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This will replace the current featured interview for this targeted profile. was contacted about a week later for an interview. What might you think is the cause of this, and how would you test it? What are the odds that would all lie and all tell the truth? If one response with more than of half of total answers being located outside of 95% confidential interval in each histogram, the response will be categorized as random fall out of mean plus tw. This concludes the round. It should be (2/3)^3, I think zen and todo is correct. Probability of the game being over in Round 1 is 10/36. Obviously we want P(X|all three say yes). public Node GetKthLargest(int k) { return GetKthLargest(ref k, this.Root); } Node GetKthLargest(ref int k, Node root) { if (root == null || k < 1) return null; var node = GetKthLargest(ref k, root.Right); if (node != null) return node; if (--k == 0) return root; return GetKthLargest(ref k, root.Left); }, recursion is not needed. i.e. P(Rain|YES)=P(Rain|YES,T)*P(T) P(Rain|YES,T)=1==> whats the probability of it raining given that they are telling the truth and have told us its raining then P(T)=(2/3)^3 its obvious. P(TTT) is 8 times more likely than P(LLL), so we have P(All same answers|TTT)=8/9, P(All same answers|LLL)=1/9. Only conditioned on the truth (raining/not raining) are they independent. And again find the largest number. For each product you have a vector of M users that bought that product. class Node: def __init__(self, value, left=None, right=None): self.value = value self.left = left self.right = right def findKthLargest(root, k): global count if root is None: return findKthLargest(root.right, k) count += 1 if count == k: print root.value return findKthLargest(root.left, k) count = 0 r = Node(10, Node(5, Node(2), Node(7)), Node(30, Node(22), Node(32))) findKthLargest(r, 3), // solution in java // main routine Node findSecondMax(Node root) { if(root == null || (root.left == null && root.right == null) return null; else { Node max = findMax(root); return (max.parent == null) ? Free interview details posted anonymously by Zoom Video Communications interview candidates. 5 days later i got an offer. Thus P(ALL YES | raining) = 2^n / (2^n + 1) = 8/9 for n=3 Notice that this corresponds exactly the bayesian answer when prior(raining) = 1/2. This will give you rows of rankings for each user; Example: "product p1's closest products p4, p600, p5, etc..." These rankings are according to purchase behavior. This can easily be solved without Bayes: There are two cases: Case 1: It is raining and all friends are telling the truth: 0.25*(2/3)^3 = 1/4*8/27 Case1: It is not raining and all friends are lying: 0.75*(1/3)^3 = 3/4*1/27 Probability: P(E) = Case1 / (Case1+Case2) = (1/4*8/27) / (3/4*1/27 + 1/4*8/27) = 2 / (11/4) = 8/11. Hypothesis: the photos are Halloween pictures. You also have corporate industries which are similar to Education. Free interview details posted anonymously by Amazon interview candidates. find the sum of the two list and subtract. I thought about this a little differently from a non-bayes perspective. 46 Kubrick Group Data Engineer interview questions and 33 interview reviews. These are those that don't feel it is logical to flip out phones each year, they wait for two years before buying a phone. Therefore the solution given ONLY THE INFORMATION GIVEN is P(Rain)=8/9, P(Dry)=1/9. Your feedback has been sent to the team and we'll look into it. (yeah yeah, unless your friends mess with you ALL the time ;). This problem requires the marginal probability of rain to solve, following Interview Candidate's answer. http://adtrk.tw/tp/rj6_ivYV_y.K. All the posts will have null parent_id. Get hired. In this special case, it does not have a parent. 2 days after the last stage, HR called to inform me that they want to proceed with my application by doing some background check. Considering the table schema to be something like this: CREATE TABLE submissions ( submission_id INT, body VARCHAR(500), parent_id INT ); SELECT DISTINCT nvl(parent_id::TEXT,'Post with no comments') AS post_id, COUNT(CASE WHEN parent_id IS NOT NULL THEN submission_id ELSE 0 END) AS number_of_comments_or_post FROM submissions GROUP BY 1; This will give results like this: post_id number_of_comments_or_post Post with no comments 8 1 10 7 11 13 8 19 9 25 7 So, the first row will give the number of posts with no comments which is 8 and remaining rows tell the number of comments per post. Free interview details posted anonymously by Top Hat interview candidates. * To find shows/movies to include in the measurement instrument, maybe do cluster analysis on large number of viewer's viewing habits. 