Question

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.

Answer 1

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: \[
\frac {5!}{20^5}
\]

Answer 2

@wio can I leave my answer like that or do I have to actually multiply them out?

Answer 3

You should multiply them out.

Answer 4

\[
5!=5\times4\times3\times2\times 1=20\times 3\times 2
\]

Answer 5

\[
\frac{20\times 3\times 2}{ 20^5} = \frac{3\times 2}{20^4} = \frac{3}{10^4\times 2^3}
\]At this point you may just want to use a calculator.

Answer 6

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?

Answer 7

I don't understand the reasoning behind that so I can't tell you what is wrong with it.

Answer 8

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

Answer 9

I meant to put (1-20)=20! at the top

Answer 10

Well, the reason we use \(20^5\) instead of \(20!/15!\) is because it didn't say we can't have repeated numbers.

Answer 11

It didn't say we can't chose something like 1 1 1 1 1

Answer 12

got it... thx