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A parking lot on campus has the probability of parking in an illegal spot and getting a ticket is .13, while the probability of finding no park and having to park illegal is .2 In a rushing to class one day you find no parking space. What is probability you get a ticket given to you for being parked in illegal spot?

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What do you know and what are you being asked (from the question)? –  Daryl Sep 30 '12 at 3:31

3 Answers 3

Here's one way to see your way through. Suppose you drive to campus 100 times. The second clause says you have to park illegally 20 times, right? The first clause says you get ticketed 13 times, right? So, of the 20 times you park illegally, you get ticketed 13, right? So, can you answer the question now?

My answer would be, get a bicycle.

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Let's assume you will only park illegally if you can't find a legal spot.

Let $A=\{$ get a ticket when you park illegally $\}$ and $B= \{$ can't find a legal space so you park illegally$\}$.

You have $P[A and B]= 0.13$ and $P[B]=0.2$. You want $P[A|B]$.

You know $P[A|B] =P[A and B]/P[B]$.

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Not exactly sure because it has been a while since I did this, but this is what I think.

Look at the formula $$P(A\ |\ B) = \frac{P(A \cap B)}{P(B)}.$$

In this scenario, we know that the probability of parking illegally AND getting a ticket is $0.13$ and that equals the $P(A \cap B)$. You also know that the probability of not being able to find a space and having to park illegally is $0.2$. This probability is your given so in $P(A\ |\ B)$ (Read as "The probability of $A$ given $B$") you would plug $0.2$ into $B$. That leaves $A$ which must mean that $A$ = the probability of getting a ticket. This leaves us with.

$$P(\textrm{Getting A Ticket}\ |\ \textrm{You Park Illegally}) = \frac{0.13}{0.2}.$$

This comes out to be $0.65$ so the probability of getting a ticket given that you park illegally is $0.65$.

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