Wtf am i missing here?

**** it, my brains just gone, i blame it on 3:42 maths. Mod delete if i haven't edited it by the morning

in fact don't delete any more

Reply 1

DJB MASTER

ermm, sorry but what is the actual question? lol.

Reply 2

abcdefghijklmnopqrstuvwxyz

DJB MASTER

ermm, sorry but what is the actual question? lol.

tbh i don't have a ******* clue lol, i just cannot think

Reply 3

Illusionary

abcdefghijklmnopqrstuvwxyz

Maybe it's because it's 3:32 or maybe it's because i'm retarded but........

probability of king given red (pack of cards) is 1/13

so.............. P(KnR)/P(R) given by (4/52)x(26/52) all divided by (26/52)

that's right isn't it?

so probability of ace of spades given black is apparently 1/26 but if i do it the same way, i get the wrong answer. And in this example it's only got AnB as draw of ace of spades and not draw of ace of spades intersect drawing a black card :s-smilie:

as it has with the example above where P(KnR) was probability of king intersect red

EDIT: in fact, the more i think about this the more my brain hurts....................... i'm 99% sure the top part is just wrong now -.-

P(AS|B)=P(ASnB)/P(B)=P(AS)/P(B)=(1/52)/(1/2)=1/26

Your problem is that P(ASnB)≠P(AS)×P(B), as AS and B are not independent events in the same way that K and R are.

Instead, it's just P(ASnB)=P(AS). :smile:

Reply 4

abcdefghijklmnopqrstuvwxyz

Illusionary

that sentence just killed me :bawling:

why are they not the same?

independent = event of one happening doesn't affect other

so what you're saying is that if you draw black, it affects AS right? hence not independant
but then is that not the same for KR. If you draw red you still affecting King red

Reply 5

Illusionary

abcdefghijklmnopqrstuvwxyz

that sentence just killed me :bawling:

Yep.

Perhaps it's easier to come at it from the other direction - if you draw not black, the probability of getting the Ace of Spades is then zero rather than 1/52 as it was before. The first event has altered the probability of the second, hence they can't be independent.

Alternatively, consider that the number of red Aces of Spades doesn't equal the number of black Aces of Spades, but the number of red kings does equal the number of black kings. :smile:

Reply 6

abcdefghijklmnopqrstuvwxyz

Illusionary

Yep.

it's getting there........... i'm at the stage where my eye lids are getting heavy, fml -.-

thanks for the help though :p:

i'll have a read of this in the morning because if i'm going to be honest, i don't think maths is happening tonight. I'll quote you again tomorrow if it still doesn't make sense lol if that's alright?

Reply 7

abcdefghijklmnopqrstuvwxyz

Illusionary

just had another read and this confused me again just as i thought i had it :s-smilie:

my notes say here that if A and B are independent

P(AlB) = P(A)

and P(AnB) = P(B)xP(AlB) = P(B)xP(A - due to independancy)?

so does that not go against what you said? because you said AS and B are not independant

Reply 8

Illusionary

abcdefghijklmnopqrstuvwxyz

just had another read and this confused me again just as i thought i had it :s-smilie:

my notes say here that if A and B are independent

P(AlB) = P(A)

so does that not go against what you said? because you said AS and B are not independant

I said that P(ASnB)=P(AS), not P(ASlB) = P(AS)

In this case, P(AS)=1/52, but P(AS|B)=1/26, hence AS and B are not independent.

While it may not be becoming clear now, it could well be crystal tomorrow when you're feeling fresher - don't let yourself get too frustrated! :smile:

Reply 9

abcdefghijklmnopqrstuvwxyz

Illusionary

how do you go about finding P(ASnB) if they're not independent? is it just the probability of the event that is dependent ie AS in this case?

Reply 10

Illusionary

abcdefghijklmnopqrstuvwxyz

how do you go about finding P(ASnB) if they're not independent? is it just the probability of the event that is dependent ie AS in this case?

P(AnB)=P(A)×P(B|A)

You're not necessarily going to be able to get a better result than that for a general situation.

Also, you'll see that if A and B are independent, such that P(B|A)=P(B), this simplifies down to P(AnB)=P(A)×P(B) which you'll be familiar with.

Reply 11

abcdefghijklmnopqrstuvwxyz

Illusionary

but i still don't get how you got to the conclusion that P(ASnB) - when they're not independant - equals P(AS)=1/52

Reply 12

Illusionary

abcdefghijklmnopqrstuvwxyz

but i still don't get how you got to the conclusion that P(ASnB) - when they're not independant - equals P(AS)=1/52

If you know that the card is the Ace of Spades, then it must also be black, so the 'black' part of the event has no effect on the overall probability.

I don't know if it helps, but if you use the expression that I just gave you'd get P(ASnB)=P(AS)×P(B|AS) = (1/52)×1 = 1/52