Conditional Probability when picking without replacement

[uteskier]

Pretty basic question Im sure, but I cant wrap my head around it.



Hey guys, so this is a data set that I was given. The question is asking me to determine the probability of picking a fiction book first, and then picking a hard cover book second. No matter what happens, the probability of picking a fiction book first will always be (72/95), however the probability of the second choice depends on what the first hard cover book was. Below I have attached a picture of a tree diagram I drew because I thought it may be helpful.





The tree diagram below is related to a different problem, which I actually understood. The problem statement wanted to know the probability of someone testing positive and actually having a disease. 3% of the population actually have the disease, 97% don't. The test they use is 99% accurate when the subject tests positive and 98% accurate when they test negative. I understand that if we want to know if someone who tests positive actually has the disease we need to take into account someone testing positive when they dont have it. Since we know what scenario we are after I understand why it ends up being .0297/(.0297+.0194) instead of .0194/(.0297+.0194)







The part that confuses me about the picking without replacement book problem is that we are not concerned about one specific scenario as in the disease question, also there was no picking with out replacement in that one so I do not know how that will play into all of it.

Let me know what you all think and if a tree diagram is the right was to go about answering this problem

 

[pka]

Lets use \(H_2\) second is hardback. \(F_1\) for fiction first, \(N_1\) for non-fiction first.
Also \(\mathscr{P}(H_2F_1)\) is the probability second is hardback and first fiction.

So \(\mathscr{P}(H_2)=\mathscr{P}(H_2F_1)+\mathscr{P}(H_2N_1)\)
\(\mathscr{P}(H_2)=\mathscr{P}(H_2|F_1)\mathscr{P}(F_1)+\mathscr{P}(H_2|N_1)\mathscr{P}(N_1)\)
Can you finish?

 

[uteskier]

Lets use \(H_2\) second is hardback. \(F_1\) for fiction first, \(N_1\) for non-fiction first.
Also \(\mathscr{P}(H_2F_1)\) is the probability second is hardback and first fiction.

So \(\mathscr{P}(H_2)=\mathscr{P}(H_2F_1)+\mathscr{P}(H_2N_1)\)
\(\mathscr{P}(H_2)=\mathscr{P}(H_2|F_1)\mathscr{P}(F_1)+\mathscr{P}(H_2|N_1)\mathscr{P}(N_1)\)
Can you finish?

My apologies, I should have proof read my original post...

I forgot to mention that we are picking without replacement. The probability of picking fiction first will always be the same. However, the fiction book that was picked first could have been hardcover OR paperback. These two scenarios affect the number of hard covered books that are left. Theres 1 less hardcover if the first book was fiction and hardcover. If the first book was fiction and paperback then the amount of hardcover books stays the same.

I can definitely plug in numbers to the equation you provided, but I think it may be incorrect for what Im trying to do. I could be wrong but I think the equation I would use is different based on the information I forgot to include originally.

 

[Steven G]

My apologies, I should have proof read my original post...

I forgot to mention that we are picking without replacement. The probability of picking fiction first will always be the same. However, the fiction book that was picked first could have been hardcover OR paperback. These two scenarios affect the number of hard covered books that are left. Theres 1 less hardcover if the first book was fiction and hardcover. If the first book was fiction and paperback then the amount of hardcover books stays the same.

I can definitely plug in numbers to the equation you provided, but I think it may be incorrect for what Im trying to do. I could be wrong but I think the equation I would use is different based on the information I forgot to include originally.

I saw in you title that this was done w/o replacement. Maybe you should find the probability of each case that you listed above and .....