2/3. Then lets go to iPod. So selling to schools will raise the sales in both other categories, and selling to instructors will raise the sales for students. Calculating the distance between u1 and u2: f(u1, u2) = intersection(u1, u2) / (len(u1) + len(u2) - intersection(u1, u2)) same with products: f(p1, p2) = intersection(p1, p2) / (len(p1) + len(p2) - intersection(p1, p2)) You do this for each of the N^2 and M^2 pairs. There is the High-Tech segment, meaning those who always want the newest and best. Before getting into the coding, think about it logically - how would you find this? You can group similar users and similar items by calculating the distance between like users and items. The wonder of the iPod, the product that sent Apple on a crash course to stardom. 109 HubSpot Developer interview questions and 85 interview reviews. This essentially means P(raining, yes) + P (non-raining, no) = 2/3. There's a whole list of them curated on Prepfully. of it raining elephants given the 3 friends said yes, it'd be still 0. Submissions can be posts, or comments to a post. The second last element in the array in the answer. why in the world would i do bayesian methods when its certain. Drop items with very low item-total correlations (either 1.5. 3. The probability in question has to depend on the probability of rain in Seattle. Any attempts to interpret this as conditional probability P(raining | yes) = 2/3 or P(yes | raining) = 2/3 are making other assumptions. if this is not, return the largest element of its left child. Here's my attempt (made easy with CASE) to exclude all the posts from the table and grouping/counting comments. Thank you. Free interview details posted anonymously by Pythian interview candidates. For B to Win 6 - Na + Nb > 8 +Na - Nb or, Nb - Na > 1 which can happen if (B,A) is (3,1),(4,1),(4,2),(5,1),(5,2),(5,3),(6,1),(6,2),(6,3),(6,4) Hence prob of B winning is = 10/36 = 5/18 For any round to ensue, the prior round has to end in a draw, or 6 - Na + Nb = 8 +Na - Nb or, Nb - Na = 1, for which prob = 4/36 = 1/9 Once equal, B's prob of winning is if total stones with B > total stones with A or 13/36 possibility of another draw = 1/6 For B to win in nth round, prob = 1/9x1/6^(n-2)x13/36, For B to Win 6 - Na + Nb > 8 +Na - Nb or, Nb - Na > 1 which can happen if (B,A) is (3,1),(4,1),(4,2),(5,1),(5,2),(5,3),(6,1),(6,2),(6,3),(6,4) Hence prob of B winning is = 10/36 = 5/18 For any round to ensue, the prior round has to end in a draw, or 6 - Na + Nb = 8 +Na - Nb or, Nb - Na = 1, for which prob = 5/36 Once equal, B's prob of winning is if total stones with B > total stones with A or 13/36 possibility of another draw = 1/6 For B to win in nth round, prob = 5/36x1/6^(n-2)x13/36. It's essential to demonstrate that you can really go deep... there are plenty of followup questions and (sometimes tangential) angles to explore. Similar to Cronbach’s alpha, calculate corrected item-total correlations. Please see our Community Guidelines or Terms of Service for more information. I would design the test in a way that certain information is asked two different ways. 3) The friends' answers are conditionally independent assuming no collusion. 12 other people were in my interview group, we were broken up and interviewed one on one with 3 different people and had a tour. How can B win in so many rounds ? The marginal probabilities are not independent, Pr(y,y,y) does not equal pr(y)^3, it equals pr(y,y,y,rain) + pr(y,y,y, no rain), the integration of the joint space over rain. But we can compare the averages of all the months by performing a hypothesis testing and rejecting the null hypothesis if the F1 score is significant. Now you keep going. Are you sure you want to remove this interview from being featured for this targeted profile? Free interview details posted anonymously by HubSpot interview candidates. You can't have all 3 says yes and have some people lying and some people telling the truth. Bayesian stats: you should estimate the prior probability that it's raining on any given day in Seattle. 