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
Step 2 : Obtain the probability that a unit does not have a type 1 defect
=
=
= 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
This is calculated as
= 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 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 :
= 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)
=
= 1 - 0.01
= 0.99
