anonymous
  • anonymous
what is the probability that your five number combination will be the winning combination in a lottery that involves the selection of random numbers from 1 through 20? the order of the five numbers is irrelevant.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
When order matters, the total number of tickets (sample space) is \(20^5\) We can remove order by dividing out the number of permutations from the sample space. The \(5\) numbers can be permuted \(5!\) ways. Thus I ways say the probability is: \[ \frac {5!}{20^5} \]
anonymous
  • anonymous
@wio can I leave my answer like that or do I have to actually multiply them out?
anonymous
  • anonymous
You should multiply them out.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
\[ 5!=5\times4\times3\times2\times 1=20\times 3\times 2 \]
anonymous
  • anonymous
\[ \frac{20\times 3\times 2}{ 20^5} = \frac{3\times 2}{20^4} = \frac{3}{10^4\times 2^3} \]At this point you may just want to use a calculator.
anonymous
  • anonymous
this is what i thought it was: 20*19*18*17*16*15! / 15! * 5*4*3*2*1 = 1,860,480 / 120 = 15,504... you can only get one winning ticket so... 1 / 15,504 is that wrong?
anonymous
  • anonymous
I don't understand the reasoning behind that so I can't tell you what is wrong with it.
anonymous
  • anonymous
total no:(1-20)=29 use combination ,u have to select only 5 no's total no of ways=20 C 5 =20! / (15! * 5!) =20*19*18*17*16*15! / (15! * 5*4*3*2*1) =15,504 since only one will be correct out of these ways therefore answer is 1/15,504
anonymous
  • anonymous
I meant to put (1-20)=20! at the top
anonymous
  • anonymous
Well, the reason we use \(20^5\) instead of \(20!/15!\) is because it didn't say we can't have repeated numbers.
anonymous
  • anonymous
It didn't say we can't chose something like 1 1 1 1 1
anonymous
  • anonymous
got it... thx

Looking for something else?

Not the answer you are looking for? Search for more explanations.