2 Paytm Labs Data Engineer interview questions and 2 interview reviews. In Python: (just numbers) def rem_elem_num(listA,listB): sumA = 0 sumB = 0 for i in listA: sumA += i for j in listB: sumB += j return sumA-sumB (general) def rem_elem(listA, listB): dictB = {} for j in listB: dictB[j] = None for i in listA: if i not in dictB: return i, How about this in python, will this work? InterviewQuery.com has it more in depth of an answer. Love your job. Segment by country and date and check for a continual rise in photo uploads leading up to October 31st and a few days after for the lag. root : findMax(root.right); }. Since they are all in agreement (all lying or all truthful), they are essentially voting as one person. You call 3 random friends of yours who live there and ask each independently if it's raining. It either is raining or it isn't. Free interview details posted anonymously by ThoughtWorks interview candidates. 4 HelloFresh Data Engineer interview questions and 4 interview reviews. Player A has 8 stones, player B has 6. That is the number of times that at least one friend would tell you the truth (i.e., 1 - probability that would all lie: 1/27). Find the largest number in the binary tree and delete it. I assumed as prior) P(3y) = all truth or all lie = 2/3 ^ 3 + 1/3 ^3 = 9/27 hence P(R | 3Y) = 8/9. 8 big data engineer ~1~null~1~ interview questions. Glassdoor has 3 interview questions and reports from Data center technician interviews. Let Na be the # player A rolls, and Nb be the number player B rolls. Short and fast. The 3 random variables are not to be treated as intrinsically independent. But apple has already got several segments that I believe work. There were 6 stages of recruitment process: Gmat > HR Interview > Excel test > 1st round of User interview (with 2 users) > Case study > 2nd round of User interview. How would you test if survey responses were filled at random by certain individuals, as opposed to truthful selections? Answer from a frequentist perspective: Suppose there was one person. Look at the mean average precision of the movies that the users watch out of the rankings. Calculate Cronbach's alpha for the survey items. Using conditional independence and bayes rule, this becomes: pr(y|rain)^3*pr(rain) + pr(y|no rain)^3(1-pr(rain)). You know that job you’re proud of? Free interview details posted anonymously by ESB interview candidates. I was contacted directly by hiring manager and had a brief discussion about my background and the role. We're growing our Respiratory Field Sales team! For me, the question is really (1 - the odds that all three of your friends are lying to you) Clearly 1 - 1/3 * 1/3 * 1/3. Shouldn't prob of B winning given it's tie at 1st round be 15/36? Select children , count(submission_id) from ( Select a.submission_id, count(b.submission_id) as children from Submissions a Left Join submissions b on On a.submission_id=b.parent_id Where a.parent_id is null Group by a.submission_id ) a Group by children. We cannot say what has caused the spike since causal relationship cannot be established with observed data. Learn about interview questions and interview process for 21 companies. Entering the second round, all stones should be equal, so the chance to draw become 1/6, and the chance for either to win is 5/12. In first round A player (8 stones) has 6 certain ways to win where the sum of both dices is (5,4 or -4,-5 when B draws). textbook example of the Bayes' formula, anything short of that I don't think will work out. Since there is 36 ways in total, i took that 50% of those can be a tie. 2. However, if the game has to go one, the probability is 15/36 (difference between A and B is equal and greater than 1). Anyway, the question is asking how you figure out the number that is missing from list B, which is identical to list A except one number is missing. I believe this is a std. Free interview details posted anonymously by HelloFresh interview candidates. Free interview details posted anonymously by Autodesk interview candidates. 1 Answer Wow... pathetically … It’s here. Free interview details posted anonymously by Deliveroo interview candidates. Jaccard distance is a common approach when building graphs of items x users relationships. 208 data engineer ~1~null~1~ interview questions. Then it's straight-forward: P(raining | Yes,Yes,Yes) = Prior(raining) * P(Yes,Yes,Yes | raining) / P(Yes, Yes, Yes) P(Yes,Yes,Yes) = P(raining) * P(Yes,Yes,Yes | raining) + P(not-raining) * P(Yes,Yes,Yes | not-raining) = 0.25*(2/3)^3 + 0.75*(1/3)^3 = 0.25*(8/27) + 0.75*(1/27) P(raining | Yes,Yes,Yes) = 0.25*(8/27) / ( 0.25*8/27 + 0.75*1/27 ) **Bonus points if you notice that you don't need a calculator since all the 27's cancel out and you can multiply top and bottom by 4. On the first round, B can win if (A,B) rolls: (1,4) , (1,5) , (1,6) , (2,5) , (2,6) , (3,6) - so there are 6 out of 36 possibilities where B wins. This is a poorly worded question. Because they all say the same thing, they must all either be lying or they must all be telling the truth. Free interview details posted anonymously by Walmart interview candidates. 