Sollicitatievragen voor Data Scientist in Utrecht

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Answer from a frequentist perspective: Suppose there was one person. P(YES|raining) is twice (2/3 / 1/3) as likely as P(LIE|notraining), so the P(raining) is 2/3. If instead n people all say YES, then they are either all telling the truth, or all lying. The outcome that they are all telling the truth is (2/3)^n / (1/3)^n = 2^n as likely as the outcome that they are not. 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. Minder

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26/27 is incorrect. 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). What you have to figure out is the odds it raining | (i.e., given) all 3 friends told you the same thing. Because they all say the same thing, they must all either be lying or they must all be telling the truth. What are the odds that would all lie and all tell the truth? In 1/27 times, they would the all lie and and in 8/27 times they would all tell the truth. So there are 9 ways in which all your friends would tell you the same thing. And in 8 of them (8 out of 9) they would be telling you the truth. Minder

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@ RLeung shouldn't you use left join? You are effectively losing all posts with zero comment. Minder

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Here is the solution. You need a left self join that accounts for posts with zero comments. 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 Minder

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select b.element from b left join a on b.element = a.element where a.element is null Minder

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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 Minder

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Can you post here your solution for the ads analysis from the takehome challenge book. I also bought the book and was interested in comparing the solutions. Also can you post here how you solved the SQL question? Minder

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for the SQL, I think both should work. Outer join between lifetime count and new day count and then sum columns replacing NULLs with 0, or union all between those two, group by and then sum. Minder

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For "MockInterview dot co": The binomial part is correct but you argue that the expected value for option 2 is not 4 but this is false. In both cases E(x) = np = 100*(4/100) = 4 and E(x) = np=100*(1/25) = 4 again. Minder

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Chance of getting exactly one add is ~7% As the formula is (NK) (0,04)^K * (0,96)^(N−K) where the first (NK) is the combination number N over K Minder

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So many answers...Here's my version: For round1, B win only if it gets 3 or more stones than A, which is (A,B) = (1,4) (1,5) (1, 6) (2, 5) (2,6) (3,6) which is 6 cases out of all 36 probabilities. So B has 1/6 chance to win. To draw, B needs to get exactly 2 stones more than A, which is (A, B) = (1,3) (2,4) (3,5) (4,6) or 1/9. 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. So the final answer is (1/6, 1/9*5/12, (1/9)^2*5/12, .....(1/9)^(n-1)*5/12) ) Minder

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I don't get it. Shouldn't prob of B winning given it's tie at 1st round be 15/36? given it's tie at 1st round, at the 2nd round Nb > Na can happen if (B,A) is (2,1), (3,1/2),(4,1/2/3), (5,1/2/3/4),(6,1/2/3/4/5), which totals 15 out of 36. Minder

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find the right most element. If this is a right node with no children, return its parent. if this is not, return the largest element of its left child. Minder

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One addition is the situation where the tree has no right branch (root is largest). In this special case, it does not have a parent. So it's better to keep track of parent and current pointers, if different, the original method by the candidate works well, if the same (which means the root situation), find the largest of its left branch. Minder

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In the vein of answers 6 and 7: SELECT a.user, COUNT(DISTINCT a.friend) AS friend_count FROM ( (SELECT accepter_id AS user, sender_id AS friend FROM ACCEPTED) UNION (SELECT sender_id AS user, acceptor_id AS friend FROM ACCEPTED) ) a GROUP BY a.user ORDER BY friend_count LIMIT 1; Minder

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I think this would be simpler select requester_id from request_accepted union all select accepter_id from request_accepted) t group by 1 order by count desc limit 1 Minder

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In both tables, concat the requestor and the recipient IDs then do a left join. Friend_requests[111,aaa,01-01-15;222,aaa,02-01-15] request_accepted[aaa,111,02-01-15] Concat and your left join is searching the second table for 111aaa & 222aaa. It finds the first one and the second one is null. You have a 50% acceptance rate. Regarding the dates, alot can be done with them but they are not strictly part of the question. The only thing that dates mean is that you could have multiple requests before an accept so use distinct. Minder

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SELECT (CAST(COUNT(r.acceptor_id) AS FLOAT) / CAST(COUNT(f.requestor_id) AS float)) AS acceptance_rate FROM friend_request f FULL OUTER JOIN request_accepted r ON (f.requestor_id=r.requestor_id AND f.sent_to_id = r.acceptor_id) WHERE f.date > (CURRENT_DATE - INTERVAL '30 day'); Minder