# Homework Help: Uniform discrete probablity distributions

1. Aug 22, 2014

### desmond iking

1. The problem statement, all variables and given/known data
An electric circuit has 5 components. It is known that one of the componenets is faulty. To detremine which one is faulty, all 5 componenets are tested one by one until the faulty component is found, The random variable X represents the number of test required to determine the faulty unit.

2. Relevant equations

3. The attempt at a solution
here's my working:
I let X = number of tests conducted until the faulty unit is found
P(X= 1) = 1/5
P(X=2) = P(good) x P( faulty ) = 4/5 x 1/5 = 0.16
P(X=3) = P(good) x P(good) x P( faulty ) = 4/5 x 4/5 x 1/5 = 0.128
P(X=4) = P(good) x P(good) x P(good) x P( faulty ) = 4/5 x 4/5 x 4/5 x 1/5 = 64/625
P(X=5) = P(good) x P(good) x P(good) x P(good) x P( faulty ) = 4/5 x 4/5 x 4/5 x 4/5 x 1/5 = 256/3125

But for the discrete probablity distributions , sum of all probability must be equal to 1 ... my sum of probability isn't equal to 1 , which is indeed wrong. why cant i do in this way?

2. Aug 22, 2014

### desmond iking

0.16 + 0.128+ 1/5 + 64/625 +256/3125 not equal to 1

3. Aug 22, 2014

### Ray Vickson

Because after observing the first one to be non-defective the remaining four are known to have one defective. That changes probabilities to 1/4 now.

Last edited: Aug 22, 2014
4. Aug 22, 2014

### desmond iking

do you mean the P(x=2) should be 4/5 x 1/4 = 4/20 ?

5. Aug 22, 2014

### Orodruin

Staff Emeritus
Yes. What do you think this implies for the remaining probabilities?

6. Aug 22, 2014

You tell me.