Uniform discrete probablity distributions

[desmond iking]

Homework Statement

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.

Homework Equations

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 probability distributions , sum of all probability must be equal to 1 ... my sum of probability isn't equal to 1 , which is indeed wrong. why can't i do in this way?

 

[desmond iking]

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

 

[Ray Vickson]

Homework Statement

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.

Homework Equations

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 probability distributions , sum of all probability must be equal to 1 ... my sum of probability isn't equal to 1 , which is indeed wrong. why can't i do in this way?

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.

 

[desmond iking]

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.

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

 

[Orodruin]

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

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

 

[Ray Vickson]

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

You tell me.