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Forum ยป This math problem has got my whole office debating each other holy **** (hard)(Srs)
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Thread: This math problem has got my whole office debating each other holy **** (hard)(Srs)

post 1461468153 09-09-2016, 10:20 AM
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Originally Posted By rootcon⏩
Bruh you need to go back to week one of a probability course where they cover dependent and independent events, there's a big difference.
Are you literally a retard?

From that page:

Best thread: http://forum.obnoxiousbrutes.com/showthread.php?t=168274783
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post 1461468173 09-09-2016, 10:20 AM
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A = Sarah
B = Mindy
C = Both
D = None
S = Output

post 1461468193 09-09-2016, 10:20 AM
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Let me break it down for you cucks.

You walk into the office every day for 100 days. 95 of those days, you won't get a blowjob from Sloot 1. Of those 95 days, 90.25 (95*0.95) days will be without blowjobs from Sloot 2. That means you're likely to get dome 9.75 days out of every 100 days you go into the office.

It can't be 5%, because what if Sloot 1 would suck you off day 1, 2, 3, 4, 5 and Sloot 2 sucks you off day 6, 7, 8, 9, 10?

It can't be 10% because what if booth sloots are programmed to blow you day 1, 15, 65, 72, and 91?

The correct probability must account for all possible scenarios, including the two above, ie. 9.75%.
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post 1461468253 09-09-2016, 10:21 AM
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Originally Posted By BigPoppaPumpin⏩
Method 1: Sarah BJ + Mindy BJ - Sarah and Mindy BJ...
I guess I get it, essentially we subtract Sarah and Mindy BJ in method 1 because we don't want to double count, whereas method 2 avoids that part on purpose and then adds it in at the end. Fuk probability, thank god i'm done with it.
post 1461470513 09-09-2016, 10:35 AM
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Even though there's a lot of bull**** in the thread to wade through, I think that on the whole it has made the world a better place.

Peace be upon you
post 1461473863 09-09-2016, 10:56 AM
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Let A=probability you get a bj from SarahONLY
Let B=probability you get a bj from MindyONLY

Then, P(A) = P(bj from sarah but no bj from mindy) = P(.05)*P(.95)= 0.0475
Likewise, P(B) = P(bj from mindy but no bj from sarah) = P(.05)*P(.95)= 0.0475
In addition, P(A and B) = P(A)*P(B) = 0.095. Note: we assume independence for these events because it wasn't explicitly stated otherwise.
Therefore, P(at least one bj) = P(A or B) = 0.0475+.0475-.095 = 0

There is ZERO chance that any miscer will get a BJ!!










JK: This is a really simple problem if we recognize that getting a bj from Sarah is not dependent on on getting a bj from the Mindy. Therefore, assuming independence

Pr(of at least a bj) = 1 - Pr(no bj) = 1 - 0.905 = 0.095 or approx. 10%

Source: Actuarial Science master race crew.
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post 1461480263 09-09-2016, 11:39 AM
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Well, 4 pages later I see why this rustled your whole office OP. After grade school people largely forget these math lessons.

I'm not sure of the answer myself but it seems like it is 5%. Probability of both is not related to the question of a single act occurring that is a different question entirely.
Posts are for fun, not to be taken seriously or as truth.

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post 1461481253 09-09-2016, 11:47 AM
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Why is this thread still going? I have already posted that this problem is not solvable without knowing how many girls there are that could POSSIBLY give a bj.


Is it just the 2 that op mentioned? 10? 100? The entire female population?

Until that parameter is known, there is no shot of solving this. If it is just the two girls? There is a 10% shot. If there are 10 girls possible? It is about 1%. 100 girls? .001%. And so on.
post 1461481323 09-09-2016, 11:47 AM
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10%
post 1461481633 09-09-2016, 11:50 AM
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Originally Posted By AmeriSwissBro⏩
Why is this thread still going? I have already posted that this problem is not solvable without knowing how many girls there are that could POSSIBLY give a bj.


Is it just the 2 that op mentioned? 10? 100? The entire female population?

