mameadows Group Title 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. 10 months ago 10 months ago

1. wio Group Title

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?

3. wio Group Title

You should multiply them out.

4. wio Group Title

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

5. wio Group Title

$\frac{20\times 3\times 2}{ 20^5} = \frac{3\times 2}{20^4} = \frac{3}{10^4\times 2^3}$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?

7. wio Group Title

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

10. wio Group Title

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

11. wio Group Title

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