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Grace [21]
3 years ago
12

In a manufactory, there are two types of spoilages. It is found that 5% of spoilages are due to transformer spoilage, 8% are due

to line spoilage and 3% involve both problems. (a) Are the two types of spoilages independent or not? Justify your answer. (b) What is the probability that: i- line spoilage given that there is transformer spoilage ii- Transformer spoilage but not line spoilage. iii- Transformer spoilage given that there is no line spoilage iv- Neither transformer spoilage nor there is no line spoilage v- Either transformer spoilage or there is no line spoilage
Mathematics
1 answer:
jok3333 [9.3K]3 years ago
8 0

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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