Question

9) Police They set up a safety roadblock, stopping cars to check for seatbelt use How many cars do they expect to stop before finding a driver whose seatbelt is not buckled estimate that 95% of drivers wear their seatbelts. Assume Independence. a) b) What is the probability that the first unbelted driver is the 6 car stopped ? c) If they stop 16 cars during the first hour, what is the probability that 6 drivers did not wear their seatbelts ? What is the probability that they will have to stop at least SIX drivers until they find a driver who is not wearing the seat belt ? d)

Answer 1

(a)

Expected value = np.

Since one unbuckled seat belt is required, so np = 1.

The probability of someone not wearing a seatbelt is 1-.95 = 0.05
n(0.05) = 1

So n = 20 cars.

(b)

This can be treated as the probability that the first five cars have belted drivers, and the 6th is unbelted.
p = 0.95 * 0.95 * 0.95 * 0.95 * 0.95 * 0.05 = 0.0387

(c)

6 out of 16 not wear seat belts

So 16C6 * (.05^6) * (.95^10) = .0000749

(d)

means 6th driver is the first one without seat belt

So first five wore seat belts

1- (0.95^5) = 0.2262