random math question

[russell]

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?

[Geno]

where's felt

[Mettalik]

where's felt

superbananaplace: i don't know

AHHHHHHHHHHH 

[Wrench]

10%

[Tri4se]

10%

I think

[russell]

10%

I think

I thought that made sense, but then I thought it might not. 

[Tri4se]

I thought that made sense, but then I thought it might not. 

Well at first I thought it was .001%, but that's getting all 5 right. You're just asking for one. So 10%

[guff]

combinatorics

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
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 

[russell]

combinatronics

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
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 

That's good

[guff]

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 

[Tri4se]

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 

Teach me to speak math.

[Travis]

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 

teach me to do math

[guff]

oh durr i overthunked it

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 

teach me to do math

okay today's class is on commutative operations in ordered fields i hope you've read chapter 4

Teach me to speak math.

okay done 

[l a c e y]

67

[Pancake Paraphernalia]

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.