anonymous
  • anonymous
@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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
I thought this question was worded badly.
saifoo.khan
  • saifoo.khan
Special question for @satellite73
anonymous
  • anonymous
\[\frac{\dbinom{5}{3}\times \dbinom{43}{2}}{\dbinom{48}{5}}\]

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anonymous
  • anonymous
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
anonymous
  • anonymous
i am assuming you can compute these numbers, right?
anonymous
  • anonymous
Yes one moment. I'm going to see if it's right.
anonymous
  • anonymous
My goodness it's right!
anonymous
  • anonymous
How did you do that? I don't understand where your numbers came from?
anonymous
  • anonymous
You did that so quickly too.
anonymous
  • anonymous
imagine that, it is right!
anonymous
  • anonymous
lol
anonymous
  • anonymous
quickly because i didn't actually compute anything, i just thought about what you need to compute i can explain more if you like
anonymous
  • anonymous
yes please, first can i tell you what i was trying to do so you can correct me?
anonymous
  • anonymous
sure but i have to go soon (ish)
anonymous
  • anonymous
alright i took 5 P 3 which is 60 multiplied by 1/48 * 1/47 * 1/46 * 44/45 * 43/44
anonymous
  • anonymous
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?
anonymous
  • anonymous
that is the probability of getting right right right wrong wrong in that order, then permute
anonymous
  • anonymous
I did 1/48 because I thought since it had to be in the right order
anonymous
  • anonymous
but you have 5 numbers to pick from so the probability that you get one on the first try is 5/48
anonymous
  • anonymous
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...
anonymous
  • anonymous
yeah but order doesn't count in this problem
anonymous
  • anonymous
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)
anonymous
  • anonymous
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
anonymous
  • anonymous
no it isn't different, just have to pick 3 out of the five numbers
anonymous
  • anonymous
it is really much easier to write it the way i did i think. get it in one shot
anonymous
  • anonymous
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
anonymous
  • anonymous
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?
anonymous
  • anonymous
it is much harder to compute one seqeunce and then permute
anonymous
  • anonymous
yeah for power balls etc that would be a win
anonymous
  • anonymous
you have to pick the correct numbers, not in the correct order
anonymous
  • anonymous
typo on first line it should have been \[\dbinom{5}{3}\times \dbinom{43}{2}\]
anonymous
  • anonymous
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"
anonymous
  • anonymous
Wait, now i do never mind, how come you multiply that by 5 Choose 3 though?
anonymous
  • anonymous
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
anonymous
  • anonymous
but there are 5 choose 2 combiniations of these rights and wrongs
anonymous
  • anonymous
so if you want to use your method take that and multiply it by 5 choose 2
anonymous
  • anonymous
Right because order doesn't matter! You're a genius! How did you become so good with probability?
anonymous
  • anonymous
Well, thanks for your help. I hope to be as quick as you are at probability someday.
anonymous
  • anonymous
you will learn it for sure good luck
anonymous
  • anonymous
thanks

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