Answer to Question #141910 in Statistics and Probability for bornface

Question

RATSA is planning to enforce speed limits along Great North road by

using speed cameras at 4 different locations, L1, L2, L3 and L4. The

speed camera at L1 is operated 50% of the time, the speed camera at

L2 is operated 30% of the time, the speed camera at L3 is operated 20%

of the time and the speed camera at L4 is operated 40% of the time.

A person who is speeding from for work has probabilities 0.2, 0.1, 0.5

and 0.2 respectively, of passing through these locations.

(i) What is the probability that the person will receive a speed ticket?

(ii) If the person received the speed ticket, what is the probability

that it was at L3 where he violated speed limit rules?

Answer 1

Consider the following events:

A: the person will receive a speed ticket,

B1: it was at L1 where he violated speed limit rules,

B2: it was at L2 where he violated speed limit rules,

B3: it was at L3 where he violated speed limit rules,

B4: it was at L4 where he violated speed limit rules.

(i) We can write

$$P(A)=P(B_1\cap A)+P(B_2\cap A)+P(B_3\cap A)+$$

$$+P(B_4\cap A)=P(B_1)P(A)+P(B_2)P(A)+$$

$$+P(B_3)P(A)+P(B_4)P(A)=$$

$$=0.5(0.2)+0.3(0.1)+0.2(0.5)+0.4(0.2)=0.31$$

The probability that the person will receive a speed ticket is $0.31.$

(ii)

$$P(B_3|A)=\dfrac{P(B_3\cap A)}{P(A)}=\dfrac{0.2(0.5)}{0.31}=\dfrac{10}{31}\approx$$ $$\approx 0.32258$$

If the person received the speed ticket, the probability that it was at L3 where he violated speed limit rules is $\dfrac{10}{31}\approx 0.32258.$