Probability question: Two cab companies operate in River City... Please help

Two cab companies operate in River City, the "Blue" and "Green", named according to the colors of the cabs they run. 85% of the cabs in River City are Blue cabs and 15% are Green cabs. A cab was involved in a hit-and-run accident at night. an eyewitness later identified the ca as Green. The court tested the witness' ability to distinguish between Blue and Green cabs under nighttime visibility conditions. it found the witness was able to identify each color correctly 80% of the time, but he confused it with the other color 20% of the time. What is the probability that the cab in the accident really was Green, as the witness claimed it was? (Draw a diagram or table and also show your calculations).

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you need to draw a 2x2 table.
on one axis (say x) you have two options: the cab was blue or green.
on the y axis, you have two options: the guy sees blue or green.

now calculate the probability of each of the four combinations:

P(was blue, said blue) = 0.85 * 0.8 = 0.68
P(was blue, said green) = 0.85 * 0.2 = 0.17
P(was green, said blue) = 0.15 * 0.2 = 0.03
P(was green, said green) = 0.15 * 0.8 = 0.12

Now, given that the witness said "green", the probability of the car actually being green is given by (conditional probability formula):

P(was green) = 0.12 / (0.12 + 0.17) = 0.41 <---- ANS

(this seemingly low prob is the result of the 20% chance of "misreading" the car color, plus the fact that blue cabs are so much more frequent than green cars !)

good judging/ color 0.15 0.85
0.2 0.03 0.17
0.8 0.12 0.68

Either he saw it green and it was green
or saw it green but was blue
or saw it blue and was blue
or saw it blue but was green

To have the cab really green
so P(1)=0.12
P(2)=0.17
their sum is 0.29 i.e 29%
but to have it green and he was right it's 12%
I showed you both cases