Question

There are 24 cars in the parking lot outside of my building. They are all either 2-door or 4-door, are either white, black, or red, and are either Ford, Toyota, BMW, or Volkswagen. No two cars are the same with respect to these properties (in other words, for example, there are no two 4-door red Toyotas). If 2 cars are chosen at random, then what is the probability that they are different in all three of these properties? (For example, a 2-door white BMW is different from a 4-door black Ford, but not from a 4-door white Volkswagen).

Answer 1

how do you do this

Answer 2

Your properties yield 24 different cars.
There are 2 different numbers of doors.
There are 3 different colors.
There are 4 different makes.
2 x 3 x 4 = 24

Answer 3

so its 24?

Answer 4

Wait.

Answer 5

but its asking for a fraction

Answer 6

yeah

Answer 7

Here you see that if you have a 2 door car and it's white, there are 4 possible colors.

Answer 8

its says all three properties have to be different but how do we solve for that

Answer 9

I'm trying to figure it out.

Answer 10

ok thx

Answer 11

Notice this picture. If the first choice is one of those three marked with lines, there are only two combinations that can be used as the second pick (4, black, VW) or (4, red, VW)

Answer 12

i see

Answer 13

Let me try this way.

Answer 14

For any first choice you make, the second choice is limited to six different choices.

Answer 15

i know

Answer 16

Since there are 24 different combinations, and after you make the first choice, you can only pick one 6 out of 24, the probability is 6/24 = 1/4

Answer 17

ok
thx

Answer 18

but its wrong

Answer 19

1/4 is wrong?
Do you have choices?
How do you know 1/4 is wrong?

Answer 20

i typed it in my hw box online and it says its wrong

Answer 21

ok

Answer 22

for aops countandprob. hw challenge probs.

Answer 23

Is it 6/23 instead of 6/24?

Answer 24

@mathstudent2000 Did you try 6/23?

Answer 25

no not yet