A secret government agency has developed a scanner which determines whether a person is a terrorist. The scanner is fairly reliable; 95% of all scanned terrorists are identified as terrorists, and 95% of all upstanding citizens are identified as such. An informant tells the agency that exactly one passenger of 100 aboard an aeroplane in which you are seated is a terrorist. The agency decide to scan each passenger and the shifty looking man sitting next to you is the first to test positive. What are the chances that this man is a terrorist?
My solution:
$P(\text{ Scanning positive } | \text{ terrorist })=0.95$
$P(\text{ Scanning negative } | \text{ normal }) = 0.95$
$P( \text{ man is terrorist })=\frac{1}{100}=0.01$
So, we are looking for the following posterior probability: $$P(\text{ man on plane is terrorist } | \text{ scanned positive })=\frac{P( \text{ scanned positive } | \text{ man is terrorist })P(\text{ man is terrorist})}{P(\text{ scanning positive })}$$
Calculating the probabilities:
$$P(\text{ scanning positive })=P(\text{ scanning positive }| \text{ man is terrorist })P(\text{ man is terrorist })+P(\text{ scanning positive } | \text{man is not terrorist })P(\text{ man is not terrorist })=0.95(0.01)+0.05(0.99)=0.0095+0.0495=0.059$$
So, $$\frac{P( \text{ scanned positive } | \text{ man is terrorist })P(\text{ man is terrorist})}{P(\text{ scanning positive })}=\frac{0.0095}{0.059}=0.161$$
Is this correct?
Your answer is right, I'm only giving a shortened version
akin to the "non-technical" way of scanning a contingency table
With $T$ = P(terrorist), $+$ = P(scans positive),
$P(T|+) = \dfrac{P(T\cap +)}{P(T\cap +)+P(T^c\cap +)}$
$= \dfrac{0.01*0.95}{0.01*0.95 + 0.99*0.05} = 0.161$