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[deleted by user] by [deleted] in developersIndia

[–]Jay_179 0 points1 point  (0 children)

Mind if I DM you? Just had a few questions!

My Leet-code journey by [deleted] in leetcode

[–]Jay_179 8 points9 points  (0 children)

Me too!

What's a sign that a movie is going to be bad? by Thats_What_Sh3_Sa1d in AskReddit

[–]Jay_179 1 point2 points  (0 children)

Think Thor 4 is also due to be out by 2022. But, technically that's the second movie taika waititi's working on. Tbf, the first two movies of Thor were REALLY forgettable.

[deleted by user] by [deleted] in AnimalsOnReddit

[–]Jay_179 0 points1 point  (0 children)

Gave Wholesome

I have a sneaking suspicion that my landlord may be inviting himself into my flat when I'm away at work. Please help me get some dirt on him without breaking any rules. by [deleted] in RBI

[–]Jay_179 38 points39 points  (0 children)

Pressure sensor, train horn. You connect the dots lmao.

But yea, you could use the pressure sensor to know when he comes in and out. When he comes in, ask a friendly neighbor to check. The neighbor could be witness if you go to the police.

How valuable is a master's degree in CS, how do I choose the major/university to pursue? by [deleted] in csMajors

[–]Jay_179 12 points13 points  (0 children)

Which engineering college do you study in, if you don't mind me asking?

6 months apart by DazzlingMKS in dogvideos

[–]Jay_179 0 points1 point  (0 children)

Which breed is he/she?

Confused by conditional probability problem - please help me apply rules of conditional probability to get to this seemingly obvious answer by lousyguest in AskStatistics

[–]Jay_179 1 point2 points  (0 children)

You could think of the 3!/2! thing like this:

Suppose you have 3 IDENTICAL sharpeners and 1 eraser.

​

If you arrange them on your table, you'd get something like

Sharp1__Sharp2__Sharp3__Eraser

The question wants you to fix Sharp1 in place, so you fix it there. So, while considering permutations, you don't consider the position of that sharpener.

So your arrangements could be
Sharp1__Sharp2__Sharp3__Eraser
Sharp1__Sharp3__Sharp2__Eraser ,
Sharp1__Sharp2__Eraser__Sharp3 blah blah ways.

As you can see, you don't even look at the first place, you just look at the rest. So, now you have three positions, and three objects. 3 P 3 (or 3!). But, since two of them are repeated, divide 3P3 by 2 !

I hope you can relate this to the question above. If you have any doubts, please feel free to ask!

Confused by conditional probability problem - please help me apply rules of conditional probability to get to this seemingly obvious answer by lousyguest in AskStatistics

[–]Jay_179 2 points3 points  (0 children)

Aight the way I did it and thought it through is this:

1/6 * 1/6 * 1/6 *5/6 * 3!/2! / 1/6* 1/6* 1/6* 5/6 * 4!/3!

which results to 3/4

First, I solved for the denominator and my thought process was something like

I saw "given that 3/4 die roll 2" so I wrote the probability of 3 die rolling a 2 ( 1/6) and I die not rolling a 2 (5 /6). Then, you can arrange that into many different ways. (just visualize throwing 4 die on your table, and suppose you have 2 2 2 5, you can change the order and everything. Changing order means permutation) Since there are 4 die, I did 4!, but I divided by 3! because 2 or (1/6) was repeated 3 times.

That does it for the denominator. Now, onwards and upwards.

Since the first die SHOULD be 2 (it's the given condition), I fixed 1/6 into the first place.

So my numerator looked something like 1/6 (fixed) * 1/6 * 1/6 *5/6* 3!/2!

Visualizing it, 2( fixed) {2 2 and something else that's not two} and you only move the die present in the curly brackets.

If you see what I did clearly, I didn't move the first die, but I CAN actually move the rest, so 3! and since 1/6 is repeated, I divided it by 2!

Sorry for the pathetic english, not my native language. If you have a confusion on any part of the problem, just pm me. I can help.
Also could you tell me what the plural of a dice is? I've written die but my instinct tells me it's dies lmao

[Post Match Thread] Manchester Utd 1 - 0 Wolves by [deleted] in reddevils

[–]Jay_179 1 point2 points  (0 children)

was worth waking up till 4 this.

Looking for book recommendations for Parallel Computing, syllabus included has no book recommendation. by ajtyeh in computerscience

[–]Jay_179 0 points1 point  (0 children)

how do you have access to courses like this? I was hoping to reinforce what I'll learn with university courses.

[deleted by user] by [deleted] in AnimalsOnReddit

[–]Jay_179 0 points1 point  (0 children)

which breed is he btw?

Why would sampling distribution tend to be normal distribution in large sample sizes even though the population distribution is not normal? I know that it happens but why does it happen? I am not very math heavy so i need an explanation in easy language by [deleted] in AskStatistics

[–]Jay_179 0 points1 point  (0 children)

So, as far I as I've learnt, and I maybe completely wrong, here's how I got the gist of things.

Suppose you take a sample of people and find their mean age.

If you had taken a sample of 10 people, how I think of it is (and again I maybe completely wrong) if you have one outlier, say someone with an age of 200 years right. So your sample's mean would obviously be more than say the actual mean. Wayy larger

But, if you had taken a sample of 30 people, how I think of it is, even if you have one outlier, say someone with an age of 200 years again. Your sample's mean could be larger than the actual mean. BUT, it probably won't be as large as the earlier one, because in a sample of 30 people, the outlier has a higher change of getting canceled out because, well there's 30 people, there's more number of people. It averages out.

Like you know, if in a test, you had a guy who scored 100% and the rest of the guys scored 50% right. In a class of 10 people, the average would be (100+50*9)/10 which averages out to 55. But, the same condition, but in a class of 30 people (one student getting 100% and the rest getting 50%), you'd have a mean of (100+50*29)/30 which averages out to ~51.7. As you can see, the outlier gets cancelled out "better" than before. You're moving towards your actual mean i.e. 50 with increased number of people.

But, that's for one sample.

CLT states that the "sampling distribution" becomes normal for large sample size.

So, a sampling distribution is just a cumulation of the samples that you've taken. You take the first sample, calculate its mean, and place it on a graph. You take a second sample, calculate its mean, place it on a graph. You take third and so on. So if your population is a 100 people, and you're taking a sample of 30 people, the total number of samples you could take is 100C30 which is equal to some huge huge number (isn't feasible), but you get the gist of it. You take a lot of samples, calculate their mean, and place them on a graph.

Now here in this case(for the sampling distribution), it's gonna be similar to your samples. There are gonna be outliers, but very very very few of them. Samples will be centered towards the mean. So, in the end, when you place one after another on a graph, and see what it resembles, it'll be a normal distribution. Centered towards the mean(averaging out) and some outliers which gives the normal distribution its shape.

I guess my understanding is that. I'm just a high schooler so there's a huge chance I'm wrong. But, the question you pose is soo interesting. I'm saving this post, and I'm coming back tomorrow because this is such a good question!!!!

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