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anonymous

  • one year ago

20 candy canes 17 cracked draw 3 what is the probability of getting exactly 1 cracked candy cane

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  1. campbell_st
    • one year ago
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    you can use a tree diagram |dw:1435124494372:dw|

  2. anonymous
    • one year ago
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    I appreciate the reply, however the tree diagram doesn't solve it mathematically. I'm confused on the formula and what numbers to use where. Thank you again!

  3. campbell_st
    • one year ago
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    then assigen probabilities |dw:1435124643358:dw| multiply along the branches that have only 1 cracked cane to get the probabilities then sum the probabilities for 3 ways you can get 1 cracked cane

  4. campbell_st
    • one year ago
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    the tree diagram does solve it mathematically... you just need to assign the probabilities... and multiply along the bracnhes for your outcomes...

  5. campbell_st
    • one year ago
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    why is it a combination... if you select a cane, there is 1 fewer to select from for the 2nd pick...and then 1 less again for the 3rd pick

  6. anonymous
    • one year ago
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    its a combination because thats how the book wants me to solve it. lol

  7. campbell_st
    • one year ago
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    really.... wow... I think the book lacks sufficient information in the question ...

  8. anonymous
    • one year ago
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    the 3/20^2 is the number of potential outcomes right?

  9. kropot72
    • one year ago
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    There are 3 positions for the cracked cane to appear. 3C1 = 3.

  10. campbell_st
    • one year ago
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    so are you doing binomial probabilities... of are you doing probabilities using counting techniques..?

  11. kropot72
    • one year ago
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    17/20 is the probability of selecting a cracked cane. 3/20 is the probability of selecting a cane that is not cracked.

  12. anonymous
    • one year ago
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    its a probability of combinations question according to the book

  13. campbell_st
    • one year ago
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    that is on the 1st choice... then what happens...do you have 1 fewer to choose from..?

  14. campbell_st
    • one year ago
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    I wonder if the tree diagram returns the same answer... that would be freaky...

  15. anonymous
    • one year ago
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    I don't understand where \[(\frac{ 3}{ 20 })^{2}\] comes into play

  16. anonymous
    • one year ago
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    I think I'm having a hard time plugging the information into the formulas. I'm not sure what info to extrapolate and plug

  17. campbell_st
    • one year ago
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    perhaps the book has the answer

  18. campbell_st
    • one year ago
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    then draw a diagram... and assign probabilities...

  19. anonymous
    • one year ago
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    @campbell_st I wish, I tried, I've been banging my head on this problem for 2 hours

  20. campbell_st
    • one year ago
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    well tree diagrams are a starting point in counting techniques and probabilities

  21. anonymous
    • one year ago
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    crap the answer he typed out just disapeared before I could get it

  22. kropot72
    • one year ago
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    Sorry, the distribution is sampling without replacement. So the hypergeometric distribution is needed. \[\large P(1\ cracked)=\frac{17C1\times3C2}{20C3}\]

  23. campbell_st
    • one year ago
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    what answer is in the book..?

  24. anonymous
    • one year ago
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    I have 4 multiple choice answers A) 0.2484 B) .3466 C) .1552 D).0447

  25. kropot72
    • one year ago
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    One of those choices is obtained by using my last posted equation.

  26. anonymous
    • one year ago
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    ok, if you don't mind, can you walk me through how you arrived at the numbers you used in your formula

  27. anonymous
    • one year ago
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    I think I have a basic understanding. 17c1 is 1 out of 17 cracked. Im not sure what the 3c2 is and the 20c3 is choosing 3 out 20.....????????

  28. kropot72
    • one year ago
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    The number of ways of choosing a cracked cane is 17. For each of these 17 choices there are 3C2 ways of choosing 2 canes that are not cracked. Therefore the number of ways of choosing 1 cane that is not cracked is given by 17 * 3C2. the number of ways of choosing 3 canes is given by 20C3.

  29. kropot72
    • one year ago
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    Therefore the probability of getting exactly one cracked cane is given by: \[\large \frac{17\times3C2}{20C3}\]

  30. campbell_st
    • one year ago
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    |dw:1435126127667:dw| so on each branch the numerators are 17 x 3 x 2 = 102 and there are 3 valid branches = 306 the denominators are 20 x 19 x 18 = 6840 the the P(1 crack) = 306/6840 calculate that probability

  31. kropot72
    • one year ago
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    Not surprisingly the tree diagram gives the correct answer also :)

  32. anonymous
    • one year ago
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    @campbell_st, somewhere along the line I came up with your equation hours ago but was off somewhere along the lines, but had it close.

  33. campbell_st
    • one year ago
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    wow... what a fluke.... the tree diagram does it without formulae... and it doesn't do your head in

  34. anonymous
    • one year ago
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    @kropot72, Thank you for your help with this, I don't know why I was having such a brain fart doing doing it, now only 25 more questions lol

  35. campbell_st
    • one year ago
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    one of the 1st rules of maths ... if you can draw a diagram... do it... it helps to make thing clear...

  36. kropot72
    • one year ago
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    @campbell_st Your result was no fluke. Please take credit where its due.

  37. campbell_st
    • one year ago
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    my point is... you don't need to memorise formulae... use a basic maths tool kit that uses diagrams... and keep going back to it.... and you'll find you use the same skills over and over...

  38. anonymous
    • one year ago
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    @campbell_st I agree with what your saying but sometimes formulas are time savers (testing etc) so thats why I try to use the formulas to beat them into my head.

  39. campbell_st
    • one year ago
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    well I hope it helps you pass your exam

  40. anonymous
    • one year ago
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    How do I read the word problem and pick out whether its a perm, combination, or probability of combination? are there keywords I should look for?

  41. kropot72
    • one year ago
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    For permutations, the order of selection matters. For example voting for official positions in an election, or the finishing order in a race. Generally the keyword 'ways' indicates that combinations apply.

  42. anonymous
    • one year ago
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    awesome thank you again!!

  43. kropot72
    • one year ago
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    You're welcome :)

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