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

Answer:

A. P(EC) = 0.4  

B. P(E∪E) = 0.75

C. P(EC ∩ FC) = 0.25

Step-by-step 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.