IQ TEST GENERAL: /pol/ edition

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>>140285806

>>140285806
420%

>>140285806

Shit, I have no clue. I felt my brain melting just reading that. and I have an IQ of 140. Then again, my verbal IQ is way higher than my spatial IQ, so my brain just fries when I see problems like this

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>>140285806
It's easily solved by using a graph.

>>140285806
25%?

>>140285806
42

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>>140285958
Try this one. pic related

>>140286083
You're on the right path

33%

Only matters when the lazy 15 minute window arrives but idk, that's just how my half drunk brain sees it

>>140285806

50%, it either happens or it doesnt

>>140285806
15%?

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>>140285806
>>140286083
If I didn't fuck it up, it's 50%.

>>140286166
A lot...

42?

>>140285806
I put it at 55.6%

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>>140286404
FUCK
wrong

It's 25%.

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>>140286237
No.

>>140285806
25%, the first group has to arrive between 0 and 15 minutes to arrive if they want to meet the second group.

If they are late:
Group A arrives after 30 minutes of meeting beginning. Group B needs 30 minutes to get there too. 15 minutes later Group A leaves and Group B still needs 15 more minutes to arrive.

43.75%

>>140285806
>It has been awhile since we have had one of these
We shouldn't have these ever, faggot
IGNORE ALL SHILL THREADS
>in all fields

>>140286404
>>140286536
Fuck, idk I messed something up.

>>140286166
364 I think

>>140285806
~43%?

>>140286166
455-364+2=93 ways to get aids

455-93=362 ways to not get aids

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>>140286823
How many times will you fags post this same copy pasta? There is more to life than muh [current conspiracy that aligns with my political views]. Learn to live a bit man.

Pic related is (You)

>>140286166
2,184

>>140285806
doesn't matters. You kill the ones in front of you and then you hunt down the others.

>>140286166
364, an 80% probability

>>140285806
what time does the train arrive at the apples Jill has left.

>>140285806
50%

>>140286166
C14,3 = 14!/3!(14-3)!
14!/3!(11)!
14*13*12/3!
2184/6 = 364

>>140286777
The trips of truth

>>140285806
71.67%

>>140285806
is it 1:8? 15 min is 1/4th of hour, then square that.

you should be prepared regardless. it makes no difference what the probability is; you should be there waiting regardless

1/4

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Now, let's see the roo fucker edition. Protip: roofuckers call the number!

>allmpairs of delays are equally likely
Wtf does that even mean

Kys OP
There can be one one delay between the first and second, how can there be A PAIR of delays

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>>140285806
0.4333...

If one group arrives from 15 to 45, the probability of meeting the other is 1/2, and from 0 to 14 and 46 to 60 it's from 15/60 to 29/60.

1/2*1/2+(1/30*1/60*(sum from 15 to 29)) = 0.4333...

>>140288201
This is what i thought as well. The question is flawed.

>>140285806
>20%
Here's why it doesn't matter which group arrives first. If the first group arrives exactly 1 hour later, they wait 15 minutes, so there are 75 minutes in play. 15 minutes of that 75 minutes will be covered by the group that's first to arrive. 1 in 5 chance or 20%

>>140285806
14.88%

>>140288296
Close but they don't have to arrive exactly on the minute

>>140285806

14/88 %

70%. You have to account for 0; so, their are 5 possible arrival times, and each arrival time has 2 available overlaps, add up possible logistical intersections under each variable, the divide the two.

>>140287694
this

>>140286166
>>140285806
Shit like this is why I love /pol/, qe learn math and probability through how not to get fucking AIDS
>captcha: Goyette
>(((Goy)))

>>140286166
Whatever 14*13*12 is

>>140286563
if group a arrives at :30 group b can arrive any time between :15 and :45

>>140285806

3/8


So first time is equally likely for all, don't have to worry about it just integrate over it

Second time needs to be +=15 mins of first time,
from 15 min to 45 min there is a 1/2 chance the constant distribution second time will show in the 30 min window available. 1/2 x 1/2 area under the graph = 0.25 = 8/32
From 0-15 and 45-60 there is a shrinking amt from 1/2 to 1/4. integrate .25 + x from 0 to 0.25 gives .09375 = 3/32

3 + 3 + 8 = 14/32 = 0.4375 chance.
Someone tell me how I'm wrong

>>140289816
oops not 3/8 that was just my guess

All probabilities are possible; Schrödinger.

