J = japanese, X = other car type 1, Y = other car type 2

There needs to be 2 types of normal cars and 1 type of Japanese car.

XXX

XXY

XYX

YXX

XYY

YXY

YYX

XXJ*

XJX*

JXX*

XJJ*

JXJ*

JJX*

XYJ*

XJY*

JYX*

JXY*

YJX*

YXJ*

YYY

YYJ*

YJY*

JYY*

JJY*

JYJ*

YJJ*

JJJ*

Those are the 27 combinations you could see. Remember that X and Y are both any other car but I put two of them there to show how there's a 2/3 chance of a non-Japanese car and a 1/3 chance of a Japanese car.

Out of those, there are 19 situations where you'd see at least one Japanese car.

Mathematically:

You have 3 cars on the road that you see. You need to consider the situations where you have either 1 J car, 2 J cars, or 3 J cars.

For the 1 J car, you need to choose which car is the J car. You do this with (3 choose 1) = 3:

??J

?J?

J??

For the 2 J cars, you choose which 2 are the J cars. Use (3 choose 2) = 3:

JJ?

J?J

?JJ

For the 3 J cars, use (3 choose 3) = 1:

JJJ

Now you use the probability part. For each of the 3 possibilities with only 1 J car, the probability of the car being J is 1/3 and the probability of the car not being J is 2/3. So with 1 J car, you use (1/3) * (2/3) * (2/3) = 4/27. Multiply this by 3 because there are 3 ways the J car could appear (as shown above). You get 12/27.

For the 2 J case you use (1/3)(1/3)(2/3) = 2/27. Multiply by 3 because there are 3 ways to find 2 J cars in a group of 3 cars. You get 6/27.

For the 3 J case you use (1/3)(1/3)(1/3) = 1/27. Multiply by 1. You get 1/27.

Add them up. You get 12/27 + 6/27 + 1/27 = 19/27.

For the 1 J car case:

JXY

JYX

XJY

YJX

XYJ

YXJ

XXJ

XJX

JXX

YYJ

YJY

JYY

Those are your 12/27.

Notice how if we hadn't paid attention to where the J went, we'd have these:

JXY

JYX

JXX

JYY

Those are your 4 cases (out of 9 - didn't put the other 5 without the J).

For the 2 J case:

JJX

JJY

JXJ

JYJ

XJJ

YJJ

Those are your 6/27.

I think the 3 J case is obvious:

JJJ

Hope this helps you visualize it.