1 Pythian Big Data Engineer interview questions and 1 interview reviews. Depending on what A rolls (1-6), and then what B rolls (1-6), you are given 36 different possibilites: and then you can make a web out of that to see in what scenarios B will win in round 1. Replace None value with previous value present in a list. Someone who bought a Macbook retina probably have enough money to buy an another expensive laptop but kids of only paid $30 for headphones probably don't. Specialist IS Business Systems Analyst (4), Bio-pharmaceutical Sales Representative (4), The Ultimate Job Interview Preparation Guide. Later he said you will have 4 rounds of skype technical interview ,which was scheduled in 3 days time. Please describe the problem with this {0} and we will look into it. SELECT recommended_page FROM (SELECT f.user1_id as users, f.user2_id as freinds, l.page_id as recommended_page FROM friendship f INNER JOIN likes l ON f.user2_id = l.user_id WHERE f.user1_id = 1 UNION ALL SELECT f.user2_id as users,f.user1_id as friends,l.page_id as recommended_page FROM friendship f INNER JOIN likes l ON f.user1_id = l.user_id WHERE f.user2_id = 1) MINUS (SELECT page_id as recommended_page FROM likes WHERE user_id = 1); select c.subID as SubmissionID, count(c.body)-1 as Counts_Comments from subm c LEFT JOIN subm b ON c.subID = b.pID where b.pID is null AND c.pID is NULL group by c.subID UNION ALL select a.pID as SubmissionID, count(a.body) as Counts_Comments from ( select *, case when pID IS NULL then 'P' Else 'C' END as P_O_C from subm)a where P_O_C = 'C' group by a.pID Order by SubmissionID; select a.user_name,b.user_name,page_liked from services_db.pages_liked a, services_db.user_friends b where 1=1 and a.user_name = b.friend_user and a.page_liked not in ( select page_liked from services_db.pages_liked c where 1=1 and c.user_name = b.user_name ) ; Given an list A of objects and another list B which is identical to A except that one element is removed, find that removed element. Instructors will be more likely to spend more on a single product, and buy software relevant to their subjects, but these decisions will influence there students to do the same, but generally students will seek a "value" product, and will buy software based on requirements. The three data structure questions are: 1. the difference between linked list and array; 2. the difference between stack and queue; 3. describe hash table. 3 Twitter Data Engineer interview questions and 3 interview reviews. Prepare for your interview. They either are all lying or all telling the truth. But probability of A and B winning are the same hence that would be 15/36. 2 Autodesk Senior Data Engineer interview questions and 2 interview reviews. B has a 10/36 chances to win in round 1: (A rolls 1 and B rolls above 4) + (A rolls 2 and B rolls above 3) + ... A has a 21/36 chance to win in round 2 using the same idea and they will tie with 5/36 chances. School on the other hand will buy a large amount of Computers and software at once, which also effect instructor and student purchases. because the iPhone acts as a iPod there is a spill of sales that goes to iPhone, although the iPod touch does offer an affordable alternatives to those who do not want an iPhone. Martin Grassi has enjoyed interning with Amgen Australia's Value, Access, and Policy team, growing his network, driving innovation, and exploring new fields. Glassdoor has millions of jobs plus salary information, company reviews, and interview questions from people on the inside making it easy to find a job that’s right for you. Although the iPod Nano does capture the convenience segment. 14 selenium automation test engineer ~1~null~1~ interview questions in Pune, India. How would you account for NULLs? I flagged Nub data scientist's answer as useful, because it shows an interesting flaw in reasoning. Most of the answers/comments made all unconditional assumptions except a few reasonings that lead to the 8/9 probability. So B has 1/6 chance to win. All question histograms will likely follow the normal distribution if they are truthful selection. 