## 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?_______

1. anonymous

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2. anonymous

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3. anonymous

1- ( P(No pasta dinner) + P(Exactly one pasta dinner))

4. anonymous

P(No pasta dinner) =$$\huge \frac{ \binom{9}{5}} {\binom{20}{5}}$$

5. anonymous

P(Exactly one pasta dinner) = $$\huge \frac{ \binom{11}{1}\times \binom{9}{4}} {\binom{20}{5}}$$

6. anonymous

how do i solve the 9 and 5 in ( 9 5 )

7. Directrix

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

0.44=44/100=%44

9. anonymous

$$\huge \binom{n}{k} = \frac{n!}{k! \times (n-k)!}$$

10. anonymous

or 1-(no pasta+1pasta)

11. anonymous

hold on, what is the total dinner? is it 20 or 40?

12. anonymous

20

13. anonymous

Directrix answer could be simplified using Vandermonde's identity.

14. anonymous

they are frozen dinners to not pizzas/.... haha typo but shouldnt matter

15. anonymous

come on!!!

16. Directrix

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

no pizza is involved pizza was supposed to be frozen dinners

18. anonymous

i got 15504 for the total possible outcomes but i was using C(20.5) for the total

19. Directrix

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

%90

21. anonymous

I see i need to break down every combination greater than 2 and add them up divide by possible outcomes?

22. Directrix

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

that will help on the exam tomorrow i havent completed this set yet of homework yet either that is due fri. morning

24. Directrix

Where is the next problem?

25. anonymous

$1-\left\{ \frac{\left(\begin{matrix}9 \\ 5\end{matrix}\right)}{\left(\begin{matrix}20 \\ 5\end{matrix}\right)}+\frac{\left(\begin{matrix}9 \\ 4\end{matrix}\right)\left(\begin{matrix}11\\ 1\end{matrix}\right)}{\left(\begin{matrix}20 \\ 5\end{matrix}\right)} \right\}$

26. anonymous

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27. anonymous

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28. anonymous

these are long ways..

29. Directrix

CW, where are you? Did you see the answer to your question about Thursday night on this site?