Answer:
a.) 0.115
b.) 0.087
Explanation:
Given,
P(A) = 20% = 0.20
P(B) = 30% = 0.30
P(C) = 50% = 0.50
Let M - major repairing during first year
P(M/A ) = 0.05
P(M/B) = 0.10
P(M/C) = 0.15
a.)
P(M) = P(M∩A) + P(M∩B) + P(M∩C)
= P(M/A).P(A) + P(M/B).P(B) + P(M/C).P(C)
= (0.05)(0.20) + (0.30)(0.10) + (0.15)(0.50)
= 0.01 + 0.03 + 0.075
= 0.115
⇒P(M) = 0.115
b.)
P(A/M) = P(A∩M)/ P(M) = P(M/A).P(A)/P(M)
= (0.05)(0.20) / 0.115
= 0.01 / 0.115 = 0.087
⇒P(A/M) = 0.087