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Recent Comments

Probability and Statistics

Basketball Practice #2

A much easier way to approach this is to realize that this is a Pólya's urn application, which makes all shots exchangeable. Therefore, if we use the conditional probability: P(make n+1 | make n) = P(make n+1 and make n) / P(make n) By exchangeability: P(make n) = P(make shot 3) = 1/2 P(make n+1 and make n) = P(make shot 3 and make shot 4) For P(make shot 3 and make shot 4): P(make shot 3) = 1/2 P(make shot 4 | made shot 3) = 2/3 (now 2 hits out of 3 total shots) So P(make shot 3 and make shot 4) = 1/2 × 2/3 = 1/3 Therefore: P(make n+1 | make n) = (1/3) / (1/2) = 2/3

william.xiao430

Transfer 3000 Apples

If the return is free, the answer would be 1000 theoretically, you can look to these 2 simulations Edinburgh → 200 miles Each 1000-apple load loses 200 apples: 1000→8001000 to 8001000→800 Three trips: 800+800+800=2400800+800+800=2400800+800+800=2400 At 200-mile pit: 2400boxed{2400}2400 200 → 400 miles Start with 2400 apples. Trip 1: 1000→8001000 to 8001000→800 Trip 2: 1000→8001000 to 8001000→800 Trip 3: 400→200400 to 200400→200 Total at 400-mile pit: 800+800+200=1800800+800+200 = boxed{1800}800+800+200=1800 400 → 600 miles Start with 1800 apples. Trip 1: 1000→8001000 to 8001000→800 Trip 2: 800→600800 to 600800→600 Total: 800+600=1400800+600 = boxed{1400}800+600=1400 600 → 800 miles Start with 1400 apples. Trip 1: 1000→8001000 to 8001000→800 Trip 2: 400→200400 to 200400→200 Total: 800+200=1000800+200 = boxed{1000}800+200=1000 800 → London Start with 1000 apples. One trip: 1000−200=8001000-200 = boxed{800}1000−200=800 Summary Position Apples 0 miles 3000 200 miles 2400 400 miles 1800 600 miles 1400 800 miles 1000 1000 miles (London) 800 So your fixed-pit strategy delivers: 800 applesboxed{800 text{ apples}}800 apples next Yes — there's a pattern emerging. With 4 pits, the spacing was 200 miles and we got: Distance Apples 0 3000 200 2400 400 1800 600 1400 800 1000 1000 800 Now let's use 5 pits, so the route is split into 6 equal segments: 1000/6≈166.67 miles1000/6 approx 166.67 text{ miles}1000/6≈166.67 miles Each loaded trip loses 166.67166.67166.67 apples. Pit 1 (166.67 miles) 3000 apples ⇒ 3 trips. Each trip delivers: 1000−166.67=833.331000-166.67=833.331000−166.67=833.33 Total: 3(833.33)=25003(833.33)=25003(833.33)=2500 Pit 2 (333.33 miles) 2500 apples. Two full trips and one half trip: 833.33+833.33+(500−166.67)833.33+833.33+(500-166.67)833.33+833.33+(500−166.67) =833.33+833.33+333.33=833.33+833.33+333.33=833.33+833.33+333.33 =2000=2000=2000 Interesting! Exactly 2000. Pit 3 (500 miles) 2000 apples ⇒ 2 trips. Each delivers 833.33. 1666.671666.671666.67 Pit 4 (666.67 miles) 1666.67 apples. One full trip and one partial trip: 833.33+(666.67−166.67)833.33+(666.67-166.67)833.33+(666.67−166.67) =1333.33=1333.33=1333.33 Pit 5 (833.33 miles) 1333.33 apples. 833.33+(333.33−166.67)833.33+(333.33-166.67)833.33+(333.33−166.67) =1000=1000=1000 Again exactly 1000! London 1000 apples for the last 166.67 miles: 1000−166.67=833.331000-166.67 = 833.331000−166.67=833.33 Final: 833.33boxed{833.33}833.33

Smallest Number to 10,000

Smallest Number to 10,000

there should be a 5^4 not 2^4 multiplied with 2^4 to result in625 (just a typo, but still, why not point it out :))

Probability and Statistics

Uniform Distribution #1

this isn't the topic discussed here right? and one can only factor the product if given independence of rv's.

Probability and Statistics

Uniform Distribution #1

By definition of a Uniform(0,1) distribution, all values are equally likely of appearing. Thus, the CDF is simply the number itself. For example, if z=0.7, the CDF is 0.7. So, P(max<=z) = P(X<=z).P(Y<=z) = z.z = z^2. Differentiate for PDF = 2z. Integrate 2z.z (EV = int(x.f(x))dx) between 0 and 1 to get E[max(X,Y)] = 2/3.

Probability and Statistics

Football or Cupcake

Correct, thanks!

Probability and Statistics

Football or Cupcake

i assume the factor 0.25 in front of the ev of merges ist just a typo, right?

Probability and Statistics

Meeting Probability

I think there is a mistake in the solution. If traders can arrive no later than 12:55, then the arrival times are uniformly distributed over a 55-minute interval. The square area should therefore be (55^2), not (60^2).

Probability and Statistics

Both Card Colors

1-2 * 26c3/52c3

Probability and Statistics

It's a biased 1d random walk and you should know by heart 1-(q/p)^i / 1- (q/p)^n

Probability and Statistics

I dont think they will want you to compute the exact result for this specific problem but be able to setup the problem up until equation 4 and recognize the structure.

A friend, who had this question in an interview, did it exactly your way. The recruiter said that your method was fine as well, it's often an early indicator if someone is more fit for a more discrationary/ screen trading role, or more quant trading role. So, both methods are good, it just shows a bit which desk might be the best fit for you.

Maybe for completion a "brute force" way of solving this going backwards that may even seem a bit more intuitive. the same thing happens three times: one fish gets discarded, then a third of whats left is taken. So if I undo the operation, each step I multiply the remaining pile by 3/2 (undoing the 1/3) and add 1 (undoing the 1). Apply this three times to the final pile and I should land back on the starting count, as long as everything stays a whole number along the way. So the problem reduces to finding the smallest possible value of the final pile that produces integers all the way back. Naturaly, we would choose 2 as a starting number as this is the smallest possible (because the starting number MUST at least be a natural number when multiplied by 3/2). - For 2: 2*3/2 + 1 = 4, then 4*3/2 + 1 = 7, then 7*3/2 + 1 = 11.5. Fails at the last step. - 3: 3*3/2 + 1 = 5.5. Fails (comment above). - 4: 4*3/2 + 1 = 7, then 7*3/2 + 1 = 11.5. Fails. - 5: 5*3/2 + 1 = 8.5. Fails. - 6: 6*3/2 + 1 = 10, then 10*3/2 + 1 = 16, then 16*3/2 + 1 = 25. Works, and it must be the minimum. So the minimum is 25 fish.

Probability and Statistics

shouldnt the formulation be what is expected number of throws... "to get" rather than "before you get"? Before you get is exclusive of HH, isnt it?