# Doubt about independent and dependent events

For example:

An airport screens bags for forbidden items, and an alarm is supposed to be triggered when a forbidden item is detected.

Suppose that $$5\%$$ of bags contain forbidden items. If a bag contains a forbidden item, there is a $$98\%$$ chance that it triggers the alarm. If a bag doesn't contain a forbidden item, there is an $$8\%$$, percent chance that it triggers the alarm.

$$P(F\cap A)=(0.05)(0.98)=0.049$$

$$P(A)=P(F∩A)+P(N∩A) =0.049+0.076 =0.125$$ $$P(F) = 0.05$$, right?

$$P(F∣A)=0.392$$ So, are these events independent or dependent?

So they are not independent right? $$P(F∣A)=0.392 \neq P(F) = 0.05$$ ? But why we an calculate $$P(F∩A)$$ as $$P(F)P(A)$$ ? Isn't this case/formula only for independent events?

But why we an calculate $$\mathbb P(F\cap A)$$ as $$\mathbb P(F)\mathbb P(A)$$ ? Isn't this case/formula only for independent events?

You are actually calculating it as $$\mathbb P(F\cap A)=\mathbb P(F)\mathbb P(A|F)$$ because the following sentence

If a bag contains a forbidden item, there is a 98% , percent chance that it triggers the alarm

means that $$\mathbb P(A|F)=0.98$$, not $$\mathbb P(A)$$ because of the part "If a bag contains a forbidden item", which means $$F$$ is on the given side of the expression.

• That's right, thanks Gunes! Good job. Apr 30, 2020 at 3:18
• Could reference me some site/course or book on probability? Maybe with lots of exercises. Apr 30, 2020 at 3:19
• I personally find Intro to Prob from Dimitri Bertsekas - MIT Press, very enlightening, examples are quite sophisticated. Apr 30, 2020 at 10:03
• Gunnes, why do you shift A and F? P(F∩A) = P(A|F)*P(F) instead of P(F∩A) = P(F|A)*P(F) , isn't this last formula the standard one? When we see the definition of conditional probability we use the second formula not the first. May 4, 2020 at 19:26
• @JoãoVitorGomes No, the first one is true. May 4, 2020 at 19:36