View Full Version : getting the probability
01-24-2006, 09:04 AM
i would like to know how many sets of 6 will i get within the numbers 1-42, 1-45, and 1-49. just wanna know the probability. hope to get an answer
Ewe! That's Tough. Maybe Am Absent Or Sick The Time My Prof Teach That Lesson.
Kidding Aside Let Me Think About It. My Brain Is Kinda Tarnish Now:(
This Work Makes Me Dumb.
01-24-2006, 10:03 AM
:rolleyes: :rolleyes: hey, that's smart!!! I have learned getting the probability wayback high school, but no longer remember...!!! :confused: I really do hope you get the answer soon!!!
God Bless!!!:p ;)
01-24-2006, 01:34 PM
:o :( my husband and i would want to know our chance of winning the lottery by using the "PROBABILITY"...we are trying to get all of the numbers we can combine but we are ending up unsuccessful. Fun stuff has the lottery number combinations but it is not showing how many combinations you can make.:confused: :confused:
01-25-2006, 03:19 AM
That is a simple formula, but it is very tough to get your mind wrapped around the problem. Even I hated math like this back in college. :)
You need a calculator with a factorial key, it is the one with a n! on it. (Though you could do it by hand, it is more work)
The formula is:
C = n! / (r! * (n-r)!)
C is the number of possible combinations
n is the total number of objects
r is how many taken at a time
So lets use your first one as an example. You have 42 numbers, and you take 6 at a time.
C = 42! / (6! *(42-6)!)
C = 5,245,786
So about 5.25 million possible combinations for picking 6 numbers from 1-42
01-25-2006, 03:34 AM
I know I said I hated that type of math, but that was actually very fun.
Who wants to work out some more? :cool:
01-25-2006, 06:08 AM
That's great Rob, you'll make a good math teacher...
01-25-2006, 04:18 PM
Oh my gosh!!! :eek: :eek: Rob you're a wizzard... I really love math but I don't excel in it. That is why I am so thankful you're there to help out. ;)
Godspeed!!! :) :)
A whale of thanks to this site!!!:cool: :p
hi! i'm having a hard time figuring out this problem. "An integer number from 1 through 100 is generated randomly. what percent is the probability that it is neither a multiple of 3 nor 5".
hope to get response from this problem.. thanks
04-24-2007, 02:15 AM
I might get this wrong, It is late, and I'm having a fraught week :) Check out the logic at least, it might put you on the right track.
There are 33 multiples of 3 between 1 and 100(inclusive).
There are 20 multiples of 5 between 1 and 100(inclusive).
6 of these multiples are comon to both groups. That gives 53 - 6 = 47 integers not allowed, so 53 integers are allowed.
Your percent of it being one of these allowed numbers is 53%, or 0.53probibility.
What does anyone else think?
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