3. (a) Assume 10% of the drivers are drunk on a friday night. You have a alcohol

Question

3.

(a) Assume 10% of the drivers are drunk on a friday night.

You have a alcohol detector that gives

+ 95% of the time for drunk people

+ 10% of the time for people that are not drunk.

If you tested some driver on a Friday night and the machine gave -, what is the probability that he was drunk?

(b) Monty hall : 3 doors, 1 winning door. You pick a door, host opens 1 irrelevant door, and offers you an option to switch. Should you switch or not?

(c) Drunk monty hall : 3 doors, 1 winning door. Let us assume the room behind the doors are dark so you can't actually see the prize from the outside even when the doors are open. This time, Monty the host is drunk, and has 40% chance to open the the door that contains the winning prize. Should you switch or not?

Explanation / Answer

3(a) Probability of the drivers to drink on friday P(D) =0.1

Alcohol detector 95% correct and 10% not drunk

Test is -ve

that means either driver is not drunk and test works succesfully OR driver is drunk and test failed

1st case

P(Not drunk) = 0.9

P( test works succesfully for not drunk) = 0.9

P (driver is not drunk and test works succesfully) = 0.9*0.9 =0.81

2ns case

P(Drunk and test failed ) = P(drunk)*P(test failed on drunk drivers)

= 0.1 * (1-0.95) = 0.1*0.05 = 0.005

So, Probability of machine gave negative = 0.81+0.005 = 0.815

Probability of drunk P(D) = 0.1

P(Drunk/Machine negative) = P(drunk intrsection Machine negative)/P(machine neagative)

= 0.005/0.815

= 0.0061

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