Probability

[livid ridge]

oh basic shit

[true lantern]

@brazen trellis do i know you from somewhere

[brazen trellis]

We're doing geometry and trigo next sesmester but as of now im teaching myself

[brazen trellis]

Where the frick were u my dude

[true lantern]

???

[livid ridge]

oh yay trigonometry

[brazen trellis]

I dmed you

You ghosted me

[true lantern]

we arent even friended lil bro

[brazen trellis]

Wait what

[livid ridge]

bro is on rider 💀

[true lantern]

currently riding 🔥

[livid ridge]

jetbrains for life

[true lantern]

w

[livid ridge]

anyway im sleeping lmk if u finished the Qn

[brazen trellis]

Alr gn

[true lantern]

gn

[brazen trellis]

huh

No i meant leb

[true lantern]

gn

[brazen trellis]

Alr...

[true lantern]

did u get it right

[brazen trellis]

Wdym

[true lantern]

the question

[brazen trellis]

Nope

Im redoing it

[true lantern]

K

[brazen trellis]

n = 7
k = 7 (more than 6 days)
p = 0.90 (90%)

x^0 = 1

,

Just gonna mark with whatever im currently doing

Cant pin

Don't really have to do much here I think

55% of 12 is 6.6, but since we're looking at whole numbers, 6.6 is rounded to 7.

So if we were to look at the odds in favor of extroverts:

Using formula: odds in favor of extroverts = # of favorable outcomes / # of unfavorable outcome

Odds against introverts = 3 / 7

Probability of finding 3 extroverts: 3/7 or simplified 0.428%

0.428% ≅ 0.43%

Same thing from Q3, this time an experimental probability.

So identify variables for formula: Experimental probability = # of successes / # of trials

Number of successes = 2
Number of trials = 10

20% of 10 is 2 so we're good. Dont have to do anything with this.

Plug in values: Experimental probability = 2 / 10
= 0.2 = 20%

[brazen trellis]

But since 20% is too vague of the probability, so we use the other formula

So identify variables:
n = 10 (total)
k = 2 (preffered variables)
p = 0.2 (20%)

And plug in values
We end up with our probability to be 0.288%, which can be rounded up to 0.3 or 30%.

0.288(28%) ≅ 0.3 (30%)

[true lantern]

bruh why math gotta be so hardj

[livid ridge]

correct

[livid ridge]

I think I made a mistake, what I meant was assume 55% of the population are introverts

so ur wrong

[livid ridge]

wrong

redo Q3 and Q4

[brazen trellis]

I dont think we can calculate with decimals since we're counting humans, but it's close to accurate

55+ 43 = 98 it's not perfect

[brazen trellis]

If you wake up, guide me through Q4 pls ty

[brazen trellis]

@true lantern idk if youre awake rn, but if you can check my work on Q3 that'll be nice

[cunning halo]

Post it and I’ll see if I can help.

[brazen trellis]

@cunning halo I've done it but it was marked wrong so wondering if anyone can check

[cunning halo]

Where is it?

[livid ridge]

your answer would be right if Jimmy wins exactly 2 prizes. But the question is atleast 2 prizes

so x >= 2

[brazen trellis]

I replied within reply

[cunning halo]

Ahh right, I didn’t see that lol

[livid ridge]

He wanted to spend his free time learning basic probability. So I'm trying to teach him how to use the binom distribution formula

[brazen trellis]

How are you awake rn

[livid ridge]

its 4 pm

[brazen trellis]

its probably 4 am

Huh

[livid ridge]

So I made some questions for him to try

[brazen trellis]

Where do you live mate

[cunning halo]

Where are the questions?

[livid ridge]

UK bro

here

[brazen trellis]

I thought you went to sleep

[livid ridge]

I went to sleep at 4 am

today

[brazen trellis]

My brain is not braining

[cunning halo]

I feel like Q3 is worded poorly. Firstly 55% of 12 is not a whole number, furthermore do you mean to say randomly choose (encounter) 3 people, and the odds of them all being extroverts?

[livid ridge]

I meant to say 55% of the global population

yes the probability that exactly 3 of the people are extraverts

[cunning halo]

Ahh ok, I think I understand now.

[brazen trellis]

I ended up with 6.6 so i assumed that was more than the introverts so i rounded to 7

[cunning halo]

Brain not working, it’s 2:50am for me rn.

[livid ridge]

but they're all the same type of question just different numbers and scenarios

you use the formula on all of them @brazen trellis

[brazen trellis]

Ye i could but i still learned a few stuff from my textbook

[livid ridge]

what did you learn

[brazen trellis]

Experimental/Theoretical Probs

Odds in favor/against

Simple formulas but since im answering these questions

[cunning halo]

would it not be 0.55^9 * 0.45^3?

[livid ridge]

I see

that doesn't really apply here

It's mostly theoretical

[brazen trellis]

Ye

wait

[livid ridge]

no

close

[cunning halo]

Did I mess up the 0.55 and 0.45?

[livid ridge]

you're missing the different combinations

[brazen trellis]

Are you looking at Q3

[cunning halo]

It’s not a combination problem is it? Didn’t you say just the odds that only 3 of the people were extroverts?