Until that parameter is known, there is no shot of solving this. If it is just the two girls? There is a 10% shot. If there are 10 girls possible? It is about 1%. 100 girls? .001%. And so on.
Are you a dumbass liberal lawyer? Serious question. Only a lawyer would read a math problem so literally and inject some unknown variables, when clearly the problem means the chance for those two girls only.
Best thread: http://forum.obnoxiousbrutes.com/showthread.php?t=168274783
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post 1461481943 09-09-2016, 11:52 AM
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You lying sack of ****, you don't work in an office

Only a group of tradies inline at the porta potty would **** up that pre high school math
post 1461482303 09-09-2016, 11:54 AM
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Originally Posted By SippinDumbass⏩
I say look at at it like this:

You are playing the blowjob lottery, and each one of the chicks is a separate ticket you've bought.

With just one ticket (chick) you have a 5% chance of winning, but with 2 tickets (chicks) you have a 10% chance of winning.

So I say 10%
Lottery tickets are not like that.

If you have 2 tickets you have 2 chances at 1:145,895,382

Not 2:145,985,382
post 1461483183 09-09-2016, 12:01 PM
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Originally Posted By Bama1970⏩
Addition Rule 1: When two events, A and B, are mutually exclusive, the probability that A or B will occur is the sum of the probability of each event.

10%



2.5%
The events are not mutually exclusive, because that would imply that you could not get bj's from both girls, when you can.
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post 1461484283 09-09-2016, 12:07 PM
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Originally Posted By dp51134⏩
Let A=probability you get a bj from SarahONLY
Let B=probability you get a bj from MindyONLY

Then, P(A) = P(bj from sarah but no bj from mindy) = P(.05)*P(.95)= 0.0475
Likewise, P(B) = P(bj from mindy but no bj from sarah) = P(.05)*P(.95)= 0.0475
In addition, P(A and B) = P(A)*P(B) = 0.095. Note: we assume independence for these events because it wasn't explicitly stated otherwise.
Therefore, P(at least one bj) = P(A or B) = 0.0475+.0475-.095 = 0

There is ZERO chance that any miscer will get a BJ!!










JK: This is a really simple problem if we recognize that getting a bj from Sarah is not dependent on on getting a bj from the Mindy. Therefore, assuming independence

Pr(of at least a bj) = 1 - Pr(no bj) = 1 - 0.905 = 0.095 or approx. 10%

Source: Actuarial Science master race crew.
Wrong. You haven't accounted for the 0.25% chance that both give a bj. Probability is 9.75%.

Source: Magna Cum Laude nuclear engineer.
"Suicide carried off many. Drink and the devil took care of the rest."
post 1461484613 09-09-2016, 12:09 PM
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Originally Posted By SippinDumbass⏩
I say look at at it like this:

You are playing the blowjob lottery, and each one of the chicks is a separate ticket you've bought.

With just one ticket (chick) you have a 5% chance of winning, but with 2 tickets (chicks) you have a 10% chance of winning.

So I say 10%
In a lottery there is only one winning set of numbers though, in this example there could be two sets of winning numbers (the scenario where you get a bj from both girls).

There are four possible scenarios here:

No BJ (95% x 95% = 90.25% chance of this happening)

BJ from girl 1 only (5% x 95% = 4.75% chance of this happening)

BJ from girl 2 only (95% x 5% = 4.75% chance)

BJ from both girls (5% x 5% = 0.25% chance)

if you sum up the last three scenarios (or just subtract scenario 1 from 100%) you get a 9.75% chance.
post 1461484623 09-09-2016, 12:09 PM
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Originally Posted By AmeriSwissBro⏩
You cannot answer this question without knowing how many possible people are in play to give you a bj. Is it just those two? The entire school? The entire world?

Need more info.
There is no reason to read more into the question than what is stated. You use the information provided.
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post 1461486183 09-09-2016, 12:20 PM
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Originally Posted By AmeriSwissBro⏩
Why is this thread still going? I have already posted that this problem is not solvable without knowing how many girls there are that could POSSIBLY give a bj.


Is it just the 2 that op mentioned? 10? 100? The entire female population?

