Question

9. Police plan to enforce speed limits by using radar traps at four different locations within the city limits. A person who is speeding on his/her way to work has probabilities of 0.2, 0.1, 0.5, and 0.2, respectively, of passing through these locations (location 1, location 2, location 3, and location 4, respectively). The radar traps at each of the locations (location 1, location 2, location 3, and location 4, respectively) will be operated 40%, 30%, 20%, and 30% of the time. If a randomly selected person received a speeding ticket on his/her way to work, what is the probability that he/she passed through the radar trap located at location 1? (10 points)

Hint: Consider the following five events:

A: a speeding person receives a speeding ticket (note: a speeding person will receive a speeding ticket if the person passes through the radar trap when operated.)

B₁: a speeding person passing through location 1,

B₂: a speeding person passing through location 2,

B₃: a speeding person passing through location 3,

B₄: a speeding person passing through location 4.

Answer 1

A: is an outcome when speeding person receives a ticket.

B1,B2,B3,B4 : indicates the probability of receiving ticket when person passing through location 1,2,3,4.

P(B1) = 0.2 , P(B2) = 0.1 , P(B3) = 0.5 , P(B4) = 0.2

Probability of speeding person receiving ticket when passes from B1 , P(A|B1) = 40% = 0.40

Probability of receiving ticket when a person passes from B2 , P(A|B2) = 30% = 0.30

Probability of receiving ticket when a person passes from B3 , P(A|B3) = 20% = 0.20

Probability of receiving ticket when a person passes from B4 , P(A| B4) = 30% = 0.30

if a person received the ticket , we need to find the probability that person went through location 1 ( B1). Hence , We need to find probability of P(B1|A), using Bayes theorem the required probability is

Hence, if a speeding person received a ticket, probability of passing him from location 1 ( B1) = 0.2963 or 29.63%