Alright, let's really weed out the retards

Alright, let's really weed out the retards.

Other urls found in this thread:

en.m.wikipedia.org/wiki/Boy_or_Girl_paradox
twitter.com/NSFWRedditImage

That's retarded. If coin A is already decided, then there is only really 1 coin flip. If you flip a coin what are the chances of heads? 50/50

50%

next question

and yes, before you ask, I'm am a genius

I'm bisexual so 50%

1/3

>2 sides of the coin
>3 possibilities

wew lad

nerve gas alternate universe

its a fucking bitcoin, you can't flip it

1/2 but the original thread answer is 1/3

If A is the event that the first coin lands heads, and B is the result that the second coin lands heads, then what we're looking for is P(A∩B|A∪B)P(A∩B|A∪B). The probability of both A and B is 1/2, so A∪BA∪B is 3/4. The probability of A∩BA∩B is P(A|B)P(B)P(A|B)P(B), which is 1/2(1/2)=1/4. Therefore, the probability is 1/4/3/4=1/3.

en.m.wikipedia.org/wiki/Boy_or_Girl_paradox
^ same thing but uses genders

l2math.

more nerve gas theorems

It's always 50/50.

50%, it either happens or it doesnt

1/2

Other thread answer is 1/3

Different questions.

1/2, 50% chance.

You are a retard, this is a conditional probability problem and the answer is 0.5

a coin has 3 sides the the two main sides and the thin side between them both. it can land on the thin side sometimes but it's a bit rare

75% of 100

0 you can'r flip a virtual currency faget

sounds about right

A= Coin A
B= Coin B
h= head
t= tail

possibilities are : At+Bt, At+Bh, Ah+Bt, Ah+Bh

there are 4 possibilities so it's 25% you retards

>Im a university of michigan proffsr

>well i myself took the plasure of building thiiss simulation

1/3
For it is
1/33d

1/3
For it is:
1/33

>Being this retarded

>given that coin A lands on heads

Except, its not a question of how many possibilities you retard. ITs a question on the chance of both being the same. There are 3 different possibilities seen as At+Bh and Ah+Bt are the same outcome. One heads and one tails. So the probability is 1/3.

>given that coin A landED on heads

You read it wrong, it nevers says coin A always lands on heads, the question is basicly if i flipped 2 coins what is the probability that they both land on heads. The previous flips doesnt impact the result as the probabilities are reseted at each coin toss

Where you're under the impression you have the remaining possibilities of HH, TH and HT, you only actually have HH and HT or HH and TH. Since coin A either had to be first OR second you can't include both HT and TH.

The Wikipedia article just confirms this.

Doesn't matter.

Coin A landed on heads.

What's the probability coin B will also land on heads?

>coin A is already on heads

At isn't a possibility as is clearly stipulated.

Only Ah + Bh and Ah + Bt are possible outcomes since we know Ah is true.

it depends which side up the coins are when flipped.

Are the coins buttered? If so what side are they buttered in? Can't answer the question until we have all the details user.

>a bit rare
The sarcasm in this reply is splitting my sides

P(AnB)=P(AnB|A)*P(A)+P(AnB|Ac)*P(Ac) Since the flips are independent =P(B)*1+0*0=1/2*1=1/2
This is basic probability and can also be shown with counting

Coin B isn't a coin, though. It's data. Can't flip data, retard.