- #1

LDC1972

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## Homework Statement

Components purchased from 2 companies.

Company A is 60% of total purchase with 2% defective parts.

Company B is 40% of total purchase with 1% defective parts.

Components from both companies are thoroughly mixed on receipt.

A/ Draw a tree diagram to represent possible outcomes when a single component is selected at random.

i/ What is the probability that this component cam from company A and is defective?

ii/ Calculate the probability the component was defective

iii/ If was defective, what is probability it was supplied by company A?

## Homework Equations

P(A U B)

## The Attempt at a Solution

Ok, this seems simple, I'd like to know if I'm doing the right thing here if possible please?

I didn't start with the tree (having never done / used this method before).

So I calculated probabilities.

i/ What is the probability that this component came from company A and is defective?

Probability supplied by A = 60% or 0.6

Probability defective if A = 2% or 0.02

Therefore;

0.6 x .02 = 0.012

So probability that this component came from company A = 0.6 (60%) and is defective = 0.012 (1.2%)

ii/ Calculate the probability the component was defective

I assume this incorporates both companies, so:

Company A = 0.6 (60%)

Company B = 0.4 (40%

As per question i/ 0.6 x 0.02 = 0.012 (1.2%)

Now for company B:

0.4 x 0.01 = 0.004 (0.4%)

Now add probability the component picked was defective = 0.012 + 0.004 = 0.016 (1.6%)

iii/ If was defective, what is probability it was supplied by company A?

Probability component from company A AND defective = 0.012 / 0.016 = 0.75 = 75% probability component was from company A and is then ALSO defective

For the tree I then drew vector lines at 45 degrees from start point, one line company A, othe company B. Company A I wrote 0.6 beside line, B I wrote 0.4 beside line. Continued B line to defective P(B U defective) = 0.4 x 0.01 = 0.004%

Continued a line to defective P(B U defective) = 0.6 x 0.02 = 0.012%

Tagged off both lines at 90 degrees mid point with company A P (U not defective) = .6 x .98 = 0.588

And line off Company B as P(U not defective) = 0.4 x 0.99 = 0.396

Seem right? Anymore infor' req'd or inpit gladly welcomed!

Many thanks

Lloyd