Guests at the white house have to clear four layers of security by the secret service: two identity checks, a dog sniff check, and a metal detection check with an x-ray machine. historically 16% of the people in line are tourists and the rest are guests. all tourists are filtered out at the identity checkpoints, so they are not going through the dog and x-ray checkpoints. the dog check results in an alarm for 20% of the people in line. guests who trigger a dog alarm also trigger an x-ray alarm with probability 1/10. guests who do not trigger a dog alarm trigger an x-ray alarm with probability 1/16. what is the probability that a guest who triggers an x-ray alarm also triggers a dog alarm?

0.6863

Step-by-step explanation:

Let A and B be the events

A = the dog check alarm is triggered

B= the x-ray alarm is triggered

As guests who trigger a dog alarm also trigger an x-ray alarm with probability 1/10, we have P(A∩B) = 0.1 .

The dog alarm can only be triggered by a guest, so we want to find P(A|B) , the probability that a guest triggers the dog alarm given that she triggered the x-ray alarm.

But

So we only need to find P(B) to find P(A|B) out.

Since 16% of people (tourists) in line are filtered out before reaching the dog alarm, only 84% of people (guests) reach the dog alarm. That is to say, 16% of 84% = 13.44% of people in line reach the dog alarm.

Since the dog check results in an alarm for 20% of the people in line, 20% of 13.44% = 2.688% of people in line (guests) trigger the dog alarm, so

P(A) = 0.02688.

Since guests who do not trigger the dog alarm trigger an X-ray alarm with probability 1/16, we have P(B|) = 0.0625.

We have then the following:

But and are disjoints sets whose union is B, so

and finally

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