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Math Challenge II-A Combinatorics Question 8.29

 
 
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Math Challenge II-A Combinatorics Question 8.29
by Zeyin Wu - Sunday, 3 October 2021, 4:59 PM
 

Question 8.29:

Pick 44 balls at once from 77 red, 66 green, and 55 yellow. What is the probability you get at least 11 red, and at least 11 green and at least 11 yellow ball? Hint: Be careful here!

What will the answer be if I DO NOT pick the 4 balls at once?

Thanks,

 
Picture of John Lensmire
Re: Math Challenge II-A Combinatorics Question 8.29
by John Lensmire - Monday, 4 October 2021, 8:19 PM
 

If we do not pick the balls at once we need to be a little careful with how we understand the problem. Here is where the terms with replacement and without replacement come into play.

(i) If we pick the 4 balls with replacement, then we put the ball back after each pick (so the same exact ball could be picked more than once). Then each of the picks is independent of the others, and this is discussed more in chapter 9.

(ii) If we pick the 4 balls without replacement, then in fact the probability is the same whether we view the balls all chosen at once or whether we pick the balls one by one (in order). If we pick each ball one by one (in order) then actually both the numerator and denominator are multiplied by $4!$ (since we're doing with order) so the final probability is not affected.

Picture of Zeyin Wu
回复: Re: Math Challenge II-A Combinatorics Question 8.29
by Zeyin Wu - Wednesday, 6 October 2021, 6:28 PM
 

Thanks, what's the answer if there is no replacement though? I tried but I can't figure it out. 

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Re: 回复: Re: Math Challenge II-A Combinatorics Question 8.29
by Dr. Kevin Wang - Thursday, 7 October 2021, 1:37 AM
 

To pick 4 balls without replacement, and with at least one of each color, there are 3 cases: (1) 2 red, 1 green, and 1 yellow; (2) 1 red, 2 green, and 1 yellow; (3) 1 red, 1 green, and 2 yellow.  These 3 cases do not overlap, so you can count them separately, and add them up.