Example — Robbery and law enforcement

Example — Robbery and law enforcement

Let $A$ = “the subject is caught by the police” and $B$ = “the subject commits the robbery”. The data: $P(A\mid B) = 0,6, \quad P(A\mid \overline{B}) = 0,1, \quad P(A) = 0,3.$ Question: what is $P(B)$?

We set up the Venn diagram with four unknowns $x, y, z, t$, where $x=P(A\cap\overline{B})$, $y=P(A\cap B)$, $z=P(B\cap\overline{A})$, $t=P(\overline{A}\cap\overline{B})$. We translate the data into a system:

x + y + z + t = 1 \\[3pt] P(A) = x + y = 0,3 \\[3pt] P(A\mid B) = \dfrac{y}{y+z} = 0,6 \\[3pt] P(A\mid \overline{B}) = \dfrac{x}{x+t} = 0,1 \end{cases}$$ From the third equation: $y = 0,6\,(y+z) \implies 0,4\,y = 0,6\,z \implies y = 1,5\,z$. From the fourth: $x = 0,1\,(x+t) \implies 0,9\,x = 0,1\,t \implies t = 9\,x$. From the second: $x = 0,3 - y = 0,3 - 1,5\,z$. Substituting into the first: $$(0,3-1,5\,z) + 1,5\,z + z + 9\,(0,3-1,5\,z) = 1,$$ that is $0,3 + z + 2,7 - 13,5\,z = 1$, whence $3 - 12,5\,z = 1 \implies z = 0,16$. Therefore: $y = 1,5\cdot 0,16 = 0,24$,   $x = 0,3 - 0,24 = 0,06$,   $t = 9\cdot 0,06 = 0,54$. $$P(B) = y + z = 0,24 + 0,16 = \boxed{0,4 = 40\%}.$$ **Check:** $x+y+z+t = 0,06+0,24+0,16+0,54 = 1$ ✓. ## Links **Topics:** [[English/Knowledge Graph/Topics/Probabilita|Probability]] **Concepts:** [[English/Knowledge Graph/Concepts/Diagramma di venn|Venn diagram]] · [[English/Knowledge Graph/Concepts/Probabilita condizionata|Conditional probability]] · [[English/Knowledge Graph/Concepts/Sistema di vincoli|System of constraints]] **Methods:** [[English/Knowledge Graph/Methods/Diagramma venn prob|Venn diagram probability]] · [[English/Knowledge Graph/Methods/Prob condizionata|Conditional probability]] **Skills:** [[English/Knowledge Graph/Skills/Calcolo probabilita|Probability calculation]] · [[English/Knowledge Graph/Skills/Risolvere sistemi|Solving systems]]