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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
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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campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1you can use a tree diagram dw:1435124494372:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.1then 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

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1the tree diagram does solve it mathematically... you just need to assign the probabilities... and multiply along the bracnhes for your outcomes...

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1why 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
 one year ago
Best ResponseYou've already chosen the best response.0its a combination because thats how the book wants me to solve it. lol

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1really.... wow... I think the book lacks sufficient information in the question ...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the 3/20^2 is the number of potential outcomes right?

kropot72
 one year ago
Best ResponseYou've already chosen the best response.1There are 3 positions for the cracked cane to appear. 3C1 = 3.

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1so are you doing binomial probabilities... of are you doing probabilities using counting techniques..?

kropot72
 one year ago
Best ResponseYou've already chosen the best response.117/20 is the probability of selecting a cracked cane. 3/20 is the probability of selecting a cane that is not cracked.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its a probability of combinations question according to the book

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1that is on the 1st choice... then what happens...do you have 1 fewer to choose from..?

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1I wonder if the tree diagram returns the same answer... that would be freaky...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't understand where \[(\frac{ 3}{ 20 })^{2}\] comes into play

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.1perhaps the book has the answer

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1then draw a diagram... and assign probabilities...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@campbell_st I wish, I tried, I've been banging my head on this problem for 2 hours

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1well tree diagrams are a starting point in counting techniques and probabilities

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0crap the answer he typed out just disapeared before I could get it

kropot72
 one year ago
Best ResponseYou've already chosen the best response.1Sorry, the distribution is sampling without replacement. So the hypergeometric distribution is needed. \[\large P(1\ cracked)=\frac{17C1\times3C2}{20C3}\]

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1what answer is in the book..?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I have 4 multiple choice answers A) 0.2484 B) .3466 C) .1552 D).0447

kropot72
 one year ago
Best ResponseYou've already chosen the best response.1One of those choices is obtained by using my last posted equation.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok, if you don't mind, can you walk me through how you arrived at the numbers you used in your formula

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.1The 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
 one year ago
Best ResponseYou've already chosen the best response.1Therefore the probability of getting exactly one cracked cane is given by: \[\large \frac{17\times3C2}{20C3}\]

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1dw: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
 one year ago
Best ResponseYou've already chosen the best response.1Not surprisingly the tree diagram gives the correct answer also :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@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
 one year ago
Best ResponseYou've already chosen the best response.1wow... what a fluke.... the tree diagram does it without formulae... and it doesn't do your head in

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@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
 one year ago
Best ResponseYou've already chosen the best response.1one of the 1st rules of maths ... if you can draw a diagram... do it... it helps to make thing clear...

kropot72
 one year ago
Best ResponseYou've already chosen the best response.1@campbell_st Your result was no fluke. Please take credit where its due.

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1my 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
 one year ago
Best ResponseYou've already chosen the best response.0@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
 one year ago
Best ResponseYou've already chosen the best response.1well I hope it helps you pass your exam

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How 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
 one year ago
Best ResponseYou've already chosen the best response.1For 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
 one year ago
Best ResponseYou've already chosen the best response.0awesome thank you again!!
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