Probability using tree diagrams (1 Viewer)

[MrBrightside]

Hi there, I'm not sure if this question is WITH replacement or WITHOUT replacement. I tried it both ways and didn't come to the correct answer. I'm assuming this question makes more sense WITHOUT replacement.

Question:
http://i.imgur.com/nfBSA.png


I did the following working out:

{Faulty = 3 / 100 Not Faulty = 97/100} - first branch
{(Faulty = 2 / 99 and Not Faulty = 97/99) comes off the Faulty branch, (Faulty = 3 / 99, Not Faulty = 96 / 99) comes off the not faulty branch} - Second branch
{(Faulty = 1 / 98 and Not Faulty = 97/98) comes off the first Faulty branch and second faulty branch...Etc Think you'll understand what I mean} - Third branch

I did P(1 is faulty) = (3/100*97/98*96/98) + (97/100*3/99*96/98) + (97/100*96/99*3/98) = 1164 / 13475

Correct answer: 84 681 / 1 000 000

Thanks

 

[deterministic]

The question suggests that for each car, there is a probability of 3/100 that it is faulty (and hence a probability of 97/100 that it is not faulty). You are not picking cars out of a group of 100.
For 1 faulty, there are three cases: (Faulty, Ok, Ok), (Ok, Faulty, Ok) and (Ok, Ok, Faulty). Each have the same probability of 3/100*97/100*97/100
So P(1 is faulty)=3/100*97/100*97/100 *3 =84 681 / 1 000 000

 

[muzeikchun852]

perms&comb would make this so much easier

 

[MrBrightside]

The question suggests that for each car, there is a probability of 3/100 that it is faulty (and hence a probability of 97/100 that it is not faulty). You are not picking cars out of a group of 100.
For 1 faulty, there are three cases: (Faulty, Ok, Ok), (Ok, Faulty, Ok) and (Ok, Ok, Faulty). Each have the same probability of 3/100*97/100*97/100
So P(1 is faulty)=3/100*97/100*97/100 *3 =84 681 / 1 000 000

Wait, so it's WITH replacement? It's always out of 100. ??