I know I used to know exactly how to figure this out, but I am 10 years removed from any math class and it's 3am so I'm just gonna post it here.

If there is a 2.7% probability of something happening, and you make 7 attempts, what is the probability of it happening 5 out of the 7 times? Obviously I know the chances are gonna be very low, but I'm trying to figure out exactly how low and I can't remember it.

7/11/2011 3:16:34 AM

we talking metric or imperial?

7/11/2011 3:22:46 AM

.027 * .027 * .027 * .027 * .027 * .973 * .973 = 1.35845264 × 10^(-8)
0.00000001358 or 0.000001358%

Good luck.

7/11/2011 3:28:59 AM

As for the 1.35845264 × 10^(-8) answer, you also need to consider which 5 of the 7 are successes--the first 5, the last 5, or any combination of 5 of the 7 trials: C(7,5)=21 different combinations.

So the answer is 21 × 1.35845264 × 10^(-8), about 2.85275 × 10^(-7)

(This assumes it happens exactly 5 of the 7 times, not 6 or all 7).

7/11/2011 3:48:49 AM

^ Valid answer.

7/11/2011 4:03:53 AM

That's like doing long division. There is a much more simple, calculus based formula:

Ignore the parentheses right? Why is this little two so small? It- it's weird. You just-- don't-- you just go by the x. The x means times, so that means four times x two, what is double four?

7/11/2011 4:26:51 AM

Sorry I wasn't 100% clear. Should have said at least 5 out of 7. 6 or 7 wins would have also resulted in a success. So given that, it's the 0.000001358% figure?

7/11/2011 1:27:19 PM

Give me any odds of what you think aliens existing is.

Let's do something large.

Let's do something like....

"There's a 1 out of 1x10^10 chance."

That means there's a one in ten billion chance of aliens existing.

Is that not rare enough for you? Okay. Let's do 1 out of 1x10^15. That's a one in one quadrillion chance.

Based on observations from the Hubble Space Telescope, there are at least 125 billion galaxies in the universe.

It is estimated that at least ten percent of all sun-like stars have a system of planets, there are 6.25×10^18 stars with planets orbiting them in the universe.

If even a billionth of these stars have planets supporting life, there are some 6.25×10^9 (billion) life-supporting planetary systems in the universe.

6,250,000,000.

Now, what are the odds that NONE of these have any life on them?

7/11/2011 1:28:10 PM

If you want at least 5, it's gonna be:

C(7,5)(s^5)(f^2) + C(7,6)(s^6)(f) + C(7,7)(s^7)

(where C is binomial coefficient, s=success=.027, f=fail=.973)

I get about 2.879242 * 10^(-7), but you might want to check me on that part

7/11/2011 5:12:31 PM

This thread makes my head hurt

7/11/2011 5:14:54 PM

cause you're a woman

now get back in the kitchen

7/11/2011 5:21:31 PM

Okay that actually made me laugh

7/11/2011 5:22:31 PM

Copy Pasta is the winner.

7/11/2011 5:23:11 PM

Do my hw for me tww.

message_topic.aspx?topic=610881
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After the 3rd and 4th posts in this thread (Hiro, Cabbage), you would have to be an idiot to not be able to figure it out. They showed you how it works for 5, repeat the procedure for 6 and 7 and then add the 3 answers.


[Edited on July 11, 2011 at 5:28 PM. Reason : ]

7/11/2011 5:26:49 PM

you'd have to be incredibly naive to expect el nacho to search

I'm AstralAdvent and i approved this message.

7/11/2011 5:54:40 PM

A probability / math question in Chit Chat you say?


WHAT ARE THE ODDS OF THAT?

7/11/2011 6:04:45 PM

So 21*((.027)⁵)*((.973)²) +7*((.027)⁶)*(.973)+(.027)⁷.

2.87924235 × 10^-7

V hahaha

7/11/2011 6:55:36 PM

Quote :

"+.027"

wut

7/11/2011 6:57:05 PM

The square root of 69 is 8 something....


(bc I've been trying to work it out )

7/11/2011 6:58:22 PM

^plagiarist.


off lil weezy nonetheless.

7/11/2011 7:22:33 PM

A new car built by my company leaves somewhere traveling at 60 mph. The rear differential locks up. The car crashes and burns with everyone trapped inside. Now, should we initiate a recall? Take the number of vehicles in the field, A, multiply by the probable rate of failure, B, multiply by the average out-of-court settlement, C. A times B times C equals X. If X is less than the cost of a recall, we don't do one.

7/11/2011 7:39:12 PM

ITT 0EPII1 (and probably aholeadventurer, I've got him blocked so I don't know for sure) try unsuccessfully to troll el nacho. el nacho also refers to himself in the third person.

