Answer to Question #141910 in Statistics and Probability for bornface

Question #141910

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?


1
Expert's answer
2020-11-03T11:25:33-0500

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(B1A)+P(B2A)+P(B3A)+P(A)=P(B_1\cap A)+P(B_2\cap A)+P(B_3\cap A)+

+P(B4A)=P(B1)P(A)+P(B2)P(A)++P(B_4\cap A)=P(B_1)P(A)+P(B_2)P(A)+

+P(B3)P(A)+P(B4)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=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 0.31.0.31.


(ii)


P(B3A)=P(B3A)P(A)=0.2(0.5)0.31=1031P(B_3|A)=\dfrac{P(B_3\cap A)}{P(A)}=\dfrac{0.2(0.5)}{0.31}=\dfrac{10}{31}\approx0.32258\approx 0.32258

If the person received the speed ticket, the probability that it was at L3 where he violated speed limit rules is 10310.32258.\dfrac{10}{31}\approx 0.32258.



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