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Propiedades de la probabilidad 28
Posted on marzo 20th, 2017 No commentsSean M y N dos sucesos tales que: P(M) = 1/2, P(N) = 1/3 y P(M∩N) = 1/4. Halla:
a) P(M/N)
b) P(N/M)
c) P(M’/N’)
d) P(MUN)
e) P(N/M’)
Solución:
Datos: P(M) = 1/2; P(N) = 1/3; P(M∩N) = 1/4
a)
P(M/N) = P(M∩N)/P(N)
P(M/N) = (1/4)/(1/3) = 3/4 = 0,75
b)
P(N/M) = P(N∩M)/P(M)
P(N/M) = (1/4)/P(1/2) = 2/4 = 1/2 = 0,5
c)
P(M’/N’) = P(M’∩N’)/P(N’)
P(M’∩N’) = P(MUN)’ = 1 – P(MUN)
P(M’∩N’) = 1 – [P(M) + P(N) – P(M∩N)]
P(M’∩N’) = 1 – P(M) – P(N) + P(M∩N)
P(M’∩N’) = 1 – (1/2) – (1/3) + (1/4)
P(M’∩N’) = (1/2) – (1/3) + (1/4) = (6 – 4 + 3)/12 = 5/12
P(N) + P(N’) = 1 → P(N’) = 1 – P(N) = 1 – (1/3) = 2/3
P(M’/N’) = (5/12)/(2/3) = 5/8 = 0,625
d)
P(MUN) = P(M) + P(N) – P(M∩N)
P(MUN) = (1/2) + (1/3) – (1/4) = (6 + 4 – 3)/12 = 7/12
e)
P(N/M’) = P(N∩M’)/P(M’)
P(N∩M’) = P(N) – P(M∩N) = (1/3) – (1/4) = 1/12
P(M) + P(M’) = 1 → P(M’) = 1 – P(M) = 1 – (1/2) = 1/2
P(N/M’) = (1/12)/(1/2) = 2/12 = 1/6
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