>>140285806
6.25%?

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>>140288389
Sure, seems like it does make a difference. Then the integral of the probability function from 0 to 1.

(integrate from 0.25 to 0.75 0.5) + (integrate 1/60*x + 15/60 from 0 to 0.25)*2 = 0.376042

It's 50%, pic related

>>140285806
15/32

>>140285806
43.75%

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>>140285806
im have ascended into the realm of the gods

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>>140286166
Number of total combinations minus number of infected combinations.

(15 choose 3) - 14*13 = 273

>>140290426
You forgot to divide by 2 for the area of triangles not squares

>>140289816
>>140290221

Seems like a continuous distribution is fair. If we're asking for exact answers then the group should leave after exactly 15 minutes

>>140290490
can you show work ur 1/32 off mine im curious
>>140290591
noice

>>140288179
do we have any information on the probability of a number being valid? any information on the probability of an invalid number giving a busy signal?

>>140290843
valid meaning some valid phone number, not the centrelink number

>>140290804
There are two of them, .125 is the total area of both triangles.

>>140290967
No it isn't it's 0.0625

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>>140285806

fuck all this nonsense, my adderall's worn off

>>140290967
.25 is the hypotenuse of the smaller trianges so .25 * .25 is the area of the larger square and you gotta divide

>>140285806
100%

>>140285806

The obvious answer seems to be 25%. That seems to simple, though, for an IQ test.

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>>140291211
>>140290967
>>140290426
You're right, just made a mental math mistake calculating the area, corrected version here.

>>140290221
>integrate 1/60*x + 15/60 from 0 to 0.25

Disregard this.

>>140289816

>integrate .25 + x from 0 to 0.25

is correct

>>140288179
Given no other information I'll just go with 50% and assume that the busy signal is a red herring.

>>140291283

no, the endpoints mean that it's a trapezoid looking thing

Some fresh OC for /pol/

75%

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>>140291936
Made a mistake on the subscripts. Here is the correct version.

>>140285806
id just camp in my window and wait

duh

>>140285806
25%
both of the groups would have to show up within the same 1/4th of an hour.

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>>140288179
Assuming the Centrelink number is always busy, this is a simple application of Bayes' theorem.

P(A|B) = P(B|A) * P(A) / P(B)

A = the Centrelink number is 1800050004
B = the number dialed is busy

P(B|A) = 1, from assumption
P(A) = .5, Holden is 50% sure the number is correct
P(B) = .01*.5 + .5*1 = .505, the probability the number dialed is busy, that is, the probability Holden dialed a random number and got a busy tone plus the probability Holden dialed the correct number and got a busy tone

Then applying the formula, P(A|B) = 99.01%, or as a fraction 100/101.

>>140285806
Let's say that the first group arrives at 4:00pm
For group 2, their arrival will be 4:00pm, 4:01pm, 4:02pm, ect till 5:00pm. Since they include 4:00pm or 0 minutes, the chance they'll arrive every minute is 1/61st chance of happening. The first group will leave in 15 minutes after arriving so they'll leave at 4:15pm or 4:00pm+15 minutes.
Visually, it should look like
>Arrive
>Not>Arrive
>Not>Not>Arrive
And so on. So it should be a 16/61 chance that you can shoot mudskins all at once without waiting in between.

0%. I'd open up as soon as I see the first group.

>>140292817
NO! This is a PhD (maybe masters) level problem that requires more sophistication than your typical Bayes Rule application. Hint: what does your posterior distribution look like? What does that mean in this context?

>>140285806
20%?

