Question

according to the general equation for conditional probalitiy, if P (A∩B') =1/6 and P(B') =7/18, what is P(A|B)

Answer 1

\[P(A|B)=\frac{P(A\cap B)}{P(B)}\] you have the numbers you need, plug them in

Answer 2

oh actually now that i read carefully you don't have those numbers, do you?

Answer 3

it tells me what they equal so i think i do

Answer 4

no you have \(P(B')\) but you need \(P(B)\)

Answer 5

theres a diffence..

Answer 6

on the other hand, if \(P(B')=\frac{7}{18}\) then you know what \(P(B)\) is right?

Answer 7

no

Answer 8

no?

Answer 9

does it convert to something

Answer 10

if something doesn't happen seven out of eighteen times on average, then how often does it happen?

Answer 11

11?

Answer 12

right
so if \(P(B')=\frac{7}{18}\) then
\[P(B)=\frac{11}{18}\]

Answer 13

another way to say this is that
\[P(B')=1-P(B)\]

Answer 14

the next number you need to know is \(P(A\cap B)\)

Answer 15

i must be missing something
do you have any other information?

Answer 16

@satellite73 sorry for late reply almost slept

Answer 17

use the rules of conditional probability to get

\[\Large P(A|B') = \frac{P(A \cap B')}{P(B')}\]

Answer 18

oooh it is \[\Large P(A|B')\]!!!!

Answer 19

then you do have the numbers you need, ignore all that stuff before and just compute
\[\frac{\frac{3}{18}}{\frac{7}{18}}=\frac{3}{7}\]

Answer 20

oh ok i aww thanks xx

Answer 21

see*