What is the probability of a defective component from two companies?

[LDC1972]

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

 

[haruspex]

All your working before attempting to construct a tree looks right.
I was not able to follow your verbal description of the tree.

 

[LDC1972]

All your working before attempting to construct a tree looks right.
I was not able to follow your verbal description of the tree.

Hi, thanks so much for your help.

I'll attach a real 'sketch' of what my tree is like:

Hopefully it is visible?

Thanks

Lloyd

 

[LDC1972]

Hi, thanks so much for your help.

I'll attach a real 'sketch' of what my tree is like:

Hopefully it is visible?

Thanks

Lloyd

Probably helps to view at 200% and terrible writing reads:

Company A and company B
And defective / not defective

Thanks again

Lloyd

 

[haruspex]

Yes, that looks ok, except you've written wrong numbers in the top and bottom lines. Factor of ten out. You previously posted

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%

Each of which starts out correct, but you forgot to move the decimal point when adding the '%'.

 

[LDC1972]

Hi,

I re-did the whole thing yesterday (watched you tube maths help).

All figures came out as my originals plus I now have a tidy tree.

Used the long winded Bayes theorem too for iii/

Still got 0.75 so all cool.

I also dropped percentages altogether and kept everything as straight figures (no units) - as the textbook showed no units on their examples. Hope that was right thing to do? As you say, the figure must lay between 0 and 1.

Thanks.

Lloyd

 

[LDC1972]

Thank you for the help!