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Probability Question.. please answer if you are mathematically great... :)

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  • Probability Question.. please answer if you are mathematically great... :)

    There is a raffle being held and I buy 3 tickets.
    The raffle barrel contains 100 tickets.
    What is the probability of NOT winning ANY prizes?
    Or Winning NO prizes. (its the same thing, right????)

    Opps.. forget to say that there was 3 prizes, 1st, 2nd, 3rd...
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  • #2
    Getting other people to do your homework for you? :-)

    Assuming tickets are not replaced in the ticket pool:

    Chance of not winning a prize on first ticket is 97/100
    Chance of not winning a prize on second ticket is 96/99 (one less ticket)
    Chance of not winning a prize on third ticket is 95/98 (two less tickets)

    So chances of winning on any of them:

    97/100 * 96/99 * 95/98

    or

    91.1811997 %

    (approximately)



    Rob
    Old Systems retired due to new car
    New system at design/prototype stage on BeagleBoard.

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    • #3
      oh boy..

      this reminds me how much I hated this topic in math..
      I understand everything completely except this, must be something that MY logic can't understand... (silly, because I usually do understand things..)

      I also remember how happy I was when we had final-exams....
      I rushed through the exam to see if there weren't any questions like these... When I found out the exam didn't have any off these question I made the exam very relaxed.... (scoring 9.4 out off 10)

      Good luck Kid !!!!
      Raas - The Netherlands
      ME: VIA epia m10000, lilliput 7', opus 150w, 80gb<br>
      GF: IBM Thinkpad 380, ext. 3.5 80gb, 40x4, PB-IR

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      • #4
        Have you considered doing your own homework? Or rather, not doing, as I did, wich resulted in my spectacular 30% in Grade 13 math...
        Player: Pentium 166MMX, Amptron 598LMR MB w/onboard Sound, Video, LAN, 10.2 Gig Fujitsu Laptop HD, Arise 865 DC-DC Converter, Lexan Case, Custom Software w/Voice Interface, MS Access Based Playlists
        Car: 1986 Mazda RX-7 Turbo (highly modded), 1978 RX-7 Beater (Dead, parting out), 2001 Honda Insight
        "If one more body-kitted, cut-spring-lowered, farty-exhausted Civic revs on me at an intersection, I swear I'm going to get out of my car and cram their ridiculous double-decker aluminium wing firmly up their rump."

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        • #5
          Rob got the math right, but stated it wrong.

          The probability of winning no prizes is the probability of not winning first AND not winning second, AND not winning third.

          So the 91% (didn't do the math, trusting rob) is the probability of not winning anything. In otherwords, you have an approximately 9% chance of winning something.

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          • #6
            I'm also trusting Rob's math...

            Originally posted by LoreleiGuy
            Rob got the math right, but stated it wrong.

            The probability of winning no prizes is the probability of not winning first AND not winning second, AND not winning third.

            So the 91% (didn't do the math, trusting rob) is the probability of not winning anything. In otherwords, you have an approximately 9% chance of winning something.
            This also assumes that the tickets are not re-entered to the barrel after being drawn.

            In other words, the chances will change if it is possible that a single ticket could win all three prizes...

            C!
            http://carpc.riposte.net
            I reserve all rights in connection to each post I author, without exception.

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            • #7
              Ok, thanx for the help, and yes, the tickets are NOT entered back into the barrel..... but, that answer isn't right.. (or so my maths teacher says, and I agree.. :P BTW, this was a question in a maths test we had, and was the only one I got wrong.. So, any more STABS (or mathematically based) at the right answer??
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              • #8
                How about you post the correct answer and we'll tell you why the problem was mistated or why your teacher is wrong.


                Rob
                Old Systems retired due to new car
                New system at design/prototype stage on BeagleBoard.

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                • #9
                  Well, I might give it a try.. (let's see if i can remember something)

                  first round you have 3/100 chance to win something..
                  assuming you haven't won anything, your chances for the second round are 3/99.. still assuming you didn't win in the 2nd round. the chances of winning something in the 3rd round is 3/98

                  so.. you have to calculate what the chances are in EACH round, and them add them together...

                  ( 3/100 + 3/99 + 3/98 ) * 100 = 9.09152752 % chance of winning something..

                  (I remember something from class that we everytime almost had the answer, but it just wasn't right.. didn't differ a lot though...
                  My answer doesn't differ that much from Rob's answer, but it's different... eg. 100-91.1811997 = 8.8188003 %)

                  If this is not correct, I can't help it.. this was my worst subject ever on math...

                  Greetz,

                  Raas
                  Raas - The Netherlands
                  ME: VIA epia m10000, lilliput 7', opus 150w, 80gb<br>
                  GF: IBM Thinkpad 380, ext. 3.5 80gb, 40x4, PB-IR

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                  • #10
                    Originally posted by Rob Withey
                    How about you post the correct answer and we'll tell you why the problem was mistated or why your teacher is wrong.


                    Rob
                    In a min or 2... and your answer is wrong, what I said was the exact question.. you just went about working out teh answer the wrong way... I think... I hope..... Or I am going to look like a bit of an idiot..
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                    • #11
                      Anyone else wanna havea stad at it?? I need some evidence that the Question was too hard..
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                      • #12
                        Time to give the answer??
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                        • #13
                          OKAY!!!!!!!!!

                          the correct answer is......

                          97/100 * 97/99 * 97/98 ....................... which equals.....

                          94.070604 %

                          about 94/100

                          WHY???

                          the TOP number doesn't change becasue there is always 3 tickets left that COULD win, but don't...

                          The bottom number goes down by 1 each time as there is 1 ticket drawn from the barrel...

                          Any questions???
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                          • #14
                            that pretty much explains why I wasn't any good at this subject..
                            ehehehheheheh
                            Raas - The Netherlands
                            ME: VIA epia m10000, lilliput 7', opus 150w, 80gb<br>
                            GF: IBM Thinkpad 380, ext. 3.5 80gb, 40x4, PB-IR

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                            • #15
                              yep........
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