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A self driving car has a computer vision component that is used to distinguish between colors of a traffic light: red, yellow and green. This component sometimes makes a wrong decision. The red light is not recognized correctly in 5% of the cases (in 3% of the cases it is mistaken for the green light and in 2% for the yellow light); the yellow light is not recognized correctly in 2% of the cases (in 1% of the cases it is mistaken for the green light and in 1% of the cases for the red light(; and the green light is not recognized correctly in 4% of the cases (in 3% of the cases it is mistaken for the red light and in 1% for the yellow light). It is known that the traffic light in a town are showing red and green 45% of the time and yellow 10% of the time.

(a) What is the probability of a wrong decision? I have found this to be: 0.45*0.04 + 0.10*0.02 + 0.45*0.05 = 0.0425

(b) Provided the system makes an error, what is the probability that a red light is not recognized?

I tried using Bayes formula for this one: P(A|B)=(P(B|A)P(A))/P(B) P(B|A) = 0.45*0.03 + 0.1*0.01 = 0.0145 P(A) = 0.05 P(B) = 0.45

I then get 0.0016, As the probability that red is not recognized given an error occurred. This seems way to low, and I can't figure out what to do here.

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1 Answer 1

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Don't use Bayes' Rule; just apply the definition of conditional probability.

(a) You have it correct: $\mathsf P(\text{Mistaken})~{= \mathsf P(\text{Red $\cap$ Mistaken})+\mathsf P(\text{Green $\cap$ Mistaken})+\mathsf P(\text{Yellow $\cap$ Mistaken})\\ =0.45\cdot 0.05+0.45\cdot 0.04+0.10\cdot0.02 \\ = 0.0425}$

(b) You want to find: $\mathsf P(\text{Red}\mid \text{Mistaken})=\dfrac{\mathsf P(\text{Red $\cap$ Mistaken})}{\mathsf P(\text{Mistaken})}$

Hrmmm.... so...

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  • $\begingroup$ So P(Red∣Mistaken)=P(Red ∩ Mistaken) / P(Mistaken) = (0.45*0.05)/(0.05+0.04+0.02) = 0.2045 $\endgroup$
    – Daniel
    Commented Aug 9, 2017 at 8:49
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    $\begingroup$ $\mathsf P(\text{Red}\mid\text{Mistaken})~{=~0.45\cdot 0.05/0.0425 \\=~ 9/17 \\\approx~0.529{\small 4}}$ $\endgroup$ Commented Aug 9, 2017 at 9:29

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