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

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

Mathematics
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you can use a tree diagram |dw:1435124494372:dw|
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!
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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the tree diagram does solve it mathematically... you just need to assign the probabilities... and multiply along the bracnhes for your outcomes...
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
its a combination because thats how the book wants me to solve it. lol
really.... wow... I think the book lacks sufficient information in the question ...
the 3/20^2 is the number of potential outcomes right?
There are 3 positions for the cracked cane to appear. 3C1 = 3.
so are you doing binomial probabilities... of are you doing probabilities using counting techniques..?
17/20 is the probability of selecting a cracked cane. 3/20 is the probability of selecting a cane that is not cracked.
its a probability of combinations question according to the book
that is on the 1st choice... then what happens...do you have 1 fewer to choose from..?
I wonder if the tree diagram returns the same answer... that would be freaky...
I don't understand where \[(\frac{ 3}{ 20 })^{2}\] comes into play
I think I'm having a hard time plugging the information into the formulas. I'm not sure what info to extrapolate and plug
perhaps the book has the answer
then draw a diagram... and assign probabilities...
@campbell_st I wish, I tried, I've been banging my head on this problem for 2 hours
well tree diagrams are a starting point in counting techniques and probabilities
crap the answer he typed out just disapeared before I could get it
Sorry, the distribution is sampling without replacement. So the hypergeometric distribution is needed. \[\large P(1\ cracked)=\frac{17C1\times3C2}{20C3}\]
what answer is in the book..?
I have 4 multiple choice answers A) 0.2484 B) .3466 C) .1552 D).0447
One of those choices is obtained by using my last posted equation.
ok, if you don't mind, can you walk me through how you arrived at the numbers you used in your formula
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.....????????
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.
Therefore the probability of getting exactly one cracked cane is given by: \[\large \frac{17\times3C2}{20C3}\]
|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
Not surprisingly the tree diagram gives the correct answer also :)
@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.
wow... what a fluke.... the tree diagram does it without formulae... and it doesn't do your head in
@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
one of the 1st rules of maths ... if you can draw a diagram... do it... it helps to make thing clear...
@campbell_st Your result was no fluke. Please take credit where its due.
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...
@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.
well I hope it helps you pass your exam
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?
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.
awesome thank you again!!
You're welcome :)

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