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 February 16th, 2018, 09:05 AM #1 Newbie   Joined: Feb 2018 From: California Posts: 20 Thanks: 1 High School Probability Question Here's the question: I am going to babysit my nephew for five days. Each day, I will take him to get a kid's meal at a restaurant. The kid's meal comes packed randomly with 1 of 3 possible toys. I would like to know the probability that my nephew gets all 3 toys in 5 trips. How do you solve this question without using formulas, but instead by understanding what is happening? Thank you! February 16th, 2018, 02:16 PM #2 Global Moderator   Joined: May 2007 Posts: 6,852 Thanks: 743 You first have to know what is the distribution of toys. Most likely they are all equally likely. In any case without formulas, you need to list all possibilities (243) and count those you accept. Otherwise you need to use the trinomial formula where each toy occurs at least once. Possible combinations are (3,1,1) and (2,2,1). Thanks from greg1313 and mjsilverfly February 18th, 2018, 10:55 AM #3 Newbie   Joined: Feb 2018 From: California Posts: 20 Thanks: 1 Thank you. Do your (3,1,1) and (2,2,1) represent "yes", "no"? I don't understand what these mean. February 18th, 2018, 11:20 AM   #4
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 Originally Posted by mjsilverfly Thank you. Do your (3,1,1) and (2,2,1) represent "yes", "no"? I don't understand what these mean.
What these are are the number of each type of toy that were chosen from a pick of 5. You'll notice the numbers in each "bag" add up to 5 in each case.

Since you don't care at all about the order the toys are selected in, the pick of 5 can be transformed into sets of 3 numbers, each being how many of each type of toy were chosen. So you end up with sets like

$(0,0,5), (0,1,4), (0,2,3) \dots (0,5,0), (0,4,1) \dots (3, 1, 1), (4, 1, 0), \dots$

the defining rule being that the 3 elements of each bag sum to 5.

You can list these all out, there are just 21 of them. Here they are

$\begin{array}{ccc} 5 & 0 & 0 \\ 4 & 1 & 0 \\ 4 & 0 & 1 \\ 3 & 2 & 0 \\ 3 & 1 & 1 \\ 3 & 0 & 2 \\ 2 & 3 & 0 \\ 2 & 2 & 1 \\ 2 & 1 & 2 \\ 2 & 0 & 3 \\ 1 & 4 & 0 \\ 1 & 3 & 1 \\ 1 & 2 & 2 \\ 1 & 1 & 3 \\ 1 & 0 & 4 \\ 0 & 5 & 0 \\ 0 & 4 & 1 \\ 0 & 3 & 2 \\ 0 & 2 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 5 \\ \end{array}$

Now you are interested in just the sets that have no $0$'s in them.

You can pick these out by hand. There are just 6 of them.

They are
$\begin{array}{ccc} 3 & 1 & 1 \\ 2 & 2 & 1 \\ 2 & 1 & 2 \\ 1 & 3 & 1 \\ 1 & 2 & 2 \\ 1 & 1 & 3 \end{array}$

and thus the probability that you choose at least 1 of each toy in 5 picks is

$p=\dfrac{6}{21}= \dfrac{2}{7}$ February 18th, 2018, 01:18 PM #5 Senior Member   Joined: Sep 2015 From: USA Posts: 2,636 Thanks: 1472 ps. I know you don't want a formula but in general if you are choosing $m$ toys of $n$ different types, then the probability of selecting at least 1 of each type of toy is given by $p = \dfrac{\dbinom{m-1}{n-1}}{\dbinom{m+n-1}{n-1}}$ In the problem at hand, $n=3,~m=5$ and we get $p = \dfrac{\dbinom{5-1}{3-1}}{\dbinom{5+3-1}{3-1}}= \dfrac{\dbinom{4}{2}}{\dbinom{7}{2}} = \dfrac{6}{21} = \dfrac{2}{7}$ I won't charge you any extra for the fomula. pps: $\dbinom{k}{j} = \dfrac{k!}{j!(k-j)!}$ $k! = (k)(k-1)(k-2) \dots (3)(2)$ Thanks from greg1313 Last edited by romsek; February 18th, 2018 at 01:23 PM. February 18th, 2018, 02:36 PM #6 Newbie   Joined: Feb 2018 From: California Posts: 20 Thanks: 1 Got it! Thank you for the great explanation, romsek! Thanks from greg1313 Tags high, probability, question, school Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post neko Differential Equations 1 May 2nd, 2014 03:02 AM tomast80 Math Books 3 December 30th, 2010 01:45 PM squerlyq Algebra 8 January 6th, 2009 10:10 AM lanvin12 Algebra 2 November 27th, 2008 06:47 PM symmetry Algebra 4 May 14th, 2007 08:28 AM

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