Let A represents the event that DVD has a video effect.
So,
P(A) = 1.5 %

Let B represents the event that DVD has a audio defect.
So,
P(B) = 0.8 %

A ∩ B represents that DVD has both audio and video defect.
So,
P(A∩B) = 0.4%

Part a)
A ∪ B represents that DVD has a video or audio defect.

P(A∪B) = P(A)+P(B)-P(A∩B) = 1.5 + 0.8 - 0.4 = 1.9%

Thus the probability that the DVD purchased by the customer has a video or audio defect is 1.9%.

Part b)
The complement of P(A∩B) will represent the probability that the DVD does not have video or audio defect.

So, the probability that DVD purchased by customer does not have video or audio defect is 1 - 0.4 = 0.96%

3 0

3 years ago

Matt wrote the number two million, four hundred three thousand, one hundred

damaskus [11]

Answer:

2,403,158

Step-by-step explanation:

8 0

4 years ago