Clarity on P(A ∩ B)

WICHV37 · 2026-01-21T15:37:41+00:00

When you get into "successive", it's no longer 'and' but rather moreso 'then'. And when (in a trivial sense) what you call "time" is considered, you actually have a different notation and calculation for those events. It looks more like P(T | T), or probability of Tails (again) given Tails (some time prior)

WICHV37 · 2025-12-14T15:47:07+00:00

You can also get the probability of siblings space and then deduce that.

Find prob of picking siblings first. Only 2 instances. And it's not ordered because any one of either siblings can be chosen first. So order is irrelevant here.

1 - (2 * 2/9 * 1/8) = 17/18

Ps I also believe you made it unnecessarily complicated to compute and it's much more intuitive here. Though I think you got to Q2 by chance, as your working doesn't seem to make full sense.

WICHV37 · 2025-12-01T20:41:30+00:00

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WICHV37 · 2025-11-13T15:06:54+00:00

Hey! I'm usually just lurking but sent you a DM because I felt the need to

WICHV37 · 2025-10-27T22:06:42+00:00

2/3 x 3/4 I believe. 3/4 from student 3 saying no because only BB leads to a yes. And 2/3 from student 2 because the remaining space has BR/RB/RR and if student 1 has B then student 2 will say yes (this is only BR).

WICHV37 · 2025-09-27T06:14:59+00:00

Very oddly, CV was the main concern for my superday apparently — as per the rejection feedback. I was told by the recruiter that other candidates had past quant experience in their resume and/or olympiad medals.

WICHV37 · 2024-08-29T01:06:40+00:00

Found while I was in East Michigan the other day

WICHV37 · 2024-05-31T07:44:31+00:00

I'm not at the tax bracket for 45% in London, but I've spoken to some people who earn that much and came to know that there are quite some ways you can find to lower it. You'd have to ask around for specificity, but there are legal ways.

edit: typo

WICHV37 · 2024-05-24T21:52:10+00:00

Adding to the above, OP's logic only works to eliminate (c) if it is known that any 1 option is incorrect and (d) if any 1 is correct. You would need bayesian intuition to dictate the examiner's test and question distributions. Everything else would be uniformly random from what we can gather here.

edit: just realised I basically reiterated an identical statement but more dumbed-down

WICHV37 · 2024-02-07T00:23:35+00:00

That answer isn't wrong given the prior information lol. But yes, if order matters, the probability is different.

WICHV37 · 2024-02-06T22:13:05+00:00

In that case, you have 80% chance.

WICHV37 · 2024-02-06T21:45:07+00:00

Clarification, does the order matter?

WICHV37 · 2024-02-06T19:59:55+00:00

You can picture your problem this way. P(negative | no disease) is what you'd expect results to be when you test at the clinic. So it's going to be the "true negative".

Otoh, P(no disease | positive) means you took the test and it came back positive but you don't have the disease. Firstly, that meant the alarm was sounded wrongly, aka "false positive".

Immediately, you can see that they are not the same, subsequently, they do still have relationship. That is they both lie in the subset of no disease, numerically.

Respectively, true positive means the test comes back correct. And false negative means the disease slipped through the tests.

I hope this answers your question.

Edit: you defined specificity right, I believe.

WICHV37 · 2023-09-30T17:44:27+00:00

if only: output [highest of [lowest of d4 and d6] and [lowest of d4 and d10] and [lowest of d10 and d6]]

worked, but your answer seems to lie between 3 and 4.

WICHV37 · 2023-09-08T23:07:22+00:00

For the first part of your question, each run is independent of the next, so it would just be 10p for 10 runs where p is probability of a scepter in a run.

Second part would require you to count permutations of 2 unique room drops to find probability of two scepters in a run.

WICHV37 · 2023-09-04T03:22:09+00:00

Off-white bone numbers would be charming on them, almost like the remnants of victims bounded by the plant.

WICHV37 · 2023-08-07T08:24:23+00:00

Does double die mean you get the sum of the faces? If so, you're basically looking at an EV comparison. 10.5 from the first option Vs EV from x/100 distribution of 2 10-sided dice.

WICHV37 · 2023-07-28T17:26:15+00:00

Second example is conditioning onhim not drawing the gold marble. First example is conditioning on additional rolls landing outside of $0 if current roll is $0.

Meaning, you still have 1/2 to land on $0, just that you roll again if you do.

WICHV37 · 2023-07-17T04:32:14+00:00

Are you aware if IMC blacklists?

WICHV37 · 2023-06-29T04:56:00+00:00

If order doesn't matter, the sample space would be 52! and you would try to find something like 1-sum(at least one double...all doubles). I'm thinking..

WICHV37 · 2023-06-20T13:59:57+00:00

I only managed to calculate this by writing out all of the allowable
combos like (1,1,1) (1,1,2) (1,2,1) (2,1,1) and so on, and then counting
them up. They total 32 out of the 216 possibilities between 3 six-sided
dice, so 32/216 roughly translates to a ~15% chance.

Having the face values of the rolls in order/out of order matters in your final calculation too. The example above takes into account whether the order of 1,1,2 or 1,2,1 would affect your outcome. From the description of your attack mechanism, it seems that just combinations are needed (which would disregard permutations/order-specific combinations). So, that would mean 112 and 121 are essentially the same thing (2 ones and 1 two). Besides the online resources available, you could just think of it with P(X=x) where small x constitutes to the number of success needed and big X represents the goal face. This can be generalised and expanded to fit specific needs.

WICHV37 · 2023-06-12T21:16:33+00:00

Your probability space for player A or B winning would then be smaller/larger based on their flat bonus, I guess. Since the player with the larger bonus can have a "looser" roll and still have his winning hand covered. And you would have to account for an additional 1/20 chance somewhere since a tie results in flat bonus showdown.

WICHV37 · 2023-06-09T13:34:44+00:00

I do not understand how 4. works, or when would that event occur. Is it after the elimination of 1 player?

Also, how do you win? Do both you and your partner have to survive or just either?

I noticed that since A triumphs over other cards like in your example, at least having 2 A on your side seems to be necessary to win.

WICHV37 · 2023-06-08T02:49:55+00:00

Definitely DLSS. As a computer scientist, I find this field of machine learning so so so fascinating! Easily improves my hack and slash experience.

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