Question

In a school machine shop, 60% 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 and P(E) = 60 and P(F) =.15, calculate:

A. P(EC).
B. P(EUE).
C. P(EC ∩ FC).

Answer 1

A. P(EC) = 0.4

B. P(E∪E) = 0.75

C. P(EC ∩ FC) = 0.25

Explanation:

Let us calculate -:

we are given with - P(E) = 60 and P(F) =.15,

Thus , (a)
`P(E^C)=1-P(E)`

`=1-0.6`

`=0.4`

(b) P(E∪F) = P(E) + P(F)

= 0.60 + 0.15

= 0.75

(c)
`P(E^C`∩
`F^C)`
`=1-P(E`∪
`F)`

`=1-0.75`

=
`0.25`

Hence , the answers are -A. P(EC) = 0.4 , B. P(E∪E) = 0.75 , C. P(EC ∩ FC) = 0.25.