Question

Can you please show your work. The State Police are trying to crack down on speeding on a particular portion of the Massachusetts Turnpike. To aid in this pursuit, they have purchased a new radar gun that promises greater consistency and reliability. Specifically, the gun advertises +- one-mile-per-hour accuracy 93% of the time; that is, there is a 0.93 probability that the gun will detect a speeder, if the driver is actually speeding. Assume there is a 1% chance that the gun erroneously detects a speeder even when the driver is below the speed limit. Suppose that 90% of the drivers drive below the speed limit on this stretch of the Massachusetts Turnpike.

a. What is the probability that the gun detects speeding and the driver was speeding? (Round to 4 decimal places).

b. What is the probability that the gun detects speeding and the driver was not speeding? (Round to 4 decimal places).

c. Suppose the police stop a driver because the gun detects speeding. What is the probability that the driver was actually driving below the speed limit? (Round to 4 decimal places).

Answer 1

Solution

Back-up Theory

If A and B are two events such that probability of B is influenced by occurrence or otherwise of A, then

Conditional Probability of B given A, denoted by P(B/A) = P(B ∩ A)/P(A)……......….(1)

1. => P(B ∩ A) = P(B/A) x P(A) or P(A ∩ B) = P(A/B) x P(B) ……………………. (1a)

P(B) = {P(B/A) x P(A)} + {P(B/A^C) x P(A^C)}…………………………………..................(2)

P(A/B) = P(B/A) x {P(A)/P(B)}……………………………..…………………...........…….(3)

Now to work out the solution,

Let A represent the event that gun detects speeding and B represent the event that driver was speeding. ……………… (4)

Trivially, A^C represents the event that gun does not detect speeding and B^C represent the event that driver was not speeding, i.e., the driving was below the speed limit. ………...................................................................................………………… (4a)

Let us first convert the given data into the above terminology.

‘there is a 0.93 probability that the gun will detect a speeder, if the driver is actually speeding.’ => P(A/B) = 0.93 ………(5)

‘there is a 1% chance that the gun erroneously detects a speeder even when the driver is below the speed limit.’ => P(A/B^C) = 0.01 …………………………………………………….......................................................................…………(6)

‘90% of the drivers drive below the speed limit on this stretch of the Massachusetts Turnpike.’ => P(B^C) = 0.9 ..………(7)

and so P(B) = 1 - 0.9 = 0.1 ………....................................................……….…………………………………………………(7a)

Part (a)

Probability that the gun detects speeding and the driver was speeding

= P(A ∩ B)

= P(A/B) x P(B) [vide (1a)]

= 0.93 x 0.1 [vide (5) and (7a)]

= 0.093   Answer

Part (b)

Probability that the gun detects speeding and the driver was not speeding

= P(A ∩ B^C)

= P(A/B^C) x P(B^C) [vide (1a)]

= 0.01 x 0.9 [vide (6) and (7)]

= 0.009 Answer

Part (c)

If the police stop a driver because the gun detects speeding, the probability that the driver was actually driving below the speed limit

= P(B^C/A)

= {P(A/B^C) x P(B^C)}/P(A) [vide (3)] …………………………………………………….. (8)

Now, vide (2),

P(A) = {P(A/B) x P(B)} + {P(A/B^C) x P(B^C)}

= (0.93 x 0.1) + (0.01 x 09) [vide (5), (7a), (6) and (7)]

= 0.102 …………………………………………………………………………………… (9)

Substituting from (6), (7a) and (9) into (8),

P(B^C/A) = (0.01 x 09)/0.102

= 0.0882   Answer

DONE