Probability that the cab is a green cab $=0.85$
Probability that the cab is a blue cab $=0.15$
The witness can correctly identify the cab colour only $80\%$ of the time
So, probability when the witness is correct means when the witness identifies blue cab $= 0.8$
& Probability when witness is wrong $= 0.2$
We know, $\bf{P(E) = \dfrac{\text{Number of favourable outcomes}}{\text{Number of all possible outcomes}}}$
The probability that the accident was caused by a blue cab $=\dfrac{0.15*0.8}{(0.15*0.8) +(0.85*0.2) }$
$\qquad= \dfrac{1.2}{2.9}$
$\qquad= 0.41$
$\qquad= 41\%$
Option (C)