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mameadows
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.
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: \[ \frac {5!}{20^5} \]
@wio can I leave my answer like that or do I have to actually multiply them out?
You should multiply them out.
\[ 5!=5\times4\times3\times2\times 1=20\times 3\times 2 \]
\[ \frac{20\times 3\times 2}{ 20^5} = \frac{3\times 2}{20^4} = \frac{3}{10^4\times 2^3} \]At this point you may just want to use a calculator.
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?
I don't understand the reasoning behind that so I can't tell you what is wrong with it.
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
I meant to put (1-20)=20! at the top
Well, the reason we use \(20^5\) instead of \(20!/15!\) is because it didn't say we can't have repeated numbers.
It didn't say we can't chose something like 1 1 1 1 1