Need help with a Bayes' Theorem exercise

Question

Need help with a Bayes' Theorem exercise:

Of all the taxi's in the city, the possibility of a taxi's colour to be green P(TaxiGreen) or red P(TaxiRed):

P(TaxiRed) = 0.1
P(TaxiGreen) = 0.9

When a taxi commits a crime, the possibilities of a witness answer being true:

P(WitnessSaysRed | TaxiRed) = 0.95
P(WitnessSaysGreen | TaxiGreen) = 0.70

A crime has been committed, taxi colour unknown, a witness says the car is red, should the police look for a green or a red car?

I assume conditional probability is conversely true:

P(TaxiRed | WitnessSaysRed) = P(WitnessSaysRed | TaxiRed)

Expanded into Bayes' Law formula:

          P(E|H)
 P(H∣E) = _______ . P(H)
           P(E)

->​

P(WitnessSaysRed | TaxiRed) 
___________________________ . P(TaxiRed)
    P(WitnessSaysRed)

->​

          0.95 
___________________________ .0.1
    P(WitnessSaysRed)

-> I am stuck for +- 2 hours on finding P(WitnessSaysRed), any suggestions?

Answer 1

Let $WR$ be the event of witness saying taxi red and $WG$ be the event of witness says taxi green. $$P(R|WR)=\dfrac{P(R).P(WR|R)}{P(WR)}$$ $$P(WR)=P(R).P(WR|R)+P(G).P(WR|G)=P(R).P(WR|R)+P(G)[1-P(WG|G)]$$ $$P(WR)=(0.1)(0.95)+(0.9)[1-0.7]=0.365$$ Here $P(WR|G)=1-P(WG|G)$ because it is assumed that on the event of crime (by any taxi, be it red or green), the witness will either answer taxi red or taxi green. Therefore in the event of crime by taxi green, if witness doesn't answer taxi green, then it is sure that he will answer taxi red, hence giving $P(WR|G)$.

Now plugging the values: $$P(R|WR)=\dfrac{(0.1)(0.95)}{0.365}=0.260$$

Answer 2

Correct me if I'm wrong but I think it's right:

P(WitnessSaysRed) * P(TaxiRed) = 0.095
+
P(WitnessSaysRed) * P(TaxiGreen) = 0.27
P(WitnessSaysRed) = 0.365

Therefore plug it into the formula in my question, trustability = 26.03%, Police should look for a green car.

Answer 3

Use the Law of Total Probability:

$$\begin{align}\mathsf P(W_R)&=\mathsf P(W_R\mid T_R)\cdot\mathsf P(T_R)+\mathsf P(W_R\mid T_G)\cdot\mathsf P(T_G)\\[1ex]&=0.95\cdot 0.1+(1-0.7)\cdot 0.9\\[1ex]&=0.095+0.27\\[1ex]&=0.365\end{align}$$

Thus

$$\begin{align}\mathsf P(T_R\mid W_R)&=\dfrac{0.095}{0.365}\\[1ex]&=\dfrac{19}{73}\\[1ex]&\approx 0.260\end{align}$$