Answer:
Step-by-step explanation:
Let us denote probability of spoilage as follows
Transformer spoilage = P( T ) ; line spoilage P ( L )
Both P ( T ∩ L ) .
Given
P( T ) = .05
P ( L ) = .08
P ( T ∩ L ) = .03
a )
For independent events
P ( T ∩ L ) = P( T ) x P ( L )
But .03 ≠ .05 x .08
So they are not independent of each other .
b )
i )
Probability of line spoilage given that there is transformer spoilage
P L/ T ) = P ( T ∩ L ) / P( T )
= .03 / .05
= 3 / 5 .
ii )
Probability of transformer spoilage but not line spoilage.
P( T ) - P ( T ∩ L )
.05 - .03
= .02
iii )Probability of transformer spoilage given that there is no line spoilage
[ P( T ) - P ( T ∩ L ) ] / 1 - P ( L )
= .02 / 1 - .08
= .02 / .92
= 1 / 49.
i v )
Neither transformer spoilage nor there is no line spoilage
= 1 - P ( T ∪ L )
1 - [ P( T ) + P ( L ) - P ( T ∩ L ]
= 1 - ( .05 + .08 - .03 )
= 0 .9
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