if I do something five times and there's a 10% chance of succeeding each time, what is the chance that I will have succeeded at least once after all five times?
where's felt
Quote from: Geno on January 31, 2009, 07:22:55 PM
where's felt
superbananaplace: i don't know
AHHHHHHHHHHH >.<
10%
10%
I think
Quote from: Tri4se on January 31, 2009, 07:47:48 PM
10%
I think
I thought that made sense, but then I thought it might not. doodthing;
Quote from: 'но',кел on January 31, 2009, 07:54:15 PM
I thought that made sense, but then I thought it might not. doodthing;
Well at first I thought it was .001%, but that's getting all 5 right. You're just asking for one. So 10%
combinatorics THE POWER OF GOD
since order doesn't count, we'll use the binomial coefficient and because you're looking for at least one success we'll sum up the probabilities of each of one through five successes bassir;
if i remember all this correctly, then i think we would be dealing with 5c1 / 10 + 5c2 / 10^2 + 5c3 / 10^3 + 5c4 / 10^4 + 5c5 / 10^5 which gives you a probability of .61051 so that's about a 61% chance akudood;
Quote from: guff on January 31, 2009, 07:59:53 PM
combinatronics THE POWER OF GOD
since order doesn't count, we'll use the binomial coefficient and because you're looking for at least one success we'll sum up the probabilities of each of one through five successes bassir;
if i remember all this correctly, then i think we would be dealing with 5c1 / 10 + 5c2 / 10^2 + 5c3 / 10^3 + 5c4 / 10^4 + 5c5 / 10^5 which gives you a probability of .61051 so that's about a 61% chance akudood;
That's good
Quote from: 'но',кел on January 31, 2009, 08:09:08 PM
That's good
actually i think i might have screwed up somewhere because after running a few quick tests in python it seems to be closer to 41% but hey i'm a hell of a lot closer than everyone else so far and combinatorics really isn't my thing akudood;
Quote from: guff on January 31, 2009, 08:12:29 PM
actually i think i might have screwed up somewhere because after running a few quick tests in python it seems to be closer to 41% but hey i'm a hell of a lot closer than everyone else so far and combinatorics really isn't my thing akudood;
Teach me to speak math.
Quote from: guff on January 31, 2009, 08:12:29 PM
actually i think i might have screwed up somewhere because after running a few quick tests in python it seems to be closer to 41% but hey i'm a hell of a lot closer than everyone else so far and combinatorics really isn't my thing akudood;
teach me to do math
oh durr i overthunked it n_u
the probability of no successes is [tex](1 - .1)^5 = .59049[/tex], and since we're looking for the probability of one or more successes, that's the same as [tex]1 - P(no successes) = .40951[/tex] which fits with the 41% i was getting in my testing okay problem solved akudood;
Quote from: Nothing on January 31, 2009, 08:14:57 PM
teach me to do math
okay today's class is on commutative operations in ordered fields i hope you've read chapter 4
Quote from: Tri4se on January 31, 2009, 08:13:33 PM
Teach me to speak math.
okay done n_u
67
at first i was liek wat then guff made good sense because all losing is .9^5 so that's how there's .59049 and so it could b one win or moar but if all lose, that's like opposite, so if probablilites of these 2 opposites you get 1 so then 1-.59049 and so there you get .40951. I see it now. hocuspocus;
100 because you won't fail
Well, whats the chance of you failing all of them? Minus that from 100% baddood;
Edit: In case you are a moron, 100 * (1 - .9^5)