Question

Police plan to enforce speed limits by using radar traps at four different​ locations, Upper L 1L1​, Upper L 2L2​, Upper L 3L3​, and Upper L 4L4. The radar traps will be operated 4040​%, 3030​%, 2020​%, and 3030​% of the​ time, respectively. If a person who is speeding on her way to work has probabilities of 0.30.3​, 0.20.2​, 0.40.4​, and 0.10.1​, ​respectively, of passing through these​ locations, what is the probability that she will receive a speeding​ ticket?

Answer 1

This person have a 29 % probability to receive a speeding​ ticket

Explanation:

The probability of receive a speeding​ ticket in one of the traps is equal to the probability of going through that trap during the operation time of the trap.

Like we have four traps, and the probabilities are independent, then the probability of receive a speed ticket in one of them, is the sum of the probability for each trap.

P=P1+P2+P3+P4

The probability of going through a trap during the operation time is obtained by multiplying the probability of passing through the​ trap by the frequency of operation

Frequency of operation:

TL1 ⇒ 40 % = 0.4

TL2 ⇒ 30 % = 0.3

TL3 ⇒ 20 % = 0.2

TL4 ⇒ 30 % = 0.3

Probability of passing through these​ locations

PL1 ⇒ 0.3

PL2 ⇒ 0.2

PL3 ⇒ 0.4

PL4 ⇒ 0.1

then,

P1= TL1*PL1

P2= TL2*PL2

P3= TL3*PL3

P4= TL4*PL4

and

P = TL1*PL1 + TL2*PL2 + TL3*PL3 + TL4*PL4 = 0.12 + 0.06 + 0.08 + 0.03 = 0.29 = 29 %

This person have a 29 % probability to receive a speeding​ ticket