Conditional Probability : Method to approaching problems? Or at least i think these problems are conditional probability. Im having trouble approaching a probability problem in general. Here is one problem that confused me.
---A survey of a magazine’s subscribers indicates that 50% own a home, 80% own a car, and
90% of the homeowners who subscribe also own a car.
b) What is the probability that a subscriber neither owns a car nor a home?


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*My attempt for this one was P(A U B)' => 1 - P(A U B)

*The probability of owning both a car and a home = (.90)(.50) = .45

*So 1 - .45 = .55 as the answer but the back of the book says .15 is the answer?


I understand my method here is wrong so what is the correct way to answer this, My second question i believe is similiar
--- The prob af an alien getting sent to Alaska is 0.8 while probability of getting sent to Maine is 0.2. Probability of Alaskan employee wearing Fur = 0.7 while for Maine it is 0.5
b) If the first person the alien sees is not wearing a fur coat, what is the probability that the alien is in Maine?


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*I approached this one as A = fur and B = Maine so given alien does not see fur what is the probability hes in Maine or P(B|A) = P(A)'P(B|A) but that's as far as i get. After that im not sure how to calculate the probabilities. The book says the awnser to this is .15
Thanks for the help.
 A: Like any math problem, the best approach is to define your terms and write down what you know and what you want before you start doing any math.
Let the random variable $A$ represent whether a subscriber owns a home  
Let the random variable $B$ represent whether a subscriber owns a car

50% own a home, 80% own a car, and 90% of the homeowners who subscribe also own a car.

This translates to:


*

*probability of a subscriber owning a home is $P(A) = 0.5$

*probability of a subscriber owning a car is $P(B) = 0.8$

*probability of a subscriber owning a car, given that they own a home is $P(B|A) = 0.9$

*probability of a a subscriber owning neither a home nor car is $P(\overline{A \cup B}) = 1 - P(A\cup B)$ (unknown)


Since we're dealing with conditional probability, I'll also add the equation $P(B|A) = \frac{P(A\cap B)}{P(A)}$ just so I can refer to it.

We need to solve for $P(\overline{A \cup B}) = 1 - P(A\cup B)$, which means expressing $p(A\cup B)$ in terms of things we know.
If $A$ and $B$ where independent, we could simply write $P(A\cup B) = P(A) + P(B)$, but we see from above that $P(B|A) \neq P(B)$--that is to say, $A$ informs us about $B$. This tells us $A$ and $B$ are not independent and we can't use this formula. (Mathematically this is equivalent to is equivalent to  $P(A \cap B) \neq P(A)P(B) $)
Instead we use the more general formula:
$p(A\cup B)=p(A)+p(B)-p(A\cap B)$
We know from conditional probability that $p(A \cap B) = P(B|A)P(A) = 0.9\times0.5 = 0.45$
And so...  $p(A\cup B)=p(A)+p(B)-p(A\cap B) = 0.5 + 0.8 - 0.45 = 0.85$
Finally... $P(\overline{A \cup B}) = 1 - P(A\cup B) = 1 -0.85 =0.15$
A: Home ownership and car ownership are not independent.  "$90\%$ of homeowners who subscribe, own a car.  Thus of a typical $100$ subscribers, $50\%$ own a home, and $.90\cdot(50)=45$ of these homeowners own a car.  Thus $80-45=35$ own a car but not a home.  This accounts for $85$ people owning a home or a car (or both), leaving $15$ who own neither.
A: for your first problem you calculate $1-p(A\cap B)$ but you want $1-p(A\cup B)$ you can use $p(A\cup B)=p(A)+p(B)-p(A\cap B)=50+80-45=85$
for your second problem $p(A|B)=p(A\cap B)/p(B)$where B is the event the first man not wearing a fur so $p(B)=0.8*0.3+0.2*0.5$ and $p(A\cap B)=0.8*0.7*0.3+0.2*0.5*0.5$
