IQ TEST GENERAL: Sup Forums edition

Brainlets It has been awhile since we have had one of these. I will start making more of them now, but for now you will find OC in the OP.

...

420%

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

It's easily solved by using a graph.

25%?

42

Try this one. pic related

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

50%, it either happens or it doesnt

15%?

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

A lot...

42?

I put it at 55.6%

FUCK
wrong

It's 25%.

No.

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%

>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

Fuck, idk I messed something up.

364 I think

~43%?

455-364+2=93 ways to get aids

455-93=362 ways to not get aids

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)

2,184

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

364, an 80% probability

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

50%

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

The trips of truth

71.67%

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

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

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...

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

>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%

14.88%

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

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.

this

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

Whatever 14*13*12 is

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

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

oops not 3/8 that was just my guess

All probabilities are possible; Schrödinger.

6.25%?

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

15/32

43.75%

im have ascended into the realm of the gods

Number of total combinations minus number of infected combinations.

(15 choose 3) - 14*13 = 273

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

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

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

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?

valid meaning some valid phone number, not the centrelink number

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

No it isn't it's 0.0625

fuck all this nonsense, my adderall's worn off

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

100%

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

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

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

Disregard this.

>integrate .25 + x from 0 to 0.25

is correct

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

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

Some fresh OC for Sup Forums

75%

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

id just camp in my window and wait

duh

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

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.

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.

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?

20%?

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

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

Most should be able to solve THIS!

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

The coin in the better option

hanako is a cute but emi is best katawa

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

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.

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

Coin, google gambler's fallacy if you think otherwise

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

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

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.

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

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

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%.

>sees flag
>sees responses
I think I know you personally.

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