Answer:
thus the probability that a part was received from supplier Z , given that is defective is 5/6 (83.33%)
Step-by-step explanation:
denoting A= a piece is defective , Bi = a piece is defective from the i-th supplier and Ci= choosing a piece from the the i-th supplier
then
P(A)= ∑ P(Bi)*P(C) with i from 1 to 3
P(A)= ∑ 5/100 * 24/100 + 10/100 * 36/100 + 6/100 * 40/100 = 9/125
from the theorem of Bayes
P(Cz/A)= P(Cz∩A)/P(A)
where
P(Cz/A) = probability of choosing a piece from Z , given that a defective part was obtained
P(Cz∩A)= probability of choosing a piece from Z that is defective = P(Bz) = 6/100
therefore
P(Cz/A)= P(Cz∩A)/P(A) = P(Bz)/P(A)= 6/100/(9/125) = 5/6 (83.33%)
thus the probability that a part was received from supplier Z , given that is defective is 5/6 (83.33%)
Whenever you solve an equation with a single variable you need to isolate the variable by itself on one side of the equal sign. Just be sure that whatever operations you perform on one side you perform on the other sides as well.
62x=1 divide both sides by 62
x=1/62
Answer:
20%
Step-by-step explanation:
or 20 because since there is 5 and 3 you ad them and you get 5, 5+5 =10 and its like 100 but its not so, 2 of 10 is like saying 20% of 100
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