoot blank #2 next. If you shot blank #2 first, you will move and shoot blank #3 next. If you shot blank #3 first, you will move and shoot the first bullet (1/3 chance of getting bullet conditional on surviving the first shot). Given then you survived the first shot with 50% chance, you started in one of those blank chambers. And since the blanks are adjacent to each other (since the bullets are all adjacent), your chances will be 2/3 for survival. And 1/2*2/3 = 1/3 None of them, I want to live. How about we just not play Russian roulette. rather not play |

Credit card insurance case: calculate break-even claim rate where response rate = 2%, fee charged = 1% of monthly balance, and average balance = $1,000 per month. 7 Answersbreak-even claim rate = 9.5% How did you come up to that answer? Thank you Can you please explain how you got to the break even claim rate? Show More Responses Sorry - Could you elaborate on the answer. Assume a 100 customers. Claims rate = x. Interest revenue = (100-x)*1000*1% = 1000 - 10x Claims paid out = 2%*1000*x = 20x Breakeven scenario: 1000 - 10x = 20x => x = 33.3 Break even claim rate = 33.3% How did you get 9.5%. Am I misunderstanding the question? Don't think there's enough info here to solve. Response rate is 2%, but you need to know how much the marketing costs to be able to determine the breakeven. Yeah, this may be why you did not pass the final round I think the break even claim rate for this insurance scheme would be 1%. (Keeping all other non-mentioned costs aside) Let's assume 100 customers are contacted: 2% response rate = 2 people buy insurance. Revenue: 2 * (1% of 1000) = $20 At break even, cost would also be $20. Let the claim rate be x% (Note: this will be a percentage of the people who have bought the insurance, i.e. x% of 2% of 100) Insurance pay-out = $1000 per claim Cost: 1000 * x% of 2 = 1000 * x/100 * 2 = 20x Now, revenue = cost at break even, hence: $20 = 20x x = 1 So the B/E claim rate would be 1%. |

### Operations Analyst at Capital One was asked...

You have the choice between using first class or third class mail for a letter you are sending out to potential customers. First class costs $0.50 per piece and reach 100% of potential customers. Third class costs $0.40 per piece and reaches 80% of potential customers. Which do you use? 10 AnswersIt provokes you to ask questions. You do not nearly have enough information to perform the analysis. You need to know the present value of each customer, how many mailings, and calculate the point of indifference in terms of how many letters you are sending, etc. First class should be used, as the cost of each delivered letter will be the same, but you will reach all of your intended audience. Example: 1000 pieces to be sent, sending first class costs $500 and reaches 1000 ($.50 per peice), and sending second class costs $400 and reaches 800 ($.50 per piece) either one is good. For example, you will reach 0.8 person by spending $0.4. If divided 0.4 by 0.8, we get that we will reach 1 person by spending $0.5. So two methods have the same effect. Show More Responses Need to know the Revenue per customer and no of customers before we can decide on either @gaurav: don't make it too complicated. they just want to know the costs of reaching the customers. peppermint is right. peppermint is not right. say you make 1000 dollars off of each customer you reach at probability .5. Assuming you send 1000 letters, first class reaches all 1000 and has total cost 500 dollars. revenue is (1000 customers reached)(.5 prob make 1000 dollars)(1000 dollars) = 500,000 and profit is 500,000-500 = 499,500. third class costs 400 dollars, reaches 800 people, gives revenue of (800)(.5)(1000)= 400,000, profit is 400,000-400=399,600. You can generalize this to a formula to determine which is better given number of letters to be sent, expected money off of customers who accept, and probability that the customers letters reach will accept. Peppermint is right. Assumption: only when the letter has reached the recipient, he becomes a potential customer. Say they send letters to x people So, Cost per Potential Customer= 0.5/x and 0.4/(0.8x) for respective cases So, cost per potential customer is same in both cases. I believe that the point of cost is moot; what is the content of the letter and how imperative is it that it be delivered to the potential customer? If it is vital, go for 100% success rate. If it is vital and digital copies can be provided, then go for a scanned copy. Digital signature documents also work. I know most prospects yield their info to cold callers. Why do math when you can just use logic? The real answer is.... Send the letter through 1st class mail if the price of the product capital one is trying to sell is above $0.90. If not, send the letter through 3rd class mail. Assumptions:- •Potentially customers = actual customers Sample calculation:- Profit = Revenue - Cost Assume all potential customers = 100 Assume Price = $1 1st class mail profit = ($1x100)-($0.5x100) = $50 3rd class mail profit = ($1x80)-($0.4x80) = $48 Assume Price = $0.80 1st class mail profit = ($0.8x100)-($0.5x100) = $30 3rd class mail profit = ($0.8x80)-($0.4x80) = $32 Assume Price = $0.90 1st class mail profit = ($0.9x100)-($0.5x100) = $40 3rd class mail profit = ($0.9x80)-($0.4x80) = $40 Must also consider response of 3rd class versus 1st class-perception |