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jackson24

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.

  • 11 months ago
  • 11 months ago

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  1. satellite73
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    okay that was wrong, i did not read carefully

    • 11 months ago
  2. satellite73
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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 \[\frac{\binom{97}{3}}{\binom{100}{10}}\]

    • 11 months ago
  3. satellite73
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    damn typo last line should be \[\frac{\binom{97}{7}}{\binom{100}{10}}\]

    • 11 months ago
  4. satellite73
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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

    • 11 months ago
  5. kropot72
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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 \[\frac{\left(\begin{matrix}97 \\ 17\end{matrix}\right)}{\left(\begin{matrix}100 \\ 20\end{matrix}\right)}\times \frac{\left(\begin{matrix}17 \\ 7\end{matrix}\right)}{\left(\begin{matrix}20 \\ 10\end{matrix}\right)}\]

    • 11 months ago
  6. njux
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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.

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

    • 11 months ago
  8. njux
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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

    • 11 months ago
  9. kropot72
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    Your result is a long way from being correct.

    • 11 months ago
  10. njux
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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.

    • 11 months ago
  11. abhi_abhi
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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?????

    • 11 months ago
  12. abhi_abhi
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    @kropot72 @njux

    • 11 months ago
  13. njux
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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,

    • 11 months ago
  14. KABRIC
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    so what is the answer

    • 11 months ago
  15. KABRIC
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    i checked it its not correct its wrong guys

    • 11 months ago
  16. KABRIC
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    kropot72 whats wrong with you

    • 11 months ago
  17. kropot72
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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: \[2\times \frac{\left(\begin{matrix}97 \\ 7\end{matrix}\right)}{\left(\begin{matrix}100 \\ 10\end{matrix}\right)}\]

    • 11 months ago
  18. JULIAKAPRI
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    well the my answer is 0.0015

    • 11 months ago
  19. JULIAKAPRI
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    and its right

    • 11 months ago
  20. JULIAKAPRI
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    NOP ITS WRONG

    • 11 months ago
  21. kropot72
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    @JULIAKAPRI Have you posted your calculation. If not would you care to do so?

    • 11 months ago
  22. kropot72
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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

    • 11 months ago
  23. hlpwntd
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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))

    • 11 months ago
  24. njux
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    @hlpwntd, does it mean that @satellite73's answer (see above) is correct?

    • 11 months ago
  25. hlpwntd
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    Looks pretty good to me

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

    • 11 months ago
  27. abhi_abhi
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    0.0015

    • 11 months ago
  28. hlpwntd
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    Looks pretty good to me, rounded or not, depending on if you had to round or not

    • 11 months ago
  29. njux
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    @hlpwntd and @abhi_abhi, thanks!

    • 11 months ago
  30. JULIAKAPRI
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    0.0015 is the correct answer

    • 11 months ago
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