A combination problem would be saying “choose 3 out of 12 what is the chance that all three are extroverts”

[cunning halo]

Yes

[cunning halo]

Wdym?

[brazen trellis]

I didnt show any work on paper so idk how you're doing this in your head 🗿

[livid ridge]

there are different ways to get 3 extraverts and 9 introverts

they can be positioned differently

try to draw a probability tree and you'll see what I mean

[cunning halo]

If it’s a probability tree then you don’t use combinatorics.

[livid ridge]

Its hard to explain it without drawing

so for example you pick the first person

its an introvert

then you pick another person

and so on

[cunning halo]

Straight up not a combinatory question.

[livid ridge]

there r multiple ways of arranging the order that you pick 3 extraverts in total

and we use n choose k to take that into consideration

[cunning halo]

That’s not the question you laid forth

[livid ridge]

well in the question I said the probability that 3 of them are extraverts. I didnt say the probability that the first 3 are extraverts

or the last 3

[brazen trellis]

lol this funny

[cunning halo]

The question is what’s the chance that only 3 of them are extroverts, what you are saying right now is different.

[cunning halo]

That’s my point.

So my solution is correct, is it not?

[livid ridge]

isnt that the same thing

[cunning halo]

No

[livid ridge]

3 of them, meaning the rest are introverts

its the same thing as "only 3 of them"

[cunning halo]

That would be a permutation, which is what my answer is.

[livid ridge]

similar to flipping a coin

you can arrange getting 2 heads as HHT THH HTH

which is 3 choose 2

[cunning halo]

That is a permutation not a combination.

[livid ridge]

3 choose 2 is 3 wdym

$\frac{3!}{2!(3-2)!}$

[dapper marlin]

lebesgue

[livid ridge]

this is a combination formula

in this case its the same as 3P2

[cunning halo]

Yea but I’m the other question, the one we are originally talking about, we are using permutations. Unless you have forgotten a key piece of information.

The question, how I perceive it is as follows:

The given chance for someone to be an introvert is 55%. In a sample size of 12 people, what is the chance that only 3 of them are extroverts?

[livid ridge]

wait I was mistaken

permutation is $\frac{3!}{(3-2)!}$

which is 6

[dapper marlin]

lebesgue

[livid ridge]

so it is a combinations question

[livid ridge]

that is how the question is perceived yes

<@&286206848099549185> help me validate. I suck at debating

[brazen trellis]

Alr im back

Whats going on

[cunning halo]

We are getting into a discussion about whether combinatorics are needed to solve question 3.

I’m saying that they aren’t under the rules he has provided, while he says they are.

[cunning halo]

I don’t think this is a reason to @ a helper. You’d just be wasting their time.

[livid ridge]

Well it is a "question" at this point

I don't really know how to explain it to you

maybe someone who had experience teaching can

[cunning halo]

You are trying to say is that if you choose 3 random people from 12 what is the probability that all 3 are extroverts, right? Which is different from the question which asks what the chance that only 3 of them are extroverts.

[livid ridge]

no we're choose the different ways to calculate 55%^9 times 45%^3

find ill draw

hold up

[cunning halo]

That isn’t how it works.

[brazen trellis]

🥱

[livid ridge]

assume its 3 people in the party and we're looking to find 2 extraverts

[cunning halo]

.

.

[brazen trellis]

Yall wanna take a break and play chess.com or something

[livid ridge]

this is the same thing as 12 people in the party, 55% of the global population is introvert and the chances that exactly 3 are extraverts

[cunning halo]

Chess is for nerds

I’m only into math

[brazen trellis]

chess is for cool people

[cunning halo]

Haven’t played chess since winning my school tournament.

[livid ridge]

but in the drawing I made, its 3 people in the party, p is the probability that a random person is an extravert and chances that there are exactly 2 extraverts

[cunning halo]

THOSE ARE DIFFERENT QUESTIONS!!!

[livid ridge]

cus aint no way am I drawing 12 branches

its just different numbers

[cunning halo]

YOU ARENT EVEN ANSWERING YOUR OWN QUESTION

[livid ridge]

if 55% of the global population is introverted. then the chances that a random person is introverted is 55%?

I dont see the difference

[cunning halo]

This isn’t what you asked

[livid ridge]

its just different numbers

damn bro is gaslighting me

this is why I hate debate clubs

[cunning halo]

This ain’t a debate.

You don’t have to use combinatorics

[livid ridge]

lmao okay then suit yourself

[cunning halo]

Because it doesn’t solve the question

[livid ridge]

it does solve the question tho

[cunning halo]

“Q3 (Medium)

There are 12 people inside a party. 55% of the people are introverts. What is the probability that you will find 3 extraverts in the party?”

[livid ridge]

if you plug in 45% to p and 55% to (1-p) you'll get the answer

[cunning halo]

You clarified stating “yes the probability that exactly 3 of the people are extraverts”

[cunning halo]

Please wait until I explain

[livid ridge]

Okay

[cunning halo]

So continuing on, that is the question you stated. Correct?

