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2 answers:
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Let 
denote the event that an HD is defective, and 
the event that a particular HD was produced at facility 
.
You are asked to compute
From the definition of conditional probabilities, the first two will require that you first find
. Once you have this, part (c) is trivial.
I'll demonstrate the computation for part (a). Part (b) is nearly identical.
(a)
Presumably, the facility responsible for producing a given HD is independent of whether the HD is defective or not, so
.
Use the law of total probability to determine the value of the denominator:
We know each of the component probabilities because they are given explicitly: 0.015, 0.02, 0.01, and 0.03, respectively. So
and thus
(b) Similarly,
(c)
You are asked to compute
From the definition of conditional probabilities, the first two will require that you first find
I'll demonstrate the computation for part (a). Part (b) is nearly identical.
(a)
Presumably, the facility responsible for producing a given HD is independent of whether the HD is defective or not, so
Use the law of total probability to determine the value of the denominator:
We know each of the component probabilities because they are given explicitly: 0.015, 0.02, 0.01, and 0.03, respectively. So
and thus
(b) Similarly,
(c)