A particular electronic component is produced at two plants for an electronics manufacturer. Plant A produces 70% of the components used and the remainder are produced by plant B. Among the components produced at plant A, the proportion of defective components is 1%. Among the components produced at plant B, the proportion of defective components is 2%. If a component received by the manufacturer is defective, the probability that it was produced at plant A is

If a component received by the manufacturer is defective, the probability that it was produced at plant A is 0.5385 = 53.85%.

Step-by-step explanation:

We use the Bayes Theorem to solve this question.

Bayes Theorem:

Two events, A and B.

In which P(B|A) is the probability of B happening when A has happened and P(A|B) is the probability of A happening when B has happened.

In this question:

Event A: Defective component

Event B: Produced at plant A.

Plant A produces 70% of the components used

This means that

Among the components produced at plant A, the proportion of defective components is 1%.

This means that

Probability of a defective component:

1% of 70%(defective at plant A)

2% of 100 - 70 = 30%(defective at plant B). So

If a component received by the manufacturer is defective, the probability that it was produced at plant A is

If a component received by the manufacturer is defective, the probability that it was produced at plant A is 0.5385 = 53.85%.

hello from mrbilldoesmath!

answer:   (7x + 6y) ( 49x^2 - 42xy + 36y^2)

discussion:

in general, the factorization of a^3 + b^3 is

a^3 + b^ 3 = (a + b) ( a^2 - ab + b^2)

in our case note that 343 = 7^3 and 216 = 6^3 so the equation becomes:

343x^3 + 216y^3 = (7x)^3 + (6y)^3

if you think of a (above) as 7x and b as 6y the factorization becomes

343x^3 + 216y^3 =

((7x + 6y) (   (7x)^2 - 7x*6y + (6y)^2)   =

(7x + 6y) ( 49x^2 - 42xy + 36y^2)

you,

mrb

(a + b) (a^2 - a b + b^2)