Fan Forum - View Single Post - The Homework Thread #6 ~ Need a Little Help? Ask Here.

Caroo · 06-15-2008, 12:58 PM

Quote:

Originally Posted by mcgoslinginlove (View Post)

ok well if i've done my math correctly there's a 99.98% chance that they can all be used although I wouldn't quote me on that This is reallycomplicated and I think I might be doing something wrong. I can't figure it out. what Math course is this? I'll ask my step mom if she's here, She's a math major.

Well, I finally understand it after a lot of hours and some help from a friend, so this is how it's done

- 2% of 60% is (2/100)x(60/100)=120/10000=1,2%
- 0,9% of 40% is (0,9/100)x(40/100)=36/10000=0,36%
so in total the (1,2%+0,36%)=1,56% of the keys can't be used.

- What's the chance they can all be used?
The joint probability of a series of events is P=P(E1) x P(E2)...x P(En)
In this case on 5 events the probability is (100% - 1,56%) x (100% - 1,56%) x (100% - 1,56%) x (100% - 1,56%) x (100% - 1,56%) = 92,44%

- What's the chance only one of them can be used?
It's the joint probability of "one can be used" and "4 can't be used" so:
(100% - 1,56%) x 1,56% x 1,56% x 1,56% x 1,56% => (5,83 e-8 )^5

- What's the chance one or more can be used?
It's 100% -the chanche that all of them can't be used so:
100%- [(1,56%)x(1,56%)x(1,56%)x(1,56%)x(1,56%)]=100% - 9,239e-10= almost 1...

Thank you anyway