# Homework Help: Probability tree diagram

1. Jul 24, 2013

### LDC1972

1. The problem statement, all variables and given/known data
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?

2. Relevant equations

P(A U B)

3. 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

Many thanks

Lloyd

2. Jul 24, 2013

### haruspex

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

3. Jul 24, 2013

### 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

#### Attached Files:

• ###### tree.jpg
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4. Jul 24, 2013

### LDC1972

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

Company A and company B
And defective / not defective

Thanks again

Lloyd

5. Jul 24, 2013

### haruspex

Yes, that looks ok, except you've written wrong numbers in the top and bottom lines. Factor of ten out. You previously posted
Each of which starts out correct, but you forgot to move the decimal point when adding the '%'.

6. Jul 24, 2013

### 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 text book 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

7. Jul 24, 2013

### LDC1972

Thank you for the help!

Last edited: Jul 25, 2013