ShinkuDragon Wrote:n = 4, 8, 12 (depending on targets)The numbers you listed are correct.
r = 0 (i want to find out the chance of it activating at least once, so i use zero for chance of fail, and then 100-answer for chance of success)
p = 0.05 (5/100)
q = 0.95 (95/100)
and just formula it out, correct? can be used for impale and so on.
i'd keep asking about the card example, but i think it's better not >.< since each new statement confuses me.
For probability, when events are completely independent, meaning that the outcome of one event does not depend on the one that happened before, you can just multiply the probabilities together.
It's much easier when you want only one thing to happen rather than combinations. For example, if you want to find the probability of all successes on 4 targets. prob = 0.05^(4). Or the probability of all failures, prob = 0.95^(4).
The binomial formula is just more general in that in can account for combinations. For example, how many ways can you have 1 success in 4 targets: (where S is success, F is fail)
SFFF, FSFF, FFSF, FFFS
That is what is captured in the 'X choose Y' term.
If you're only looking for all successes or all fails, there's only one possible combination.
The card picking example can get quite complex. There's a difference between:
1. Picking and then replacing (all events are independent, 1/4 chance to get hearts each time.)
2. Picking, but not replacing (events are now conditional upon what happened before: if you get hearts the first time, then the next card only has a 12/51 chance, not 13/52 = 1/4 like before; but if you didn't get hearts, the next card has a 13/51 = 1/4 chance.)
3. Picking all 5 cards at once. (In this case, the probability of picking hearts for the cards is not 1/4 for each card. The hypergeometric distribution needs to be used)
And @above question: I assume that he means that with 12 targets, the chance of getting a 'success' are pretty good. ( prob of at least one success = 1 - prob(all fails) = 1 - 0.95^12 = 0.46 )