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An airport security has two checkpoints. Let A be the event that the first checkpoint isbusy, and B be the event that the second checkpoint is busy. Assume that Pr(A) = 0.2,Pr(B) = 0.4 and Pr(A ∩ B) = 0.08. Find the probability that neither of the two checkpointsis busy

Question

An airport security has two checkpoints. Let A be the event that the first checkpoint isbusy, and B be the event that the second checkpoint is busy. Assume that Pr(A) = 0.2,Pr(B) = 0.4 and Pr(A ∩ B) = 0.08. Find the probability that neither of the two checkpointsis busy

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Solution

Para encontrar la probabilidad de que ninguno de los dos puntos de control esté ocupado, primero necesitamos entender los eventos complementarios de A y B.

  1. La probabilidad de que el primer punto de control no esté ocupado es Pr(Ac) Pr(A^c) . Esto se calcula como: Pr(Ac)=1Pr(A)=10.2=0.8 Pr(A^c) = 1 - Pr(A) = 1 - 0.2 = 0.8

  2. La probabilidad de que el segundo punto de control no esté ocupado es Pr(Bc) Pr(B^c) . Esto se calcula como: Pr(Bc)=1Pr(B)=10.4=0.6 Pr(B^c) = 1 - Pr(B) = 1 - 0.4 = 0.6

  3. La probabilidad de que ninguno de los dos puntos de control esté ocupado es la probabilidad de la intersección de los eventos complementarios Pr(AcBc) Pr(A^c \cap B^c) . Dado que Pr(AB) Pr(A \cap B) es la probabilidad de que ambos puntos de control estén ocupados, podemos usar la fórmula de la probabilidad conjunta para eventos complementarios: Pr(AcBc)=1Pr(AB) Pr(A^c \cap B^c) = 1 - Pr(A \cup B)

  4. Para encontrar Pr(AB) Pr(A \cup B) , usamos la fórmula de la unión de dos eventos: Pr(AB)=Pr(A)+Pr(B)Pr(AB) Pr(A \cup B) = Pr(A) + Pr(B) - Pr(A \cap B) Pr(AB)=0.2+0.40.08=0.52 Pr(A \cup B) = 0.2 + 0.4 - 0.08 = 0.52

  5. Ahora, podemos encontrar Pr(AcBc) Pr(A^c \cap B^c) : Pr(AcBc)=1Pr(AB)=10.52=0.48 Pr(A^c \cap B^c) = 1 - Pr(A \cup B) = 1 - 0.52 = 0.48

Por lo tanto, la probabilidad de que ninguno de los dos puntos de control esté ocupado es 0.48.

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