Answer:
a) P(A) = 0.09
b) P(A') = 0.91
c) P(B|A) = 0.4
d) P(B/A') = 0.05
e) P(A ∩ B) = 0.036
f) P(A ∩ B') = 0.054
g) P(B) = 0.0815
Step-by-step explanation:
Let A mean batch is made from two different lots and B mean batch requires additional processing
It is given in the question that
P(A) = 0.09
P(B|A) = 0.4
P(B|A') = 0.05
a) P(A) = 0.09 (given)
b) P(A') = 1-0.09 = 0.91
c) P(B|A) = 0.4 (given)
d) P(B|A') = 0.05 (given)
e) P(A ∩ B) = P(B|A) × P(A) = 0.4 × 0.09 = 0.036
f) P(A ∩ B') = P(A) - P(A ∩ B) = 0.09 - 0.036 = 0.054
g) P(B) = P(A ∩ B) + P(A' ∩ B)
P(B) = P(B|A) P(A) + P(B|A') P(A') = 0.4 × 0.09 + 0.05 × 0.91 = 0.0815
QED!