Expected value = np.
Since one unbuckled seat belt is required, so np = 1.
The probability of someone not wearing a seatbelt is 1-.95 =
n(0.05) = 1
So n = 20 cars.
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
6 out of 16 not wear seat belts
So 16C6 * (.05^6) * (.95^10) = .0000749
means 6th driver is the first one without seat belt
So first five wore seat belts
1- (0.95^5) = 0.2262
9) Police They set up a safety roadblock, stopping cars to check for seatbelt use How many cars d...