# Probability - proving independence/dependence

1. Jun 3, 2010

### jasper10

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

A film is defective either when the level of sensitivity is wrong (defect D1) or when the colours are faulty (defect D2). 2% of all films made have at least one of these two defects. 1% of all films have defect D1 and 0.2% of all films have both defect D1 and D2.

Are the events "to have defect D1" and "to have defect D2" independent?

3. The attempt at a solution

P(D1) = 0.01
P(D1 ∩ D2) = 0.002
P(D1 ∪ D2) = 0.02

If independent, then: P(D1 ∪ D2) = P(D1) + P(D2) = 0.02

hence P(D2) = 0.02 - P(D1) = 0.02 - 0.01 = 0.01

P(D1 ∩ D2) = 0.01 x 0.01 = 0.0001 which is not 0.002

Is this correct? How do I prove that they are DEPENDENT, as all i have done is rejected their independency!

Thank you very much!

2. Jun 3, 2010

### HallsofIvy

Go back to your textbook and read the definitions of "dependent" and "independent" events!

3. Jun 3, 2010

### Yousef

They're independent. It's just a tree with two branches for D1 and two branches each for D2.

4. Jun 3, 2010

Again, look at the definition of independence - and, since you used it, look at the correct form for the Addition Rule of probability.
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"They're independent. It's just a tree with two branches for D1 and two branches each for D2."
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Makes no sense here.

5. Jun 3, 2010

### Yousef

Why not?

[PLAIN]http://dl.dropbox.com/u/704818/Tree.png [Broken]

Last edited by a moderator: May 4, 2017
6. Jun 3, 2010

### cronxeh

Pr(D1 ∪ D2) = P(D1) + P(D2) - P(D1 ∩ D2)
0.02 = 0.01 + P(D2) - 0.0002
P(D2) = 0.0102

P(D1 ∩ D2) = 0.0002 != 0.01 * 0.0102 -> D1 and D2 are dependent.

PD1(D2) = P(D2)/P(D1) = P(D1 ∩ D2)/P(D1) = 0.0002/0.01 = 0.02