>>140293146
61st minute doesn't exist in the problem that's after the hour. You can start counting at 0 or at 1 it's 60 mins regardless
If one group arrives after 15 minutes it's possible the other showed before, which changes chances.
If the first group to arrive arrives at minute 59 they still leave at minute 60

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Let's go with an old but classic version. pic related

Most should be able to solve THIS!

>>140292499
1 - (2/3)*exp(-t)

>>140294050
The coin in the better option

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>>140294050
hanako is a cute but emi is best katawa

>>140288179
1 out of 999999,9999 assuming all the numbers are used?

>>140293702
Could you explain how it's more complicated than that? We're talking about a discrete space and are only interested in the probability mass at a single point. It doesn't matter what the prior or posterior distributions look like at the other points, so long as each of them has a 0.01 probability of being busy.

>>140294050
P(win coin game) = 1 - .5*.5 = .75
P(win dice game) = 1 - (2/3)^3 = 19/27 = .7037

>>140294050
Coin, google gambler's fallacy if you think otherwise

>>140294472
oh, I plugged in lambda_1 = 1 and lambda_2 = 2

>>140294863
>>140294922

depends on how much faith you have in holden's belief of a 50% chance

>>140290818
Yeah I made an arithmetic mistake, it's 14/32.

From 0 to .25, take the integral .25-x, from .25 to .75 integrate x, from .75 to 1 integrate 1 - x.

so 3/32 + 8/32 + 3/32 = 14/32.

>>140294050
The correct answer is give waifu all the cake she wants on her cakeday.

>>140295154
no yeah it's just 50% guess i got tricked

>>140295763

only if you completely trust holden's judgment.

Let’s say the first group arrives at x, where x is in (0, 60) minutes

Then the probability the second group arrives within fifteen minutes is

Max(1, 15/(60-x))


To find the probability we just have to integrate over (0, 1) and divide by 60 (integral length) to get the average probability.

The max function takes the value of 1 when x is 45 or above, so its the integral of 15/(60-x) from 0 to 45, plus 15, all divided by 60.

The result is approximately 59.6%.

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>>140286777
>>140287445
>>140290843
>>140290940
>>140291735
>>140294472
>>140294736
>>140295013
>sees flag
>sees responses
I think I know you personally.

>>140285806
What formula do I use to solve this quickly without charting a graph?

>>140292817
>P(B|A) = 1, from assumption

how?

>P(A) = .5, Holden is 50% sure the number is correct
use this prior at your own risk

>P(B) = .01*.5 + .5*1 = .505, the probability the number dialed is busy, that is, the probability Holden dialed a random number and got a busy tone plus the probability Holden dialed the correct number and got a busy ton

this is given as .01

>>140294050
The coin flip is the superior option. It is 50/50

>>140296004
I highly doubt this is possible. 3 boxes have 5 doors. 5 doors means you can exit the room twice and enter the room twice, plus one extra exit/enter provided that you start/finish in that room.

There are 3 rooms with 5 doors, and only 2 of those 3 rooms you can choose to start/end in... so you'll always have an extra door at the end.

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>>140297211
>>>/sci/

>>140296004
Easy, there are no doors there.

>>140296910
>>P(B|A) = 1, from assumption
>how?
In absence of other information in the question, this seems reasonable given how I could see it being a joke among Aussies that the Centrelink number is always busy.

>>P(A) = .5, Holden is 50% sure the number is correct
>use this prior at your own risk
Presumably the author of the problem wanted a clean solution. In absence of other information of how much the reader trusts Holden, 50% is all we can go with, otherwise all answers will differ depending on each poster's prior.

>>P(B) = .01*.5 + .5*1 = .505, the probability the number dialed is busy, that is, the probability Holden dialed a random number and got a busy tone plus the probability Holden dialed the correct number and got a busy ton
>this is given as .01
It's .01 for an arbitrary number, but the suspected Centrelink number was dialed, not an arbitrary number. P(B) = P(A ∩ B) + P(~A ∩ B) = .5 + .005

>>140292499
here:
1 - (lambda_2/(lambda_1 + lambda_2))*exp(-lambda_1*t)

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>>140296004
>>140297722
You never said I couldn't draw the line through walls...

