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For example: enter image description here

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?

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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.

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  • $\begingroup$ That's right, thanks Gunes! Good job. $\endgroup$ Commented Apr 30, 2020 at 3:18
  • $\begingroup$ Could reference me some site/course or book on probability? Maybe with lots of exercises. $\endgroup$ Commented Apr 30, 2020 at 3:19
  • $\begingroup$ I personally find Intro to Prob from Dimitri Bertsekas - MIT Press, very enlightening, examples are quite sophisticated. $\endgroup$
    – gunes
    Commented Apr 30, 2020 at 10:03
  • $\begingroup$ 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. $\endgroup$ Commented May 4, 2020 at 19:26
  • $\begingroup$ @JoãoVitorGomes No, the first one is true. $\endgroup$
    – gunes
    Commented May 4, 2020 at 19:36

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