Q:

Police plan to enforce speed limits by using radar traps at four different locations within the city limits. The radar traps at each of the locations L1, L2, L3, and L4 will be operated 40%, 30%, 20%, and 30% of the time. If a person who is speeding on their way to work has probabilities of 0.2, 0.1, 0.5, and 0.2, respectively, of passing through these locations, what is the probability that they will receive a speeding ticket?

Accepted Solution

A:
Answer:The probability that the person gets a speeding ticket is 0.27Step-by-step explanation:The probability that the person receives a speeding ticket is the probability that the person passes through any of the speed limits and the radar is operating at that time.Let [tex]P(L_1)[/tex] is the probability that the person passes through radar [tex]L_{1}[/tex] and it is operating at that time is [tex]P(L_{1})=P(1)\times P(2)[/tex]Where P(1) is the probability of person passes through [tex]L_{1}[/tex]P(2) is probability that the radar is operating [tex]P(L_1)=0.2\times 0.4=0.08[/tex]Similarly the probabilities are calculated for other radars in the similar manner as [tex]P(L_2)=0.1\times 0.3=0.03[/tex][tex]P(L_3)=0.5\times 0.2=0.1[/tex][tex]P(L_4)=0.2\times 0.3=0.06[/tex] Thus the reuired probability of the reuired event is [tex]P(E)=P(L_1)+P(L_2)+P(L_3)+P(L_4)\\\\P(E)=0.08+0.03+0.1+0.06=0.27[/tex]