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anonymous

  • 4 years ago

There are 20 frozen pizza: 11 Pasta, 5 Chicken, and 4 Seafood dinners. The student selects 5 of them. What is the probability that at least 2 of the dinners selected are pasta dinners?_______

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  1. anonymous
    • 4 years ago
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    |dw:1328166063969:dw|

  2. anonymous
    • 4 years ago
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    |dw:1328166168847:dw|

  3. anonymous
    • 4 years ago
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    1- ( P(No pasta dinner) + P(Exactly one pasta dinner))

  4. anonymous
    • 4 years ago
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    P(No pasta dinner) =\( \huge \frac{ \binom{9}{5}} {\binom{20}{5}} \)

  5. anonymous
    • 4 years ago
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    P(Exactly one pasta dinner) = \( \huge \frac{ \binom{11}{1}\times \binom{9}{4}} {\binom{20}{5}} \)

  6. anonymous
    • 4 years ago
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    how do i solve the 9 and 5 in ( 9 5 )

  7. Directrix
    • 4 years ago
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    There are 20 pizzas. Five are to be chosen. 11 are pasta, 9 are not pasta. The number of ways to choose 5 pizzas with at least 2 being pasta can be done by calculating the number of pizzas with exactly 2, 3, 4, and 5 respectivvely pasta toppings Two Pastas ==> C(11,2) times C(9,3) = Three Pasta ==> C(11,3) times C(9,2) = Four Pasta ==> C(11,4) times C(9,1) = Five Pasta ==> C(11,5) times C(9,0) =

  8. anonymous
    • 4 years ago
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    0.44=44/100=%44

  9. anonymous
    • 4 years ago
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    \( \huge \binom{n}{k} = \frac{n!}{k! \times (n-k)!} \)

  10. anonymous
    • 4 years ago
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    or 1-(no pasta+1pasta)

  11. anonymous
    • 4 years ago
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    hold on, what is the total dinner? is it 20 or 40?

  12. anonymous
    • 4 years ago
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    20

  13. anonymous
    • 4 years ago
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    Directrix answer could be simplified using Vandermonde's identity.

  14. anonymous
    • 4 years ago
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    they are frozen dinners to not pizzas/.... haha typo but shouldnt matter

  15. anonymous
    • 4 years ago
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    come on!!!

  16. Directrix
    • 4 years ago
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    Two Pastas ==> C(11,2) times C(9,3) = 4 620 Three Pasta ==> C(11,3) times C(9,2) = 5 940 Four Pasta ==> C(11,4) times C(9,1) = 2 970 Five Pasta ==> C(11,5) times C(9,0) = 462 Total Desirable Outcomes: 13 992

  17. anonymous
    • 4 years ago
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    no pizza is involved pizza was supposed to be frozen dinners

  18. anonymous
    • 4 years ago
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    i got 15504 for the total possible outcomes but i was using C(20.5) for the total

  19. Directrix
    • 4 years ago
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    The number of possible outcomes is C(20,5) = 15 504 P(two or more pasta toppings) = 13 992 / 15 504 = 583 / 646 = .902 ish

  20. anonymous
    • 4 years ago
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    %90

  21. anonymous
    • 4 years ago
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    I see i need to break down every combination greater than 2 and add them up divide by possible outcomes?

  22. Directrix
    • 4 years ago
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    There are quicker ways for computation but for understanding, I thought this case approach would be the best way to explain it. Sorry about the pizza mishap. Wow! What would a pasta pizza taste like? :=)

  23. anonymous
    • 4 years ago
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    that will help on the exam tomorrow i havent completed this set yet of homework yet either that is due fri. morning

  24. Directrix
    • 4 years ago
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    Where is the next problem?

  25. anonymous
    • 4 years ago
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    \[1-\left\{ \frac{\left(\begin{matrix}9 \\ 5\end{matrix}\right)}{\left(\begin{matrix}20 \\ 5\end{matrix}\right)}+\frac{\left(\begin{matrix}9 \\ 4\end{matrix}\right)\left(\begin{matrix}11\\ 1\end{matrix}\right)}{\left(\begin{matrix}20 \\ 5\end{matrix}\right)} \right\} \]

  26. anonymous
    • 4 years ago
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    |dw:1328167831526:dw|

  27. anonymous
    • 4 years ago
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    |dw:1328167862822:dw|

  28. anonymous
    • 4 years ago
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    these are long ways..

  29. Directrix
    • 4 years ago
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    CW, where are you? Did you see the answer to your question about Thursday night on this site?

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