i might have to refresh, but i'd be happy to try.

David you're an angel... My mother will be shipping you a thank you gift if you can get me through this class =)...

Lexi:

I used to love statistics in Uni. Don't remember much of it except the basics tho. If David can't help, and if it's not advanced stuff, happy to give it a shot. 

David:

Let's see if you're worth anything is statistics. Here's the question:

A pilot has a 10% chance of having an accident each time he flies. If he flies 10 times, what's the probability that he had 0 accidents during all the flights?

Nir

it's after midnight on a friday, but my gut instinct is 100% - (0.1^10) as each flight is independent from the other.

hi i would like to help im preparing phd in econometrics and used to do statistics studies during 8 years so let me know what do you want exactly thanks Moh

Lexi:

Given Zonki's answer, I'd stay well clear of him in matters of statistics! 

Zonki:

Let's say you got it wrong cos it's after midnight on a Friday. Give you the benefit of the doubt...(-:

Zinnour:

Yalla, let's see if you can answer the question from post 3 in this thread...

Nir

Hi NIr,

the probability that the pilot dont have an accdient is 9/10 in every time so duing 10 times he will have the probability of 9/10 * 9/10 * 9/10*......... 10 times which give a final probability of 0,348 chance that the pilot dont have any accident i jope its true.

ok, ok. i got my parentheses wrong..... it's not 1 - 0.1^10 but (1-0.1)^10......

my bad.

Fridaynight, pfffft

Anyway, the airline is to be avoided, probability is about 0.3

0.9^10 = 34.9% chance of having zero accidents for 10 flights.

Correct, 34.9%.

And here's another not-hard one: in a coin toss of 3 times, what's the probability of getting just 1 "tails"?

ok, long answer, assuming all crashes to be unrelated, equiprobable and a crash is deemed "a succes" the probability for k successes in n trials is:

Pr(K=k)  = comb(n, k) *p^k * (1-p)^(n-k)

http://mathworld.wolfram.com/BinomialDistribution.html

Nir !

Lucky you, you are the winner of a tv show and are presented with 3 rooms shielded of by curtains. Behind one door is a lovely car, the two others contain goats.

You choose one at random. The gameshow host now opens one of the curtains displaying a goat, and he offers you the oppertunity to either stick with your original chosen room, or switch.

Should you switch? And if so, why?

OH GEEZ! Nir, I may pop into your office on Monday.. I came by this week to say hello and bring you a kit kat but I must have missssed you!

and Zinnour, Im gonna spend the next couple of days trying to get myself acquainted with the stuff better, but I will definately be looking you up and perhaps we can work out an arrangement. 

I have an extensive social science background (Poli Sci and Philosophy) and my post grad is in International Relations.. so this finance part of the MBA is brutal.

P.S. you boys need a bed time!

Lexi, 

Us boys don't need bed time, us boys need a girl to drag us away from the PC and into the bed!

Nir its not statistics what are you talking about here??? its may be physics or whatttttttttttt hehehehhe

Probably biology ;)

yes may be it seems logical humainnnnnnnnnn biologyyyyyyyyyyyyyyy

That's an unfair one; even famous mathematicians got that one wrong.

Here's an easy one: You are late for a job interview and need to find your socks in the dark so you don't wake up your partner. In the sock drawer there are 5 blue pairs and 8 black pairs. How many socks do you need to take out of the drawer to make sure you have at least one matching pair?

@ Lexillent: I also need to refresh my stats but need to do so anyway because I'm following a course in Advanced Statistics in December as part of my PhD. Let me know if/how I can help you.

It's better to change: Suppose you choose room 1 without switching afterwards. Then you can only win, if the car is really in room 1. If you switch, however, you win in two cases, the car being in room 2 or 3. You basically make use of the knowledge of the gameshow host to double your winning probability from 1/3 to 2/3.

90%, as only the last flight counts.  Had he had an accident in any of the precedent flights, he wouldn't have made it to the next flight... ;)

3/8 (37.5%), as you get only 1 tail in these tosses:

- tail, head, head

- head, tail, head

- head, head, tail

which have 1/8 of probability each.

Hmmm, difficult. Well, in statistics the model assumptions are the most important.  
From extensive life experience I know that socks are *evil*. From this I deduce

1) Socks decay and disappear over time, e.g sock half life

2) Socks are either in known places OR have unknown color, or the Heisenberg Uncertainty Principle for Socks

3) Black and Red socks, or spin-up/spin-down socks are inverse quantum entangled. Selecting a red sock immediately causes the other half of the pair to turn black.

The only optimal strategy is thus settling for the first two that fit, hence the number of to pick is 2, of unknown color, if and only if they haven't disappeared. And they never match in color.

No, there is no information to suggest any of the accidents would be fatal, so the 10% per flight holds for all flights, until we have more information.  An accident could be a mere scratch or broken headlight up to and including the plane and pilot being borked completely ;)

I'd say they evolve and make their getaway..

I just found this great website, with videos that explain math and other stuff. Statistics is near the bottom of the page:

http://www.khanacademy.org/

Oh, ive been googling and youtubing all day.. Zonker is lovely help, as always.  Youtube is the biggest help when it comes to learning excel within, who knew people posted 10000000s of videos.  I just get confused on why something is something, thats where I need explainations =)... its a connecting of dots in such a way ive never been good at unfortunately.  I can do it in words wonderfully, but numbers are like a russian alphabet to me. I guess perfection isnt possible!