Question

Assume that each time a metal detector at an airport signals, there is a 25% chance that the cause is
the change in the passenger’s pocket. During a given hour, 15 passengers are stopped because of a signal
signal from the metal detector.
a) Find the probability that at least 3 persons will have stopped due to change in their pockets.
b) If 15 passengers are stopped by the detector, would it be unusual for none of those to have been
stopped due to change in the pockets? Explain based on the probability of this occurring.

Answer 1

The number of successes 'X' in 'n' number of independent and identically distributed Bernoulli trials follows binomial distribution. Binomial distribution has two parameters, 'n' and 'p', where 'n' is the number of trials, and 'p' is the probability of success in each trial. The probability that a passenger has change in the pocket is p=0.25, q=1-p=0.75

a) P(x≥3)=1-P(x=0)-P(x=1)-P(x=2)

P(x=0)=$(^{15}_0)0.25^00.75^{15-0}=0.013$

P(x=1)=$(^{15}_1)0.25^10.75^{14}=0.0668$

P(x=2)=$(^{15}_2)0.25^20.75^{13}=0.1559$

P(x≥3)=1-0.013-0.0668-0.1559=0.7643

b) According to the founded probability P(x=0)=0.013 -if none of those passengers have been

stopped due to change in the pockets it would be unusual