P(outcome) = (#matches) / (#totalchoices)
There are 34 total cars, so
P(silver) = 16/34
P(blue) = 8/34
P(red) = 10/34
P(silver,~silver) = 16/34 * 18/33
In a parking lot , there are 16 silver cars , 8 blue cars, and 10 red cars. A car leaves the parking lot . What is the probability that it is
A: a silver car?
B: a blue car?
D: a red car ?
C: suppose that the first car that leaves is a silver car . What is the probability that the second car that leaves is not a silver car?
3 answers
34 cars
silver = 16
blue= 8
red = 10
in percent:
100* 16/34 =
100* 8/34 =
100* 10/34 =
=============================
33 cars left
15 are silver
fraction silver = 15/33
so fraction not silver = 18/33
so 100 * 18/33 in percent
silver = 16
blue= 8
red = 10
in percent:
100* 16/34 =
100* 8/34 =
100* 10/34 =
=============================
33 cars left
15 are silver
fraction silver = 15/33
so fraction not silver = 18/33
so 100 * 18/33 in percent
C is not clear about if they mean probability of both or not/if