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In a school machine shop, 75% of all machine breakdowns occur on lathes and 15% occur on drill presses. Let E denote the event that the next machine breakdown is on a lathe, and let F denote the event that a drill press is the next machine to break down. With P(E) = 0.75 and P(F) = 0.15, calculate the following. (Enter your answers to two decimal places.) (a) P(E C)

Respuesta :

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

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