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

Answer:

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

Step-by-step explanation:

We define as success to: missing metal carrying detection.

p=0.01

P(x)=$\frac{n!}{x!(n-x)!} p^x(1-p)^{n-x}$

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