Question

in a major study by the statistics classes at Washington High School, city parking spaces were examined for compliance with the requirement to put money in the parking meters. Overall, the students found that 76% of metered parking places had meters that had NOT expired, and 24% had meters that were expired. If the traffic officer in charge of ticketing cars with expired meters checks meters at random, what is the probability he will find an expired meter BEFORE the 3rd one

Answer 1

Answer:

0.4224 = 42.24% probability he will find an expired meter BEFORE the 3rd one

Step-by-step explanation:

For each car, there is a 24% probability that it had expired parking meters.

What is the probability he will find an expired meter BEFORE the 3rd one?

Three possible outcomes:

Two expired(each with 24% probability)

First expired(24% probability) and second not(76% probability)

First not(76% probability) and second expired(24% probability). So

$p = 0.24^2 + 2*0.24*0.76 = 0.4224$

0.4224 = 42.24% probability he will find an expired meter BEFORE the 3rd one