A certain town has two taxi companies, the Green Taxi Co (cars coloured green) and the Blue Taxi Co (cars coloured blue). 10% of taxis are the Green and 90% are the Blue. There was an accident on a dark evening, and the witness claimed a Green cab was involved. On future testing, they discovered that under the conditions of that evening, there was an 80% chance of correct identification of the cab colour (regardless of colour) and 20% chance of erring in colour identification. (assume neither company is more accident prone than the other)

a. Before actually hearing the testimony of this witness, what is the probability that the witness will claim that it was a Blue Cab involved in the accident that night?

b. If we now have the testimony of the witness, who declares that the cab was Green, what is the probability that it actually was Green?

I used a probability tree to solve this question but apparently, it was actually related to Bayes theorem. How I did (a) was use the probability tree and got $P(Witness~claims~Blue) = (0.9)(0.8)+(0.1)(0.2)=0.74$. For (b), $P(Actually~green) = (0.1)(0.8)+(0.1)(0.2)=0.10$.

However, the answer for (b) was $\frac{.08}{.26}$ and I was supposed to use the Bayes Theorem. Can someone please explain why the Bayes is to be used instead of a sum of probabilities?


3 Answers 3


Given the events:

  • $T_b$: witness think it's blue
  • $T_g$: witness think it's green
  • $I_b$: is blue
  • $I_g$: is green

So we have:

  • $P(I_g) = 0.1$
  • $P(I_b) = 0.9$
  • $P(I_g|T_g) = P(I_b|T_b) = 0.8$
  • $P(I_g|T_b) = P(I_b|T_g) = 0.2$

In (a), you wanted $P(T_b)$, so you used $P(T_b) = P(T_b|I_b)P(I_b) + P(T_b|I_g)P(I_g)$

In (b), you wanted $P(I_g|T_g)$, so you use $P(I_g|T_g) = P(I_g)P(T_g|I_g)/P(T_g)$, where $P(T_g) = 1-P(T_b)$ (the answer you got on (a))

Note the difference between the two problems, in the second you wanted a conditional probability, so that's why you used Bayes Theorem.

  • $\begingroup$ I can't really grasp the difference between conditional probability and Bayes Theorem. I looked it up many times but I can't seem to understand. $\endgroup$ Commented Apr 22, 2013 at 2:35
  • $\begingroup$ The conditional probabilities the wrong way round (see answer below) $\endgroup$
    – Ray
    Commented Aug 23, 2023 at 10:25
  • $\begingroup$ @Ray I don't even recall answering this as it was more than a decade ago. Anyway, I've change edited the question to make it a community wiki, fell free to edit it and improve it. $\endgroup$ Commented Aug 28, 2023 at 12:10

Because Bayes Theorem combines prior information with collected data to create a posterior probability. think about what you know (the prior probability of the cab being green = 10%) combined with 'collected' information, (that the witness identified the car as being green). Use Bayes Theorem:

$P(B|A) = \frac{P(A|B)P(B)}{P(A)}$

You started the problem with a certain belief that the cab would be green (10%, since 10% of all cabs are green). You have new information. how should that affect your belief that it's 10 % likely that the cab is green?

  • $\begingroup$ Keep in mind that what you're trying to find out isn't $P(Actually Green)$ but rather $P(Actually Green|Witness claims Green)$ $\endgroup$ Commented Apr 22, 2013 at 2:19
  • $\begingroup$ But then can't I just use the conditional probability instead of Bayes? I am slightly confused about which one to use sometimes. $\endgroup$ Commented Apr 22, 2013 at 2:30
  • $\begingroup$ I'm not sure what you're asking. You needed to use Bayes because you were looking for a conditional probability in part (b). Bayes gives you the conditional $P(Actually Green|Witness Claims Green)$. $\endgroup$ Commented Apr 22, 2013 at 2:42

Came across this when preparing a class. The accepted answer above has the conditional probabilities the wrong way round. If it is blue we identify it correctly 80% of the time etc. So we should have:

\begin{equation} P(T_g|I_g)=P(T_b|I_b)=0.8 \end{equation}

and \begin{equation} P(T_b|I_g)=P(T_g|I_b)=0.2 \end{equation}

For (b) then we have: \begin{equation} P(T_g) =P(T_g|I_b)P(I_b) = P(T_g|I_g)P(I_g) = (0.2)(0.9) + (0.8)(0.1) = 0.26 \end{equation}

\begin{equation} P(I_g|T_g)=\frac{P(T_g|I_g)P(I_g)}{P(T_g)} = \frac{(0.8)(0.1)}{0.26} = \frac{0.08}{0.26} \end{equation}


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