>>140298002
>In absence of other information in the question, this seems reasonable given how I could see it being a joke among Aussies that the Centrelink number is always busy.

haha, i'm not familiar with centrelink, so fair enough. it makes more sense now.

>>140298518
>haha, i'm not familiar with centrelink, so fair enough. it makes more sense now.
Nope. Some Bayesian sophistication is required.

>>140299034

are you allowing the "number is always busy" assumption?

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>>140296004
>>140297722
>>140298491
The line doesn't need to both enter *and* exit. It just needs to go through the doors.

>>140299472
And I just realized I missed one. I'm a fucking dumbass

>>140299034
I think there may be something lost in translation between how you intended that problem and how people are interpreting it.

>>140299472
>so autistical he doesn't realize his own 'tism

>>140299472
you missed one in the middle

>>140298002
>It's .01 for an arbitrary number

i.e., P(B).

otherwise P(A|B) means "probability of centrelink number is _ given that centrelink number is busy" = P(A) under your first assumption

>>140299472
Let me reword

To go through all of the doors in a room if there are 5 doors:

2 doors must be used for exiting the room
2 doors must be used for entering the room
The remaining door must be used to exit/enter the room.

This means that 5 door rooms must start/finish in the 5 door room to use all 5 doors.

But there are three rooms with 5 doors... so you'll always have a 5 door room which you didn't start or finish in with only 4 of the 5 doors used... like you did in your attempt.

>>140299940
B = dialed number is busy, not that any given number is busy

P(A|B) is probability the centrelink number is ### given ### is busy. It's NOT the probability the centrelink number is ### given the centrelink number is busy.

>>140294050
The better games is the dice one. She "wins" the cake and gets to eat it without feeling guilty for not letting me have it.

>>140286777
This is what I got too, is it right?

>>140300313
>B = dialed number is busy

which was the centrelink number.

>>140300912
No, we're not absolutely sure it's the centrelink number. Our prior is that it's the centrelink number with probability .5, and based on the fact that it's busy we update our prior to .9901.

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>>140285806

Odds are 50% of meeting for all values between 15 and 45. Odds are 25% at 0 and 1 hour.

Uggggh, I'm sure there's a good equation for this; but I haven't studied probability in 20 years so I'll have to just bodge a method together.

Average of 50% and 25% is 37.5% so that should be the average probability of the values between 45 and an hour and 0 and 15.

So 50% odds of 50% and 50% odds of 37.5%....87.5/2 = 43.75%

So 43.75% odds?

>>140301239

oh, duh. that makes sense.

>>140293702
>Hint: what does your posterior distribution look like? What does that mean in this context?

concerning P(B|A), just because we dialed the number and it was busy doesn't mean this is =1.

is that what you mean?

>>140302693

i.e. dialing the number gave us a single data point.

>>140285806
0.4375

>>140287445
Answer to the q is correct but probability is actually lower. You forget both maja and sven will be servicing the bulls.

>>140288179

gonna be pissed if i look in the archive tomorrow and OP hasn't shown the answer.

>>140294000
1. 4:00
2. 4:01
3. 4:02
4. 4:03
5. 4:04
6. 4:05
7. 4:06
8. 4:07
9. 4:08
10. 4:09
11. 4:10
12. 4:11
13. 4:12
14. 4:13
15. 4:14
16. 4:15
17. 4:16
18. 4:17
19. 4:18
20. 4:19
21. 4:20
22. 4:21
23. 4:22
24. 4:23
25. 4:24
26. 4:25
27. 4:26
28. 4:27
29. 4:28
30. 4:29
31. 4:30
32. 4:31
33. 4:32
34. 4:33
35. 4:34
36. 4:35
37. 4:36
38. 4:37
39. 4:38
40. 4:39
41. 4:40
42. 4:41
43. 4:42
44. 4:43
45. 4:44
46. 4:45
47. 4:46
48. 4:47
49. 4:48
50. 4:49
51. 4:50
52. 4:51
53. 4:52
54. 4:53
55. 4:54
56. 4:55
57. 4:56
58. 4:57
59. 4:58
60: 4:59
61. 5:00
These are all the possible times group 2 can show up mate.

