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February 16th, 2018, 08:05 AM  #1 
Newbie Joined: Feb 2018 From: California Posts: 17 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: 6,704 Thanks: 670 
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).

February 18th, 2018, 09:55 AM  #3 
Newbie Joined: Feb 2018 From: California Posts: 17 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  
Senior Member Joined: Sep 2015 From: USA Posts: 2,370 Thanks: 1274  Quote:
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: 2,370 Thanks: 1274 
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 $p = \dfrac{\dbinom{m1}{n1}}{\dbinom{m+n1}{n1}}$ In the problem at hand, $n=3,~m=5$ and we get $p = \dfrac{\dbinom{51}{31}}{\dbinom{5+31}{31}}= \dfrac{\dbinom{4}{2}}{\dbinom{7}{2}} = \dfrac{6}{21} = \dfrac{2}{7}$ I won't charge you any extra for the fomula. pps: $\dbinom{k}{j} = \dfrac{k!}{j!(kj)!}$ $k! = (k)(k1)(k2) \dots (3)(2)$ 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: 17 Thanks: 1 
Got it! Thank you for the great explanation, romsek! 

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