Respuesta :
Answer:
The probability is 0.357
Step-by-step explanation:
Given that,
The system has at least one type of defect,
Suppose, A certain system can experience three different types of defects.
Let [tex]A_{i}[/tex] (i = 1,2,3) denote the event that the system has a defect of type i.
Suppose that the following probabilities are,
[tex]P(A_{1})=0.11[/tex]
[tex]P(A_{2})=0.08[/tex]
[tex]P(A_{3})=0.05[/tex]
[tex]P(A_{1}\cup A_{2})=0.13[/tex]
[tex]P(A_{1}\cup A_{3})=0.13[/tex]
[tex]P(A_{2}\cup A_{3})=0.11[/tex]
[tex]P(A_{1}\cap A_{2}\cap A_{3})=0.01[/tex]
[tex]P(A_{1}\cap A_{2})=0.06[/tex]
[tex]P(A_{1}\cap A_{3})=0.03[/tex]
[tex]P(A_{2}\cap A_{3})=0.02[/tex]
[tex]P(A_{1}\cup A_{2}\cup A_{3})=0.14[/tex]
We need to calculate the probability that it has exactly one type of defect
Using given data
[tex]P=\dfrac{P(A_{1}\cap A_{2}'\cap A_{3}')}{P(A_{1}\cup A_{2}\cup A_{3})}+\dfrac{P(A_{1}'\cap A_{2}\cap A_{3}')}{P(A_{1}\cup A_{2}\cup A_{3})}+\dfrac{P(A_{1}'\cap A_{2}'\cap A_{3})}{P(A_{1}\cup A_{2}\cup A_{3})}[/tex]
[tex]P=\dfrac{P(A_{1})-P(A_{1}\cap A_{2})-P(A_{1}\cap A_{3})+P(A_{1}\cap A_{2}\cap A_{3})}{P(A_{1}\cup A_{2}\cup A_{3})}[/tex] + [tex]\dfrac{P(A_{2})-P(A_{1}\cap A_{2})-P(A_{2}\cap A_{3})+P(A_{1}\cap A_{2}\cap A_{3})}{P(A_{1}\cup A_{2}\cup A_{3})}[/tex]+ [tex]\dfrac{P(A_{3})-P(A_{1}\cap A_{3})-P(A_{2}\cap A_{3})+P(A_{1}\cap A_{2}\cap A_{3})}{P(A_{1}\cup A_{2}\cup A_{3})}[/tex]
P = [tex]\dfrac{P(A_{1})+P(A_{2})+P(A_{3})-2P(A_{1}\cap A_{2})-2P(A_{1}\cap A_{3})-2P(A_{2}\cap A_{3})+3P(A_{1}\cap A_{2}\cap A_{3})}{P(A_{1}\cup A_{2}\cup A_{3})}[/tex]
Put the value into the formula
[tex]P=\dfrac{0.11+0.08+0.05-2(0.06)-2(0.03)-2(0.02)+3(0.01)}{0.14}[/tex]
[tex]P=0.357[/tex]
Hence, The probability is 0.357