7/11/2011 10:26:34 PM

El Nachó, doesn't even bother to say Thank You. Fuck El Nachó.

7/12/2011 6:33:04 AM

bahahahah

7/12/2011 8:06:21 AM

^^^ it wasn't trolling. if after hiro's and cabbage's posts, you really couldn't figure out how to calculate for 6 successes, you really are stupid.

7/12/2011 9:54:22 AM

Yep. Because not being able to figure out what's going on here:

Quote :

"C(7,5)(s^5)(f^2) + C(7,6)(s^6)(f) + C(7,7)(s^7) (where C is binomial coefficient, s=success=.027, f=fail=.973)"

At a quick glance means I'm stupid.

Or maybe it means I have absolutely no desire to relearn something I have no use for and haven't studied for over 10 years. You're easily one of the least intelligent people that posts on this site, so you calling anyone (aside fron maybe geniusxboy) is pretty damn amusing.

However this reply probably means that successful troll was successful, so to that.

7/12/2011 2:41:14 PM

jesus you fucking moron, i am talking about these posts

Quote :

"Hiro .027 * .027 * .027 * .027 * .027 * .973 * .973 = 1.35845264 × 10^(-8) 0.00000001358 or 0.000001358% Good luck. Cabbage As for the 1.35845264 × 10^(-8) answer, you also need to consider which 5 of the 7 are successes--the first 5, the last 5, or any combination of 5 of the 7 trials: C(7,5)=21 different combinations. So the answer is 21 × 1.35845264 × 10^(-8), about 2.85275 × 10^(-7) (This assumes it happens exactly 5 of the 7 times, not 6 or all 7)."

you can try reading next time:

Quote :

"After the 3rd and 4th posts in this thread (Hiro, Cabbage), you would have to be an idiot to not be able to figure it out. They showed you how IN ENOUGH FUCKING DETAIL it works for 5, repeat the procedure for 6 and 7 and then add the 3 answers."

Quote :

"You're easily one of the least intelligent people that posts on this site"

because you administered iq tests to everybody and have access to everybody's results?


[Edited on July 12, 2011 at 2:48 PM. Reason : fat asshole doesn't know the meaning of "3rd" and "4th", and says i have low intelligence!]

7/12/2011 2:47:52 PM

My estimation of your (complete lack of) intelligence is based on years of stupitity you've pumped into this site as hard as possible. Apparently your estimation of my intelligence is based on the fact that I have no desire to relearn how to do something I forgot how to do several years ago. It's not that I can't figure our how to do it, it's that I simply don't want to.

7/12/2011 2:57:21 PM

Quote :

"[Edited on July 12, 2011 at 2:48 PM. Reason : fat asshole doesn't know the meaning of "3rd" and "4th", and says i have low intelligence!]"

7/12/2011 2:59:03 PM

They are two different answers you spectacular retard. If I didn't remember how to do it in the first place, how am I supposed to know which one is the correct way. You're still obsessed with thinking I'm asking people how to obtain the answer myself. I was just asking for an answer, of which so far I've received 3 different ones. None of them from you, I'll add, since you're apparently so smart, it shouldnt be that hard for you to throw your two cents in. Unless you just happen to be mindlessly trolling. Nah, that couldn't be right.

7/12/2011 3:03:41 PM

This thread is now about internet fist fights.

7/12/2011 3:04:12 PM

I'm fine with that. I have no use for the original info anymore anyway.

7/12/2011 3:05:41 PM

Well in that case, this thread is now about Spiderman.

7/12/2011 3:09:26 PM

Answer:

The plane takes off.

7/12/2011 3:10:18 PM

Quote :

"I was just asking for an answer, of which so far I've received 3 different ones."

YOU. CAN'T. BE. THAT. STUPID.

You have received only one fucking answer, first by Cabbage, and then repeated by Spontaneous.

Remember this:

Quote :

"Sorry I wasn't 100% clear. Should have said at least 5 out of 7. 6 or 7 wins would have also resulted in a success."

Yeah, just 1 fucking answer after that clarification by you... but you are still counting the previous ones even though you phrased the question wrong . Amnesia? Stupidity? Troll thread?

[Edited on July 12, 2011 at 5:59 PM. Reason : obviously me pointing out your stupidity pissed you off enough! ]

7/12/2011 5:49:06 PM

Any time I get annoyed by you, all I have to do is remember that you think it's normal to put shit stained underwear from other dudes up to your face and smell it and I remember exactly what kind of nutjob I'm dealing with. There's no point in even trying to reason with someone who is that deficient.

Kinda puts it all into perspective.

7/12/2011 9:58:43 PM

7/12/2011 10:00:42 PM