Complete Question
The complete question is shown on the first uploaded image
Answer:
a) 0.88
b) 0.02
c) 0.01
d) 0.99
Step-by-step explanation:
Step one: State the given parameters
[tex]P(D_{1} ) = 0.12[/tex] [tex]P(D_{2} ) = 0.07[/tex]
[tex]P(D_{3} ) = 0.05[/tex] [tex]P (D_{1} U D_{2} ) = 0.13[/tex]
[tex]P(D_{1}n D_{2}n D_{3}) = 0.01[/tex] [tex]P(D_{1} U D_{3}) = 0.14[/tex]
Step 2 : Obtain the probability that a unit does not have a type 1 defect
[tex]P(\frac{}{D_{1} })[/tex] = [tex]1 -P(D_{1} )[/tex]
= [tex]1 - 0.12[/tex]
= 0.88
Step 3 : Obtain the probability that a unit has both type 2 and 3 defect?
The probability of the unit having both type 2 and type 3 defect is denoted as [tex]P(D_{2} n D_{3} )[/tex]
This is calculated as
[tex]P(D_{2}n D_{3}) =P(D_{2} ) + P(D_{3}) - P(D_{2} U D_{3})\\\\ = 0.07 + 0,05 - 0.13[/tex]
= 0.02
Therefore P(D_{2} n D_{3} ) = 0.02
Step 4 : Obtain the probability that the unit has both a type 2 and type 3 ,but not a type 1 defect
Let [tex]P(\frac{}{D_{1}} n D_{2} n D_{3} )[/tex] denote the probability that the unit has both a type 2 and type 3 ,but not a type 1 defect.
This can be calculated as follows :
[tex]P(\frac{}{D_{1}} n D_{2} n D_{3} ) = P(D_{2} n D_{3}) - P(D_{1} n D_{2}nD_{3})[/tex]
= 0.02 - 0.01
= 0.01
Step 4 : Obtain the probability that a unit has at most two defects
P(at most 2 defects) = 1 - P(all three defects)
= [tex]1- P(D_{1} n D_{2}nD_{3})[/tex]
= 1 - 0.01
= 0.99
