if P(A) is independent of P(BnC) is P(A) independent of P(B) and of P(C) when P(B) and P(C) are not necessarily independent?
is this the exact question as written? it makes no sense, \[P(A)\] is a number, as is \(P(A\cap B)\) numbers are not sets, they are numbers and as such it makes so sense to ask if two numbers are "independent"
Thats not the question asked but, I guess i mean A is independent of the intersection of B n C. Does that make A independent of B and A independent of C?
Here is the problem I'm trying to solve
\[P(A\cup B)=P(A)+P(B)-P(A\cap B)\]
you have all the numbers on the right, plug them in
I don't have P(A) or P(B) though
ooh i see lets see if we can find them, give me a minute i bet a venn diagram might work well
Yea this questions really hard I think
I tried doing something like that but I couldn't get enough things to cancel out.
one second it is totally doable we need to start in the center with \[P(O\cap E\cap F)\]
since you are told \(O\) and \(E\cap F\) are independent, that means \[P(O\cap E\cap F)=P(O)\times P(E\cap F)=0.5\times .125=.0625\]
yes i got that
now i am stuck again damn
oh maybe i have an idea now
nicee
this is taking a lot of paper, maybe i am missing something obvious
thats why i was asking my original question, because I thought I was missing something like that
damn i keep getting one more variable than i can get equations!!
ok i am stuck
I made the assumption i was asking about and got an answer i think is too low.
nvm i made a mistake
ok i fixed it and i got .675 as my answer which seems pretty legit
question 26 is identical except for the numbers there does seem to be another assumption made that we did not use
Thats what i used
perhaps it was in the wording i took "the need for orthodontal work is indpendent for the need of filling and extraction" to mean \[P(O|E\cap F)=P(O)\] but they seem to have interpreted it differently as independent one at a time
yikes
yes i interpreted it that way as well. what im wonderig is if they are both true. ill let you know if i figure it out. g2g though thanks for your help :)
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