And the clarification was initially meant to be included in the original question, so it can also be taken as fact.

So the chance that someone is introvert is 0.55

[livid ridge]

yes

[cunning halo]

And we have 12 people, correct?

Yes or no

[livid ridge]

The only clarification was "Assume 55% of the global population are introverts" I did correct that when he first answered the question

[livid ridge]

yes

[brazen trellis]

Are you really expecting 100% accuracy

[brazen trellis]

This is already good enough 😭

[cunning halo]

So we have 12 people, and we only want to know the possibility of having only 3 extroverts in a group of 12.

Now each individual has. 0.55 chance of being an introvert, and 0.45 chance of being an extrovert.

0.45^3 * 0.55^9

The order does not matter because the initial question does not ask/specify order.

[livid ridge]

its the same thing with this tho?

[cunning halo]

No

[livid ridge]

I think I get your point

[cunning halo]

The real tree diagram would have 2^12 branches I think.

[waxen pier]

This is not P(x = 3), x ~ Bi(12, 0.55) -What did you mean by this

[livid ridge]

you're trying to say "If we know that 55% of the population are introverted. What is the probability that within the 12 people there are 3 extraverts in total?"

but my question is more of "What is the probability that there are 3 extraverts when you go around asking people if they are introvert or extravert"

[waxen pier]

Well that depends on how many people you ask haha

[cunning halo]

But your question included the 55%. So what did you mean by that?

[livid ridge]

I told you, assuming that 55% of the global population are extraverts

but I think I get what you mean now

let me revise the question

[cunning halo]

12 people, chance of introvert = 0.55, chance of only 3 of them being extroverts = 0.55^9 * (1-0.55)^3

[waxen pier]

you're missing a 12 choose 3 on the front

[cunning halo]

Idk what you wanted to know

[brazen trellis]

You guys done it mathmatically while i done it logically 🗿

[cunning halo]

There’s no reason.

[waxen pier]

those two things are the same? logic is a branch of mathematics

[brazen trellis]

Idk why they debating

[livid ridge]

Q3 Assuming that there are roughly 55% introverts in the global population. What is the probability that when you randomly ask 3 people in a party of 12 people, all 3 of them are extraverts?

[brazen trellis]

98% is still accurate

[livid ridge]

you only start approximating when the numbers are large mate

12 is small

when its 1000 you dont use binomial distribution, you use poisson which is an approximation to the binomial distribution

[waxen pier]

central limit theorem kicks in at n =2 everybody knows this

[brazen trellis]

Can we move on pls

Or are we still debating

[livid ridge]

nope

[livid ridge]

this

do this

[icy tendon]

Not all non introverts are extroverts

Some ppl just chilling

[livid ridge]

lmaooo like me

[brazen trellis]

Alr

[brazen trellis]

@livid ridge wait so how come this wrong

I’m eating rn

[livid ridge]

so this is basically assuming half of the room are introverts, and then assuming that each "introvert" has a 3/7 chance of being an introvert

tbh i dont rlly understand ur logic here

[brazen trellis]

I meant extrovert

"Odds against introvert"

i think

I dont know how to put it

[brazen trellis]

Uh

Im not gonna ping you leb since youre asleep but

I did it 2 ways

Variables:
n = 3
k = 3
p = 0.45 (45%)

resending 2nd approach 1 sec

n = 12
k = 3
p = 0.45 (45%)

I attempted twice because off from reading the question it confused me what variables are which

n = # of trials
k = # of successes
p = probability

[brazen trellis]

For 2nd attempt, i put n = 12, because the question maybe reffered to out of the 12, since that's how many ppl are in the party

1st attempt I put n = 3, since it asks for the probability of 3 extroverts

So both answers from attempt 1 & 2, I got 9%

[brazen trellis]

I still think this is still accurate

[brazen trellis]

I can show calc history if needed

[true lantern]

bro which one is question 3

[brazen trellis]

Question got changed

[brazen trellis]

@true lantern @livid ridge

Check my work I finished

Leb if you’re asleep sorry

[true lantern]

100% because no introvert is going to a party

youre welcome @brazen trellis

[true lantern]

but fr though i think its just 55%^3?

but idk if 55% of people are introverts translates into 55% chance of any given person to be an introvert

[livid ridge]

in this specific case it gives a good approximation because the probability is quite uniform (55 to 45) and the sample size is relatively small

but as I said again, you don't approximate when it comes to small variables

when they give you more variables like 800 or something then you logic won't apply

but if you had to choose between good approximation and true result which one would you choose

given that both requires similar computational complexity

[brazen trellis]

When it comes to physics and probability it's good approximation

There's never perfect values

[brazen trellis]

True though

And wtf happened to my chess.com account

[livid ridge]

ye i dont see u

[true lantern]

ur just dogwater lil bro

[brazen trellis]

,

20 Y/N Q's
0.7 (Y)
0.1 (N)
0.2 (N/A)
n = 20
k = 19
p = 0.7

Couldnt figure out the variables so i just assumed and plug in values
n = 20
k = 19
p = 0.7