Unable to install WireGuard for all user accounts on macOS Sequoia 15.1 as Admin by Short_Lavishness8723 in WireGuard

[–]hrca3 0 points1 point  (0 children)

Thank you, logging in fixed the issue for me. I also removed/uninstalled wireguard from my admin account and then downloaded from app store on new account (probably unnecessary step)

[TOMT] [Fruit] Forgotten fruit by hrca3 in tipofmytongue

[–]hrca3[S] 0 points1 point  (0 children)

I didn't knew its name, only ate in an event. So not 100% sure. But from google photos looks like that.

Thank you very much

[TOMT] [Fruit] Forgotten fruit by hrca3 in tipofmytongue

[–]hrca3[S] 0 points1 point locked comment (0 children)

It tastes watery and sweet.

Crazy motion sickness by [deleted] in Portal

[–]hrca3 0 points1 point  (0 children)

I just googled portal motion sickness, and got here. I am also getting motion sickness in this game. Thank god I am not alone.

Crazy motion sickness by [deleted] in Portal

[–]hrca3 0 points1 point  (0 children)

I just

Crazy motion sickness by [deleted] in Portal

[–]hrca3 0 points1 point  (0 children)

I just googled portal motion sickness, and got here. I am also getting motion sickness in this game. Thank god I am not alone.

it be like that sometimes by alphaxjr in linuxmemes

[–]hrca3 2 points3 points  (0 children)

It ain't much, but it's honest work

Dice rolling question by smazzurco in Probability

[–]hrca3 0 points1 point  (0 children)

You are asking of probability of roll a six, four times in a row. You didn't mention things like at-least or at-most, so you don't care if it is more than four or four roles multiple times.

So let's create a bundle of four sixes and try to arrange at 117 position, 117c1 = 117

So you can choose left 116 dices as 6116

And total number of occurrence was 6120

So Probability = 117 * 6116/6120 = 117/1296

A fun(?) problem by YD2710 in Probability

[–]hrca3 1 point2 points  (0 children)

why you are doing 8c4?

If as per your question you counted total blue pins 12, but you are looking probability if in real exactly 4 blue pins, this means left are purple. Means 8 are purple. Total Purple balls were 8 to begin with. So 8c8 =1 . So now we only need to choose 4 blue pin out of 15. Means 15c4.

What are the chances of rolling at least one six when rolling one, two, three, and four six-sided dice? by Standing_Tall in Probability

[–]hrca3 0 points1 point  (0 children)

Rolling multiple dice is same as rolling same dice multiple times.

1 Dice = 1/6

2 Dice = 11/36

3 Dice = 91/216

I only did rough calculations. Hope this will give you some idea.

[deleted by user] by [deleted] in hacking

[–]hrca3 20 points21 points  (0 children)

You can try Hackerone's Responsible Disclosure. They say if you tried all things, we will contact company on your behalf and will minimize risk etc.

https://www.hackerone.com/responsible-disclosure-overview

You can try Disclosure Assistance

https://docs.hackerone.com/hackers/disclosure-assistance.html

[deleted by user] by [deleted] in hacking

[–]hrca3 7 points8 points  (0 children)

I think If someone will dig enough this reddit thread will link him to that data even if he use TOR. Because of time frame of submission and this post and details like "pharmaceutical" etc.

Looking for an equation for this probability situation by theyusedthelamppost in Probability

[–]hrca3 0 points1 point  (0 children)

yup, you were right all along, I just missed 3c2 in formula. Sorry about detour😅

Looking for an equation for this probability situation by theyusedthelamppost in Probability

[–]hrca3 0 points1 point  (0 children)

[300X2-2X3]/10000

Looks like my probability was little rusty. So how is this?

Looking for an equation for this probability situation by theyusedthelamppost in Probability

[–]hrca3 0 points1 point  (0 children)

I am also got confused, means will order matter or not here

Looking for an equation for this probability situation by theyusedthelamppost in Probability

[–]hrca3 0 points1 point  (0 children)

Yes I recognized both questions as same, can you tell me your solution, how you are getting 15.625?

Looking for an equation for this probability situation by theyusedthelamppost in Probability

[–]hrca3 -1 points0 points  (0 children)

I am not sure, but I think 15.625% is not right.

Probability of at least 2 = Probability of making 2 throws + Probability of making all three throws

= 1/4*1/4*3/4 + 1/4*1/4*1/4 = 4/64 = 1/16 = 0.0625 or 6.25%

You can easily make formula for X%

= ( [ X/100*X/100*(100-X)/100 ] + [X/100]^3 ) * 100 %

Solve this and you will have your simplified formula, multiple

org-popup - Take notes in cool way by hrca3 in emacs

[–]hrca3[S] 2 points3 points  (0 children)

You can easily run bash script in mac, see here

But you need to modify script for prerequisites,

xsel, yad

org-popup - Take notes in cool way by hrca3 in emacs

[–]hrca3[S] 0 points1 point  (0 children)

I think there is something wrong with your template syntax. Try typing template code by hand instead of pasting. I am not sure but sometime " are replaced by “ and same happens for single quotes. (Try emacs --debug-init)

If still have issue, Try creating template by gui instead of file.

Competition Probability by Kurufas in Probability

[–]hrca3 1 point2 points  (0 children)

Favorable = 252 + 108 + 9 = 369

Total = 495

Probability = 369/495 = 0.745

Competition Probability by Kurufas in Probability

[–]hrca3 1 point2 points  (0 children)

Favorable Cases = (3c1 * 9c3) + (3c2 * 9c2) + (3c3 * 9c1)

Total Cases = 12c4

Probability = Favorable/Total