I'm trying to do the following homework problem and am stuck trying to figure out how to manipulate the conditional probabilities given to me.
P(A) = .75
P(B|A) = .9
$P(B|A^ c )$ = .8
$P(C|A\cap B)$ = .8
$P(C|A \cap B^ c )$ = .6
$P(C|A^c \cap B)$ = .7
$P(C|A^ c \cap B^ c )$ = .3
Given the following information I'm trying to get the following:
1.) P(A intersection B intersection C)
2.) P(B intersection C)
3.) P(C)
4.) P(A|B intersection C)
I was able to figure out the first one to be P(A) * P(B|A) * P(C|A intersection B) = (.75)(.09)(.08) = .054
I'm not sure how to get the rest however and am stuck.
Note: I will go back and edit this once I figure out how to format the symbols correctly.
EDIT:
I found the answers but don't really know how to get to them. The answers are as follows:
P(C) = .74
P(B n C) = .68
P(A|B n C) = .7941
Here goes...
$P(A^c) = 1 - P(A) = 1 - \frac34 = \frac14$
$P(A \cap B) = P(B|A)P(A) = \frac 9{10} \times \frac34 = \frac{27}{40}$
$P(A^c \cap B) = P(B|A^c)P(A^c) = \frac 8{10} \times \frac 14 = \frac 8{40}$
$P(B) = P(A \cap B) + P(A^c \cap B) = \frac{27}{40} + \frac8{40} = \frac{35}{40}$
$P(B^c) = 1 - P(B) = 1 - \frac{35}{40} = \frac5{40}$
$P(A \cap B^c) = p(A) - p(A \cap B) = \frac34 - \frac{27}{40} = \frac3{40} $
$P(A^c \cap B^c) = P(B^c) - P(A \cap B^c) = \frac5{40} - \frac3{40} = \frac2{40}$
$P(A \cap B \cap C) = P(C|A \cap B)p(A \cap B) = \frac8{10} \times \frac{27}{40} = \frac{27}{50} $
$P(A^c \cap B \cap C) = P(C|A^c \cap B)p(A^c \cap B) = \frac7{10} \times \frac8{40} = \frac7{50}$
$P(A \cap B^c \cap C) = P(C|A \cap B^c)p(A \cap B^c) = \frac6{10} \times \frac3{40} = \frac{18}{400}$
$P(A^c \cap B^c \cap C) = P(C|A^c \cap B^c)p(A^c \cap B^c) = \frac3{10} \times \frac2{40} = \frac6{400}$
$P(C) = P(A \cap B \cap C)+ P(A^c \cap B \cap C)+P(A \cap B^c \cap C)+P(A^c \cap B^c \cap C) = \frac{34}{50} + \frac{24}{400} = \frac{148}{200} $
$P(B \cap C) = P(A \cap B \cap C) + P(A^c \cap B \cap C) = \frac{27}{50} + \frac7{50} = \frac{34}{50}$
$P(A|B \cap C) = P(A \cap B \cap C)/P(B \cap C) = (\frac{27}{50})/(\frac{34}{50}) = \frac{27}{34}$
Required answers are given respectively by 8th, 13th, 12th and 14th of these equations. Note there is some reliance 0n $P(X) = P(X \cap Y) + P(X \cap Y^c)$ which is used to calculate $P(B)$, $P(A \cap B^c)$, $P(A^c \cap B^c)$, $P(C)$ (using an extended form) and $P(B \cap C)$.