Answer to Question #142074 in Statistics and Probability for chilekwa

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

(i) p(the person will receive a speed ticket) = 1 - p(will not receive a speed ticket)

p(will not receive a speed ticket) = p(will not receive 1st ticket)*p(will not receive 2nd ticket)*p(will not receive 3rd ticket)*p(will not receive 4th ticket)

p(will not receive n^th ticket) = 1 - p(will receive n^th ticket)

1-((1-0.5*0.2)*(1-0.3*0.1)*(1-0.2*0.5)*(1-0.4*0.2)) = 0.277

P.S. We cant just sum the probabilities of receiving n^th ticket

For example, if there were 2 tickets with 100% probability of receiving each, summation of probabilities would give 200% probability of receiving a ticket, and it is not correct.

(ii) (0.2*0.5)/(0.5*0.2+0.3*0.1+0.2*0.5+0.4*0.2) = 0.323