Answer:
(a) 0.0028 (b) 0.7143
Explanation:
First of all, let's define the following events:
I: a container is filled incorrectly
L: a process is operated at a low speed
H: a process is operated at a high speed
S: the sample space
Then, we have according to the original text that
P(I | L) = 0.001, P(I | H) = 0.01, P(H) = 0.2, P(L) = 0.8
and we are seeking the following probabilities (a) P(I) and (b) P(H | I)
(a) P(I) = P(I∩S) = P[I∩(L∪H)] (because L∪H=S)
= P[(I∩L)∪(I∩H)] (distributive laws)
= P(I∩L)+P(I∩H) (because L and H are mutually exclusive)
= P(I | L)P(L)+P(I | H)P(H)
= (0.001)(0.8)+(0.01)(0.2) = 0.0028
On the other hand,
(b) P(H | I) =
![(P(I\cap H))/(P(I))](https://img.qammunity.org/2020/formulas/mathematics/college/oo9tc89aqj2omqc67kr6fcib6bdau2w2tk.png)
=
![(P(I | H)P(H))/(P(I))](https://img.qammunity.org/2020/formulas/mathematics/college/ppfxieqhwo2v55s1xhq39f6cun7slxz01n.png)
=
![((0.01)(0.2))/(0.0028)](https://img.qammunity.org/2020/formulas/mathematics/college/z1yevqqz3d550j74az7mgturjaf3oxvu87.png)
= 0.7143