Until that parameter is known, there is no shot of solving this. If it is just the two girls? There is a 10% shot. If there are 10 girls possible? It is about 1%. 100 girls? .001%. And so on.
The number of girls is irrelevant, we assume the rest of the female population has 0% chance of giving the OP a blowy, which seems like a fair assumption given the office he works in.
post 1461486303 09-09-2016, 12:21 PM
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This thread is still going? Why?
Order Will Be Restored
post 1461487263 09-09-2016, 12:27 PM
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I think we're all ignoring the stupidity of being 5% sure of giving a blowjob.

How could you even quantify being 5% sure of anything.

Sloots gon sloot
Look, i know i don't really know you and all, and i know you probably hear this like everyday, but your just so perfect to me. The few hours we talked were really great even if the convo was stale, your really pretty and chill, and your country thats just perfect :'D i really dont know how to explain it, but i think i have feelings for you somehow. i never felt like this with someone i just met, but i felt the need to get it off my chest. sorry for being all weird just idk how to explain it D:
post 1461487363 09-09-2016, 12:28 PM
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Originally Posted By ss367⏩
that would be 10 out of 200 times dumass
Lol. No it is out of 100. But it does not work that way. It is 9.75% including a .25% chance of getting both.
post 1461487473 09-09-2016, 12:29 PM
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Originally Posted By TrettinR⏩
None of these



This

Edit: .25% chance you get 2 blowjobs!
For the sake of the problem 9.75 rounds up to am even 10% dumbass.

You are a perfect example of why engineers and comp Sci guys are so lost in everyday life situations.
U
post 1461487693 09-09-2016, 12:31 PM
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0.350 %

in 4 pics of sarah and mindy
Don't hit at all if it is honorably possible to avoid hitting; but never hit softly.
post 1461487943 09-09-2016, 12:33 PM
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Originally Posted By leoslayer1⏩
For the sake of the problem 9.75 rounds up to am even 10% dumbass.

You are a perfect example of why engineers and comp Sci guys are so lost in everyday life situations.
You're literally saying 1/4% chance is meaningless. Do you work for the IRS? Did you lose $20 billion due to a rounding error?
Best thread: http://forum.obnoxiousbrutes.com/showthread.php?t=168274783
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post 1461488093 09-09-2016, 12:34 PM
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Originally Posted By TrettinR⏩
OP, what kind of office is it?
curry delivery service
225OHP/315B/420S/505D

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post 1461488593 09-09-2016, 12:37 PM
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1/20 days sarah will blow you
1/20 days mindy will blow you

1/400 chance they happen on the same day which is all that really matters.. we wait for this
post 1461489083 09-09-2016, 12:41 PM
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Originally Posted By DoItForCuddles⏩
compsci pleb reporting
95% chance Sarah doesn't give you a bj
95% chance Mindy doesn't give you a bj
Assumin independent 0.95^2= 0.9025 chance you get no bj
Giving 9.75% chance of bj

Where did I **** up
We can close this thread now.
Never settle
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post 1461491493 09-09-2016, 12:56 PM
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its 5%
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post 1461492313 09-09-2016, 01:02 PM
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the real answer is 0%
post 1461494203 09-09-2016, 01:16 PM
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This thread gives me bad memories of my probability class. I was thrilled to get 40 percents on those exams.
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post 1461509383 09-09-2016, 03:06 PM
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Originally Posted By TitanFall⏩
I'll put it in misc terms:

Sarah has a 5% chance to give you a blowjob today.
Mindy has a 5% chance to give you a blowjob today.

What's the % chance you get a blowjob today?

One camp says 10%
Another says 5%
Another says 25%

What's correct?
its

total outcomes S= suck N= No suk

Sarah Mindy
s s
s n
n s
n n

For both cases S has a 5% chance and n a 95%, and we are assuming being sucked by one does not affect other. Otherwise known as independent variables

1s

(.05)(.95) + (.95)(.05) = 0.095
9.5% chance you get 1 blow job from either sarah or mindy


2s
(.05)(.05) = 0.0025

.025% chance you get a blowjob from both sarah and mindy

Which leaves us with about 90% that we stay FA


Sauce: Just did my stats for college
Squat 405x1/405x1
Bench 305x1/315x1
Deadlift 535x1/585x1

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