Propiedades de la probabilidad 28

Sean 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