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jackson24 Group Title

QPROBLEM 7 : 2.0 POINTS A school is running a raffle. There are 100 tickets, of which 3 are winners. You can assume that tickets are sold by drawing at random without replacement from the available tickets. Teacher X buys 10 raffle tickets, and so does Teacher Y. Find the chance that one of those two teachers gets all three winning tickets.

  • one year ago
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  1. satellite73 Group Title
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    okay that was wrong, i did not read carefully

    • one year ago
  2. satellite73 Group Title
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    lets compute the probability that teacher X gets all three winners. there are \(\binom{100}{10}\) ways to pick 10 out of 100. if it includes the three winning tickets then there are \(\binom{3}{3}=1\) way for them to be included and \(\binom{97}{7}\) ways to choose the other 7, so the probability that teacher X gets all three is \[\frac{\binom{97}{3}}{\binom{100}{10}}\]

    • one year ago
  3. satellite73 Group Title
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    damn typo last line should be \[\frac{\binom{97}{7}}{\binom{100}{10}}\]

    • one year ago
  4. satellite73 Group Title
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    this is the same as the probability that teacher Y gets all the winners, and since these events are clearly disjoint (they can't both have all the winners) you can get the probability that one or the other gets all the winners doubling the above answer

    • one year ago
  5. kropot72 Group Title
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    The probability that the 20 tickets bought between the 2 teachers contain the 3 winners is \[\frac{\left(\begin{matrix}97 \\ 17\end{matrix}\right)}{\left(\begin{matrix}100 \\ 20\end{matrix}\right)}\] Given that the 20 tickets bought between the 2 teachers include the 3 winners, the probability that one of the teachers has the 3 winners is \[\frac{\left(\begin{matrix}17 \\ 7\end{matrix}\right)}{\left(\begin{matrix}20 \\ 10\end{matrix}\right)}\] The required probability is \[\frac{\left(\begin{matrix}97 \\ 17\end{matrix}\right)}{\left(\begin{matrix}100 \\ 20\end{matrix}\right)}\times \frac{\left(\begin{matrix}17 \\ 7\end{matrix}\right)}{\left(\begin{matrix}20 \\ 10\end{matrix}\right)}\]

    • one year ago
  6. njux Group Title
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    What is the easiest way to calculate the probability for this problem in decimals? Is there some online calculator that could simplify and accelerate the process of caclulation? Thanx.

    • one year ago
  7. kropot72 Group Title
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    A calculator with the factorial function will be helpful. For example \[\left(\begin{matrix}97 \\ 17\end{matrix}\right)=\frac{97!}{17!80!}=\frac{97\times 96\times 95\times 94\times ....\times 89\times 88\times 87\times 86\times 85\times 84\times 83\times 82\times 81}{17!}\]

    • one year ago
  8. njux Group Title
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    I have found online caclulator with factorial function, and, using the above given formula for the probability that 20 tickets bought between the 2 teachers contain the 3 winners I got the number 0,7045. Could someone just please double-check if I got the correct answer? Thx

    • one year ago
  9. kropot72 Group Title
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    Your result is a long way from being correct.

    • one year ago
  10. njux Group Title
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    Well, what is then the correct answer? And could you please explain step by step how you got it? Take some other numbers if you want, I just want to understand the principle. Many thanx.

    • one year ago
  11. abhi_abhi Group Title
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    \[\frac{\left(\begin{matrix}97 \\ 17\end{matrix}\right)}{\left(\begin{matrix}100 \\ 20\end{matrix}\right)}\times \frac{\left(\begin{matrix}17 \\ 7\end{matrix}\right)}{\left(\begin{matrix}20 \\ 10\end{matrix}\right)}=0.000742115\] m using excel for calculation... is that correct ans?????

    • one year ago
  12. abhi_abhi Group Title
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    @kropot72 @njux

    • one year ago
  13. njux Group Title
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    @abhi_abhi I have no idea, I am still trying to figure out the easiest way to do all those factorial calculations you have already done. I hope someone will be able to confirm whether you have the right answer or not. Best,

    • one year ago
  14. KABRIC Group Title
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    so what is the answer

    • one year ago
  15. KABRIC Group Title
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    i checked it its not correct its wrong guys

    • one year ago
  16. KABRIC Group Title
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    kropot72 whats wrong with you

    • one year ago
  17. kropot72 Group Title
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    The calculation by @abhi_abhi gives the same result as I get (0.000742). Have you tried the result of the method posted by @satellite73 as follows: \[2\times \frac{\left(\begin{matrix}97 \\ 7\end{matrix}\right)}{\left(\begin{matrix}100 \\ 10\end{matrix}\right)}\]

    • one year ago
  18. JULIAKAPRI Group Title
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    well the my answer is 0.0015

    • one year ago
  19. JULIAKAPRI Group Title
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    and its right

    • one year ago
  20. JULIAKAPRI Group Title
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    NOP ITS WRONG

    • one year ago
  21. kropot72 Group Title
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    @JULIAKAPRI Have you posted your calculation. If not would you care to do so?

    • one year ago
  22. kropot72 Group Title
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    My previous attempt at a solution is not correct. I believe the following method gives a valid solution. Assuming that the teachers' tickets are drawn first, the probability that the 20 tickets bought between the 2 teachers contain the 3 winners is \[\frac{\left(\begin{matrix}97 \\ 17\end{matrix}\right)}{\left(\begin{matrix}100 \\ 20\end{matrix}\right)}=0.00705\] Given that the 20 tickets bought between the 2 teachers include the 3 winners, the probability that one of the teachers has the 3 winners is can be found from the binomial distribution. The probability of a random ticket being a winner is 3/20. the probability of exactly 3 tickets out of a sample of 10 tickets is given by \[P(3\ winners\ out\ of\ 10\ tickets)=\left(\begin{matrix}10 \\ 3\end{matrix}\right)(0.15)^{3}(0.85)^{7}=0.129834\] The probability that one of the two teachers gets all three winning tickets is therefore 0.00705 * 0.129834 = 0.000915

    • one year ago
  23. hlpwntd Group Title
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    N=100, G=3, n=10, g=3 probability teacher X gets the 3 winning tickets is: ((G,g)(N-G,n-g)/(N,n)) probability teacher Y gets the 3 winning tickets is: ((G,g)(N-G,n-g)/(N,n)) probability either teacher X or teacher Y gets the 3 winning tickets is: ((G,g)(N-G,n-g)/(N,n))+((G,g)(N-G,n-g)/(N,n))

    • one year ago
  24. njux Group Title
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    @hlpwntd, does it mean that @satellite73's answer (see above) is correct?

    • one year ago
  25. hlpwntd Group Title
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    Looks pretty good to me

    • one year ago
  26. njux Group Title
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    @hlpwntd, the result I got after doing the calculation is 0,00148423. How does it look to you? :) Thx.

    • one year ago
  27. abhi_abhi Group Title
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    0.0015

    • one year ago
  28. hlpwntd Group Title
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    Looks pretty good to me, rounded or not, depending on if you had to round or not

    • one year ago
  29. njux Group Title
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    @hlpwntd and @abhi_abhi, thanks!

    • one year ago
  30. JULIAKAPRI Group Title
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    0.0015 is the correct answer

    • one year ago
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