Sample Size 5 pins, know prob of defect 3/5. What's the prob of finding FIRST defect on the 2nd pin? 3rd pin?

Question

Full Q: A quality control engineer samples 5 from a large lot of manufactured firing pins and checks for defects. Unknown to the inspector, 3 of the 5 sampled firing pins are defective. The engineer will test the 5 pins in a randomly selected order until a defective is observed (in which the case the entire lot will be rejected). Let y be the number of firing pins the QC engineer must test. Find and graph the probability distribution of y,

My understanding: y should be 1, 2, and 3, since by the 3rd pin, the defect will definitely be it (if not on the first two) because of the 3/5 probability and sample size of 5. The prob that the first item drawn is a defect is p(y=1) = 3/5 -- this is correct.

For p(y=2) I figured (2/5)(3/) -- wrong! same for p(y=3)...
The answers given are: p(2)=3/10 , p(3)=1/10
I don't understand how......

....mmh, conditional prob didn't work or may be I didn't do it right.

Here's my new interpretation:
p(y=2)=p(2nd pin is the first defect)=(1/2!)(3/5) = 3/10
=> 1 = only one way to get 2nd pin as defect, 2! = Number of possible selections
=> 3/5 = probability of defect in total sample

p(y=3)=p(3rd pin is the first defect)=(1/3!)(3/5) = 1/10
... same rationale

Am I on the right path?

Answer 1

P(y = 1) = 3/5
P(y > 1) = 2/5

For the second test, 3/4 of the pins are wrong.

P(y = 2 | y > 1) = 3/4
P(y > 2 | y > 1) = 1/4

Finally, P(y = 3 | y > 2) = 1.

So
P(y = 1) = 3/5
P(y = 2) = 2/5 * 3/4 = 3/10
P(y = 3) = 2/5 * 1/4 * 1 = 1/10

Answer 2

HI there.....this sounds like a conditional probability question!!
To find the 1st defect on the second pin tested then the first pin tested would obviously have to be a non defect pin!! So P(y=2) = P(y=2 given y does not =1)

that should do it!! Cheers