1) In Recruiter interview, questions like difference between Linked list and Queue, Group by and having etc 2) In coding, you will giving simple questions like 2 set of lists containing words, find the set of words which exist in one and not the other. 1 Top Hat Data Engineer interview questions and 1 interview reviews. 9 Deloitte Data Engineer interview questions and 5 interview reviews. Depends on the kind of elements in the lists. Then you rank each row of the euclidean matrices for the product matrix and the users matrix. P(rain)=P(rain|truth,truth,truth)*P(truth,truth, truth)+P(rain|lie,lie,lie)*P(lie,lie,lie) notice that truth does not mean yes it is raining, it simply corresponds to them telling the truth. rounded to a percentage point), the posterior would be in the interval (0,075, 1). HR was pretty careless and never got back with the result. That is one way but also clustering algorithms can help in doing it in a more efficient ways. So, the probability that it is actually raining in Seattle is 3 * (8/27) = 8/9. With only the information from the interview question, we have to assume that friends are either all lying or all telling the truth. In my case this question was like: 'you have a table Submissions with the submission_id, the body, and the parent_id. select k.post_id, count(submission_id) -1 from (select submission_id, case when parent_id is null then submission_id else parent_id end as post_id from submissions) t group by post_id, select t.post_id, count(t.submission_id) -1 from (select submission_id, case when parent_id is null then submission_id else parent_id end as post_id from submissions) t group by post_id, select parent_id as post, count(parent_id) as num_of_comments from submissions group by parent_id union select submission_id as post, 0 as num_of_comments from submissions where parent.id=null, select comments_count, count(submission_id) as post_count from ( select submission_id, count( distinct parent_id) as comments_count from Table A group by submission_id )A group by comments_count, I think all of the Posts are missing Parent_ID. Other than that, basic assessment question on Why Amgen, availability to join, commute, etc were asked. And in 8 of them (8 out of 9) they would be telling you the truth. So overall B wins in 2 rounds has prob 5/36*15/36. What you have to figure out is the odds it raining | (i.e., given) all 3 friends told you the same thing. Learn about interview questions and interview process for 9 companies. Do they continue playing after round 1 ? 2 ThoughtWorks Senior Data Engineer interview questions and 2 interview reviews. Interview Questions. If they're numbers, sum(A) minus sum(B) will give the missing element. Replace None with previous value SQL: questions on Sales, Promotion, Product tables. And the equal to have another round is 6/36=1/6. Build a 6x6 table of possible rolls and outcomes. With the further assumption that the prior probability is measured with limited precision (e.g. How would you go and make a histogram of number of posts per comment_count?' You need a left self join that accounts for posts with zero comments. B has to throw one greater than A. All these supposed answers are missing the point, and this question isn't even worded correctly. Learn about interview questions and interview process for 7 companies. This is your opportunity to Live, Win, and Thrive with a world-renowned organization—a... – More. Python questions: 1. Isn't the answer 2/3. Interview Questions. I believe the greatest segment for the iPod would be parents wanting to get a gift for kids / something to keep kids entertained. Whoever has more stones at the end of the round wins and the game is over. All 3 friends tell you that "Yes" it is raining. In 1/27 times, they would the all lie and and in 8/27 times they would all tell the truth. You are compiling a report for user content uploaded every month and notice a spike in uploads in October. You are effectively losing all posts with zero comment. It's raining if any ONE of the friends is telling the truth, because if they are telling the truth then it is raining. P(rain)=P(rain|3xtruth)*P(3xtruth) and the probability of the truth is (2/3)^3 and the probability of rain if they are telling the truth is 1. In the absence of further information, the only correct answer is the posterior probability of rain p is in the interval (0, 1). How do you take millions of users with 100's of transactions each, amongst 10k's of products and group the users together in a meaningful segments?

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