Question

The probability that Louis parks in a no-parking zone at the donut shop and gets a parking ticket is 0.08, and the probability that Louis cannot find a legal parking space and has to park in the no-parking zone is 0.18. On Tuesday, Louis arrives at the donut shop for work and has to park in a no-parking zone. Find the probability that he will get a parking ticket.

Answer 1

It is given that the  probability that Louis parks in a no-parking zone at the donut shop and get a parking ticket is 0.08 i.e. P(No parking zones and ticket) 0.08

the probability that Louis cannot find a legal parking space and has to park in the no-parking zone is 0.18 i.e. P(No parking zones)=0.18

We have to find that the probability that he will get a parking ticket i.e. P(Tickets)

Since it is clear that the events No parking zones and getting a ticket is both independent events. So, we can write as,

P(No parking zones and ticket)= P(No parking zones) * P(Tickets)

⇒0.08 = 0.18*P(Tickets)

⇒P(Tickets) = 0.08/0.18 = 0.44