>>140286083
Simple. 1 - (15C3 / 1
>>140307516
but i thought time is continuous variable. Can you estimate using discrete like that?

>>140285806
18.75%

61.666666667%

>>140285806
The two groups will add up to 6 women, one old man and two hundred journalists. Not much kebab to remove

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>>140311810 was just me shitposting
Actual solution:
0≤y≤x≤1
Find the probability that x-y≤.25
In the range y=0 through y=.75, x=25/(100-10y)
In the range y=.75 through y=1, x=1
The probability is (75n+25)% where n is the average of 25/(100-10y) in the range [0,.75]
The average value of a function is f(avg)=[1/(b-a)]∫f(y)dy on the interval [a,b]
n=f(avg)[25/(100-10y)]=[1/(.75-0)]∫[25/(100-10y)] on the interval [0,.75]
n=(4/3)∫[1/(-.4y+4)] on the interval [0,.75]
∫[1/(-.4y+4)]=(1/-.4)ln|-.4y+4|=-2.5ln|-.4y+4|

∫[1/(-.4y+4)] on [0,.75] = -2.5[ln|-.4(.75)+4| - ln|-.4(0)+4|] = -2.5[ln3.7 - ln4]
n=(4/3)(-2.5)[ln3.7-ln4]
Probability=[75(4/3)(-2.5)(ln3.7-ln4)+25]%=[250(ln4-ln3.7)+25]%≈44.490385367%

I haven't used calculus since I was 17 and I'm mid 20s now, I hope I remember enough to do this right.

>>140285806
Wouldn't this answer depend on how long the first group's delay was?

If the first group's delay was 45 minutes, and they were going to wait 15 minutes, then the answer would be 100% probability they would meet the second group, because it would arrive within those next 15 minutes.

>>140315554
Obviously, you fucking idiot, that's the whole point of the question. It would be fucking 5 year old tier if we knew the first group's delay.

>>140294050
Assuming the cake is vegan.

First game: Probability is 75%.
Second game: Probability is (1/3) + (1/3)(2/3) + (1/3)(1-2/9-1/3)=(1/3)(4/9)
Probability is 1/3+2/9+4/27=19/27
I did that really quick and didn't proof check my work so I don't even know if I got that right. But assuming I did, the coin is a better option.

>>140316495
And yes I did this the long way, I don't give a fuck, I'm almost asleep

>>140315551
Bit off, see >>140291548 or >>140301948. Not sure where the mistake is since I don't feel like looking through all your algebra.

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>>140296004

>>140285806
43.75%

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>>140285806
21
https://www.youtube.com/watch?v=BzVXbeASRiQ

>>140294050
>not letting your waifu just eat the cake
You don't love her anon. do you?

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>>140296004
Based line through the doors

>>140316961
Wow breddy gud

>>140287257
>knees lock
>not going below 90
disgusting

>>140286166
Assuming the order doesn't matter, 364
If the order does matter, 2184

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Try an actual IQ test on for size.

>>140285806
Too lazy to write anything down so gonna guess 7/16

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0% I'd kill them as they come.

>>140285806
40%?

I used a 5 number binary system where each number represented getting there at 0,15,30,45,60, there were 10 options possible, 4 were next to eachother.

>>140286166
There's only one good way to do this and it's shoot the refugee that has AIDs. Fuck it, might as well shoot the whole lot to be sure you get the right one.

>>140285806

The right answer for this one is to plant a bomb at the meeting site. You tip the FBI off about one group while they're en-route to the meet and detonate the other bomb remotely when the second group arrives. Everyone will assume it's a suicide bomb and the news will hush up about it while the feds shuttle their group of Muslims to Guantanamo Bay and you get off scot free.

>>140318212
2 ez

its the square with the white dot under it