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

The events are  independent.Their conditional probability is equal to the  probability of an event .

2) The probability that there is line spoilage given that there is transformer spoilage=0.08

- Transformer spoilage but not line spoilage=0.046

Transformer spoilage given that there is no line spoilage = 0.05

- Neither transformer spoilage nor there is no line spoilage =0.03

v- Either transformer spoilage or there is no line spoilage  = 0.97

Step-by-step explanation:

Let A be an event that the spoilage is due to transformers so

P(A) = 5% = 5/100 or 0.05

Let B be an event that the spoilage is due to line so

P(B) = 8% = 8/100 or 0.08

Let C be an event that the spoilage is due to both then

P(C) = 3% = 3/100 or 0.03

1)  Independence

For events to be independent their joint probability is given by

P(A) = P(A∩B)/ P (B)  By Law of independence

P(A∩B)= P(A/B)P(B)

P(A∩B) = 5/100 *8/100=40/10000= 4/1000

P(A) = 5/100  

P(A/B) =P(A∩B)/ P (B)

= 4/1000/8/100= 1/20  

P(A)= P(A/B)

5/100 is equal to  1/20

The events are independent.Their conditional probability is equal to the probability of an event

2) The probability that there is line spoilage given that there is transformer spoilage

P(B/A)= P(A∩B)/ P (A) = 4/1000/5/100= 4/50= 2/25=0.08

Which is again equal to probability of event B.

- Transformer spoilage but not line spoilage

P(A∩1-B)= 5/100*92/100= 460/10000=0.046

Transformer spoilage given that there is no line spoilage

P(A/1-B) =P(A∩1-B)/ P (1-B)

= 5/100*92/100/92/100= 5/100 = 0.05

- Neither transformer spoilage nor there is no line spoilage

1-P(A) +P(1-B)= 1-(5/100+92/100)= 3/100= 0.03

v- Either transformer spoilage or there is no line spoilage  

P(A)+P(1-B)= 5/100+92/100= 97/100= 0.97

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step-by-step explanation:

-6k+7k

(-6+7)k

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