Answer:
0.25
Explanation:
Data provided in the question:
Percentage of machine breakdowns occur on lathes = 75% = 0.75
Percentage of machine breakdowns occur on drill presses = 15% = 0.15
P(E) = 0.75
P(F) = 0.15
Here,
E denote the event that the next machine breakdown is on a lathe,
F denote the event that a drill press is the next machine to break down
Now,
(a) [tex]P(E^C)[/tex] = 1 - P(E)
on substituting the value of P(E), we get
[tex]P(E^C)[/tex] = 1 - 0.75
or
[tex]P(E^C)[/tex] = 0.25