+Answer:
Explanation:
Lets A be the machine fills to specification
Lets B be the machine fills to under fill
Lets C be the machine overfill
Lets P(A) = 0.990
Lets P(B) = 0.001
A
See P(A) + P(B) + P(C) = 1
⇒ 0.990 + 0.001 +P(C) = 1
⇒ 0.991 +P(C) = 1
⇒ P(C) = 1 - 0.991
which gives P(C) = 0.009
B
Now we got to find the probability that the machine fill to specification and underfill is equals to 0 because a machine can't fill to specification and underfill at the same time.
so need to find P(B')
⇒ P(B) + P(B') = 1
⇒ 0.001 +P(B') = 1
⇒ P(B') = 1 - 0.001
⇒ P(B') = 0.999
C
Now we got to find the probability that machine overfill or undefill
we will find then P ( B ∩ C)
⇒ P ( B ∩ C) = P (B) + P(C)
⇒ P ( B ∩ C) = 0.001 + 0.009
⇒ P ( B ∩ C) = 0.01
which means the probability to overfill or underfill is 0.01
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