Need some help with Probability

I'm not quite sure how to calculate this, so I'd appreciate a little help.

30% scrolls pass at a 30% rate, fail at a 35% rate, and break at a 35% rate.
The % chance of passing 4 30% scrolls in a row is 0.0081, or 0.81%.

What is the chance of passing at least 4 30%s, in the first 5 slots? That is, 5 slots are used, one of them can fail but not break, but 4 must pass? 6 slots? 7 slots?

For all of the above questions, how many scrolls would it take to achieve this result, on average? How many equips would it take on average?

I'm not quite sure how to calculate this, so I'd appreciate a little help.

I'd say base it upon the Binomial Distribution with 0.3 as success and 0.35 (fail w/o break) as failure.

What is the chance of passing at least 4 30%s, in the first 5 slots? That is, 5 slots are used, one of them can fail but not break, but 4 must pass?

So 1.6605%, 302 scrolls on average.

6 slots?

So 2.071575%, 290 scrolls on average.

7 slots?

So 2.04109875%, 294 scrolls on average.

For all of the above questions, how many scrolls would it take to achieve this result, on average?

I added the scroll averages, but I'm not 100% sure on whether they're correct or not.

How many equips would it take on average?

Not 100% sure how to go about this either, it'd take more time I think as you'd have to factor in the chance of breaking and the number of slots an item has.

All in all, my answers (my calculations probably as I only had Windows' Calculator handy) may be wrong, lol, but just trying to give a general idea. I hate probability.

I'd say base it upon the Binomial Distribution with 0.3 as success and 0.35 (fail w/o break) as failure.

Oh pomegranate, that works?

When I was trying to calculate this crap a year or two ago, I thought it was some special case, couldn't figure it out, and resorted to Excel-bashing. (Which does get the probabilities quite nicely, but required a bit more setup.)

Ok, it doesn't conform to the rules of a binomial distribution because p + q =/= 1, but the principle of how it works still applies. The probability of getting 4 scrolls to work and 1 to fail but not break is still P[work] * P[work] * P[work] * P[work] * P[fail w/o break] and since you could have the scrolls appearing in a different order, times it by 5C4.

Ok, it doesn't conform to the rules of a binomial distribution because p + q =/= 1, but the principle of how it works still applies. The probability of getting 4 scrolls to work and 1 to fail but not break is still P[work] * P[work] * P[work] * P[work] * P[fail w/o break] and since you could have the scrolls appearing in a different order, times it by 5C4.

That's a way to say it. It's also possible to show through trinomial distribution that the other factors evaluates to 1 and thus fall away, "becoming" the binomial distribution.