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 February 16th, 2018, 08:05 AM #1 Newbie   Joined: Feb 2018 From: California Posts: 5 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, 01:16 PM #2 Global Moderator   Joined: May 2007 Posts: 6,444 Thanks: 565 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, 09:55 AM #3 Newbie   Joined: Feb 2018 From: California Posts: 5 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, 10:20 AM   #4
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Quote:
 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, 12:18 PM #5 Senior Member     Joined: Sep 2015 From: USA Posts: 1,781 Thanks: 919 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 12:23 PM.
 February 18th, 2018, 01:36 PM #6 Newbie   Joined: Feb 2018 From: California Posts: 5 Thanks: 1 Got it! Thank you for the great explanation, romsek! Thanks from greg1313

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