Probability of Exactly 4 Defective Relays from 5 Random Select

[pluto31]

1. Homework Statement [

A stockpile of 40 relays contain 8 defective relays. If 5 relays are selected random, and the number of defective relays is known to be greater than 2, what is the probability that exactly four relays are defective?

Homework Equations

Calculus of probability

The Attempt at a Solution

I've been try to find the solution but I got confused. Can you guys help me?

This is my solution :

the probability from defective relays is : 8/40, the probability from 5 random is : 5/40 and the probability of 4 defective from 5 random select is : 4/5.
so my solution is ((5/40)/(8/40))x(4/5) = 25/32 =0.7815

is this right or my solution is wrong ... thanks

 

[Infinitum]

Hi pluto31! Welcome to physics forums!

Your solution is wrong. The question has already told you that the number of defective relays is known to be more than 2. Apply this condition to your solution, too

 

[Ray Vickson]

1. Homework Statement [

A stockpile of 40 relays contain 8 defective relays. If 5 relays are selected random, and the number of defective relays is known to be greater than 2, what is the probability that exactly four relays are defective?

Homework Equations

Calculus of probability

The Attempt at a Solution

I've been try to find the solution but I got confused. Can you guys help me?

This is my solution :

the probability from defective relays is : 8/40, the probability from 5 random is : 5/40 and the probability of 4 defective from 5 random select is : 4/5.
so my solution is ((5/40)/(8/40))x(4/5) = 25/32 =0.7815

is this right or my solution is wrong ... thanks

Do you understand that this is a conditional probability problem? Suppose D = number of defects in 5 chosen relays. For now, suppose you are able to compute the 6 probabilities $p_0 = P(D=0), \; p_1 = P(D=1), \ldots, p_5 = P(D=5).$ Can you express the solution to the problem in terms of these $p_i$?

Now you have the problem of computing the $p_i$ for the relevant values of i. Can you see how to do that?

RGV