Question

@satellite73
I have a probability question for you :D
A lottery consists of 48 possible numbers to be selected. 5 numbers are chosen for the winning lottery combination. If one ticket is purchased, what is the probability the ticket has exactly 3 numbers correct?

Answer 1

I thought this question was worded badly.

Answer 2

Special question for @satellite73

Answer 3

\[\frac{\dbinom{5}{3}\times \dbinom{43}{2}}{\dbinom{48}{5}}\]

Answer 4

number of ways to pick three out of the 5 correct ones times number of ways to pick 2 out of the 43 incorrect ones divided by the number of ways to pick 5 out of 48

Answer 5

i am assuming you can compute these numbers, right?

Answer 6

Yes one moment. I'm going to see if it's right.

Answer 7

My goodness it's right!

Answer 8

How did you do that? I don't understand where your numbers came from?

Answer 9

You did that so quickly too.

Answer 10

imagine that, it is right!

Answer 11

lol

Answer 12

quickly because i didn't actually compute anything, i just thought about what you need to compute
i can explain more if you like

Answer 13

yes please, first can i tell you what i was trying to do so you can correct me?

Answer 14

sure but i have to go soon (ish)

Answer 15

alright i took 5 P 3 which is 60 multiplied by 1/48 * 1/47 * 1/46 * 44/45 * 43/44

Answer 16

ok i think i see what you are doing, but shouldn't it be
5/48 * 4/47 * 3/46 * 43/45 * 42/44 using your method?

Answer 17

that is the probability of getting right right right wrong wrong in that order, then permute

Answer 18

I did 1/48 because I thought since it had to be in the right order

Answer 19

but you have 5 numbers to pick from so the probability that you get one on the first try is 5/48

Answer 20

Like 5 12 3 32 47
1/48 you get a 5, then 1/47 you get a 12, then 1/46 you get a 3...

Answer 21

yeah but order doesn't count in this problem

Answer 22

but it's the lottery
5 12 3 32 47
is different from
12 47 3 5 32
isn't it? (I've never played the lottery before)

Answer 23

just get 3 out of 5
probability first one is good is 5/48
second good given first is good 4/47
third is good given first and second are 3/46
fourth is not good given the above 43/45
fifth is not good give the above 42/ 44

Answer 24

no it isn't different, just have to pick 3 out of the five numbers

Answer 25

it is really much easier to write it the way i did i think. get it in one shot

Answer 26

you pick 5 numbers, three are right, two are wrong. by counting priniciple number of ways you can do this is
\[\dbinom{5}{3}\times \dbinom{4}{2}\] then divide by total number of ways you can pick 5 out of 48 which is
\[\dbinom{48}{5}\] and you are done

Answer 27

Oh so it isn't different. I did not know that, so in the lottery if my card said 10 12 13 11 9
and the draw was 12 13 11 9 10 I would win?

Answer 28

it is much harder to compute one seqeunce and then permute

Answer 29

yeah for power balls etc that would be a win

Answer 30

you have to pick the correct numbers, not in the correct order

Answer 31

typo on first line it should have been
\[\dbinom{5}{3}\times \dbinom{43}{2}\]

Answer 32

alright then, that helps, i don't understand where this came from though
"ok i think i see what you are doing, but shouldn't it be
5/48 * 4/47 * 3/46 * 43/45 * 42/44"

Answer 33

Wait, now i do never mind, how come you multiply that by 5 Choose 3 though?

Answer 34

this line
5/48 * 4/47 * 3/46 * 43/45 * 42/44
computes one particular sequence of right and wrong
right right right wrong wrong

Answer 35

but there are 5 choose 2 combiniations of these rights and wrongs

Answer 36

so if you want to use your method take that and multiply it by 5 choose 2

Answer 37

Right because order doesn't matter! You're a genius! How did you become so good with probability?

Answer 38

Well, thanks for your help. I hope to be as quick as you are at probability someday.

Answer 39

you will learn it for sure
good luck

Answer 40

thanks