anonymous
  • anonymous
20 candy canes 17 cracked draw 3 what is the probability of getting exactly 1 cracked candy cane
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
campbell_st
  • campbell_st
you can use a tree diagram |dw:1435124494372:dw|
anonymous
  • anonymous
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!
campbell_st
  • campbell_st
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

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More answers

campbell_st
  • campbell_st
the tree diagram does solve it mathematically... you just need to assign the probabilities... and multiply along the bracnhes for your outcomes...
campbell_st
  • campbell_st
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
anonymous
  • anonymous
its a combination because thats how the book wants me to solve it. lol
campbell_st
  • campbell_st
really.... wow... I think the book lacks sufficient information in the question ...
anonymous
  • anonymous
the 3/20^2 is the number of potential outcomes right?
kropot72
  • kropot72
There are 3 positions for the cracked cane to appear. 3C1 = 3.
campbell_st
  • campbell_st
so are you doing binomial probabilities... of are you doing probabilities using counting techniques..?
kropot72
  • kropot72
17/20 is the probability of selecting a cracked cane. 3/20 is the probability of selecting a cane that is not cracked.
anonymous
  • anonymous
its a probability of combinations question according to the book
campbell_st
  • campbell_st
that is on the 1st choice... then what happens...do you have 1 fewer to choose from..?
campbell_st
  • campbell_st
I wonder if the tree diagram returns the same answer... that would be freaky...
anonymous
  • anonymous
I don't understand where \[(\frac{ 3}{ 20 })^{2}\] comes into play
anonymous
  • anonymous
I think I'm having a hard time plugging the information into the formulas. I'm not sure what info to extrapolate and plug
campbell_st
  • campbell_st
perhaps the book has the answer
campbell_st
  • campbell_st
then draw a diagram... and assign probabilities...
anonymous
  • anonymous
@campbell_st I wish, I tried, I've been banging my head on this problem for 2 hours
campbell_st
  • campbell_st
well tree diagrams are a starting point in counting techniques and probabilities
anonymous
  • anonymous
crap the answer he typed out just disapeared before I could get it
kropot72
  • kropot72
Sorry, the distribution is sampling without replacement. So the hypergeometric distribution is needed. \[\large P(1\ cracked)=\frac{17C1\times3C2}{20C3}\]
campbell_st
  • campbell_st
what answer is in the book..?
anonymous
  • anonymous
I have 4 multiple choice answers A) 0.2484 B) .3466 C) .1552 D).0447
kropot72
  • kropot72
One of those choices is obtained by using my last posted equation.
anonymous
  • anonymous
ok, if you don't mind, can you walk me through how you arrived at the numbers you used in your formula
anonymous
  • anonymous
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.....????????
kropot72
  • kropot72
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.
kropot72
  • kropot72
Therefore the probability of getting exactly one cracked cane is given by: \[\large \frac{17\times3C2}{20C3}\]
campbell_st
  • campbell_st
|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
kropot72
  • kropot72
Not surprisingly the tree diagram gives the correct answer also :)
anonymous
  • anonymous
@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.
campbell_st
  • campbell_st
wow... what a fluke.... the tree diagram does it without formulae... and it doesn't do your head in
anonymous
  • anonymous
@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
campbell_st
  • campbell_st
one of the 1st rules of maths ... if you can draw a diagram... do it... it helps to make thing clear...
kropot72
  • kropot72
@campbell_st Your result was no fluke. Please take credit where its due.
campbell_st
  • campbell_st
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...
anonymous
  • anonymous
@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.
campbell_st
  • campbell_st
well I hope it helps you pass your exam
anonymous
  • anonymous
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?
kropot72
  • kropot72
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.
anonymous
  • anonymous
awesome thank you again!!
kropot72
  • kropot72
You're welcome :)

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