Question

Suppose an airport metal detector catches a person with metal 99% of the time. That is, it misses detecting a person with metal 1% of the time. Assume independence of people carrying metal. What is the probability that the first metal-carrying person missed (not detected) is among the first 50 metal-carrying persons scanned?

Answer 1

The probability of missing the first metal-carrying person is P(x≤50)=0.395

Explanation:

We define as success to: missing metal carrying detection.

p=0.01

P(x)=
`(n!)/(x!(n-x)!) p^x(1-p)^(n-x)`

We look for the probability when all metal carring people is detected so x=0

P(x≤50)=1-P(x=0)=
`1 - 0.99^(50)`=0.395