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anonymous

  • 5 years ago

In a sample of 26 hand-held calculators, 20 are known to be nonfunctional. If 6 of these calculators are selected at random, what is the probability that exactly 4 in the selection are nonfunctional? Round to the nearest thousandth. A. 0.667 B. 0.316 C. 0.769 D. 0.300 E. 0

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  1. anonymous
    • 5 years ago
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    Draw a probability tree. I would, but I don;t have the patience, and hate statistics with a passion.

  2. amistre64
    • 5 years ago
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    I know you have a 76.9% chance of picking one the first time..... but what to do after than I would have to read up on :)

  3. amistre64
    • 5 years ago
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    I would assume our initial options would lead to these b,b,b,b,b,b b,b,b,b,b,g b,b,b,b,g,g b,b,b,g,g,g b,b,g,g,g,g b,g,g,g,g,g g,g,g,g,g,g

  4. anonymous
    • 5 years ago
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    \[\frac{20 \times 19 \times 18 \times 17 \times 6 \times 5}{26 \times 25 \times 24 \times 23 \times 22 \times 21} \times ^6 C_4\]

  5. anonymous
    • 5 years ago
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    Well I tried that but that doesn't get me close to any of the solutions?

  6. anonymous
    • 5 years ago
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    Of course it does, I just told you it.

  7. anonymous
    • 5 years ago
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    Do you know what 6C4 means?

  8. anonymous
    • 5 years ago
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    \[^NC_R \equiv \frac{n!}{n!(n-r)!}\]

  9. anonymous
    • 5 years ago
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    Yes lol sorry I think I just plugged in my calc wrong was all! Thanks a bunch!

  10. anonymous
    • 5 years ago
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    You have to explain to me why that is the answer now, though.

  11. amistre64
    • 5 years ago
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    20/26; then the odds go to 19/25, then to 18/24 .....

  12. anonymous
    • 5 years ago
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    My definition of NCR has n! on the bottom when it should be r!, apologies to anyone confused.

  13. anonymous
    • 5 years ago
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    well the equation you just gave me is saying the number of combinations of n elements taken r times

  14. amistre64
    • 5 years ago
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    thats last bit is combinatorical .... i think

  15. anonymous
    • 5 years ago
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    Ya I'm pretty sure it is combinatorical just not positive on how to solve it via the way my book is showing me lol

  16. anonymous
    • 5 years ago
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    Sort of - You need to pick 4 non functioning, and 2 functioning. The first one gives you the overall odds of each in turn, and the second gives you the different orders you can do it in.

  17. anonymous
    • 5 years ago
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    uh ok I'm confused again.... Can you show me on http://www.dabbleboard.com/draw?b=Guest644716&i=0&c=3327a9e1158d904a38fbad735c24ed28cad7bcd9

  18. anonymous
    • 5 years ago
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    No, sorry, I have stuff to do. Draw a probability tree.

  19. anonymous
    • 5 years ago
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    ok Thanks for helping though! Much appreciated... I'll mess around with it until I figure it out:)

  20. anonymous
    • 5 years ago
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    amistre64....what were you trying to say when you posted...20/26; then the odds go to 19/25, then to 18/24? if I keep working my way down what does that do?

  21. anonymous
    • 5 years ago
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    amistre64....what were you trying to say when you posted...20/26; then the odds go to 19/25, then to 18/24? if I keep working my way down what does that do?

  22. anonymous
    • 5 years ago
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    Let's say you choose FFFFNN, the odds are: (20/26 x 19/25 x 18/24 x 17/23) x (6/22 x 5/21) However, you could actually do this in any order, so you have to multiply it by 6C2

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