Question

There are an unlimited amount of toy cars that come in the colors red, blue, yellow and green. If a kid receives three cars at random, what is the probability that two will be the same color and one will be different?

Answer 1

kind of weird since we don't know the ratios of the numbers of cars
just because there are an infinite amount of each, doesn’t mean they are in equal proportion does it?

Answer 2

we can pretend they are i guess, but it doesn't say it

Answer 3

how might you go about setting up an equation for something like this? i think i'm getting thrown off by the fact that they are of unlimited quantity and i don't have any numbers to begin from.

Answer 4

that is the problem
you can't really do it without assuming they are in equal proportion, one third of each color
if you get to assume that, then i guess we can do it without too much difficulty (famous last words)

Answer 5

if we assume they are in equal proportion, then the probability of any one particular pick is
\[\frac{1}{3}\times \frac{1}{3}\times \frac{1}{3}=\frac{1}{27}\]
then we can count how many ways to pick 2 of one and one of the other
it is not that many (only 6) so we can list them

Answer 6

for example
RR B
BB R
RR Y
YY R
BB Y
YY B

Answer 7

unless i missed any

Answer 8

oh damn, that is wrong, sorry
order doesn't count
lets to it an easier way

Answer 9

the probability they are all blue is \(\frac{1}{27}\) same for all yellow and all red
we can now compute
\[1-\frac{3}{27}\]

Answer 10

why do we say 1/27 as the probability instead of 1/3

Answer 11

for example all blue
first one is blue and second one is blue and third one is blue
\[\frac{1}{3}\times \frac{1}{3}\times \frac{1}{3}\]

Answer 12

same for all red and all yellow

Answer 13

therefore
probability of all blue is \(\frac{1}{27}\)
probability of all red is \(\frac{1}{27}\)
probability of all yellow is \(\frac{1}{27}\)

since these are clearly disjoint events, the probability that they are any one color is
\[\frac{1}{27}+\frac{1}{27}+\frac{1}{27}=\frac{3}{27}=\frac{1}{9}\]

Answer 14

and so the probability you get two different colors is
\[1-\frac{1}{9}\] if my reasoning is correct

Answer 15

do you mean two of the same color?

Answer 16

oh this makes sense. nevermind! thank you very much!

Answer 17

should i multiply by on more though since there are four colors?