Dumb Probability Question
 

The question is not dumb, i just don't like math.

A box contains two defective Christmas tree lights that have inadvertently mixed with eight nondeffective lights. IF the lights are selected one at a time without replacement and tested until both defective lights are fund, what is the probability that both defective lights will be found after three trials.

I think that you work it out like this to find the answer, but i am not sure:
(2/10)*(2/9)*(1/8)=(1/180)

Thanks

Exactly three trials, or by the time the third trial takes place?

Exactly three trials, and by the time you finish them you found both of the defective lights

I think it should be:

(2/10)(2/9)(1/8) + (2/10)(1/9)(1/8) = 1/120

As you can either get the light first or second, but the last one must be deffective.

first try
you get defective/nondefective/defective

(2/10) (8/9) (1/8) = 1/45

second try
nondefective/defective/defective

(8/10) (2/9) (1/8) = 1/45

probability = 1/45 + 1/45 = 2/45


there couldn't be defective/defective/nondefective because according to you, it must be exactly 3 trials, so why would you want to try again after you get 2 defectives.

however.. if you want to consider defective/defective/nondefective
then it should be
2/45 + (2/10) (1/9) (8/8) = 1/15

:) Hope this helps.

Well darn, I was wrong.

Good work Itch...

fuiyoh..good