How Do You Solve These Probability and Combination Problems?

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Homework Help Overview

The discussion revolves around problems related to probability and combinations, specifically focusing on committee selection and drawing counters with replacement. Participants are exploring how to correctly apply combinatorial principles to various scenarios involving constraints on selections.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss different methods for calculating the number of ways to form committees under specific conditions, such as including or excluding certain individuals.
  • There is confusion regarding the application of combinatorial formulas, particularly in relation to the constraints given in the problems.
  • Some participants express uncertainty about the correct interpretation of the problems, especially regarding the use of replacement in probability calculations.

Discussion Status

The discussion is ongoing, with participants seeking clarification on their approaches and the correctness of their calculations. Some have received feedback on their interpretations, but there is still a lack of consensus on the correct methods to apply for certain problems.

Contextual Notes

Participants are working under the constraints of homework guidelines and are referencing textbook problems, which may lead to discrepancies in understanding. There are also mentions of past exam papers that may not align with textbook examples.

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Homework Statement


1/ A committee of 6 is selected from 10 people, of whom A and B are two. How many committees can be form excluding A if B is included ?
2/ A committee of 5 is to be chosen from 6 boys and 4 girls. How many committees can be form containing a particular boy X but excludes a particular girl Y?
3/ In a box there are 4 black counters and 5 white counters. 4 counters are withdrawn one at a time. (the first counter being replaced before the second counter is drawn). Find the probability that two black counters and two white counters are drawn, in any order.

2. Homework Equations :
Premutation and Combination


The Attempt at a Solution


1/ I do this two way.
1st: 8C5 (no B but A) + 8C5 (no A but B) + 8C6 (no B no A) = 140
2st: 10C6 - 8C4 (A and B included) = 140
I have no question for this one. However, the question comes when i do n.o 2

2/ I do the same thing because it looks very similar to the question n.o 1. I made 10C5 - 8C3. However the answer is 8C4 :|. And of course the answers are not the same number form the calculator. Please check for me :(

3/ I do (4C2 x 5C2) / 9C4. And i know the answer is not correct as the book :(

Please help me :( The exam is coming !
 
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Sarina3003 said:

Homework Statement


1/ A committee of 6 is selected from 10 people, of whom A and B are two. How many committees can be form excluding A if B is included ?
2/ A committee of 5 is to be chosen from 6 boys and 4 girls. How many committees can be form containing a particular boy X but excludes a particular girl Y?
3/ In a box there are 4 black counters and 5 white counters. 4 counters are withdrawn one at a time. (the first counter being replaced before the second counter is drawn). Find the probability that two black counters and two white counters are drawn, in any order.

2. Homework Equations :
Premutation and Combination


The Attempt at a Solution


1/ I do this two way.
1st: 8C5 (no B but A) + 8C5 (no A but B) + 8C6 (no B no A) = 140
2st: 10C6 - 8C4 (A and B included) = 140
I have no question for this one. However, the question comes when i do n.o 2

2/ I do the same thing because it looks very similar to the question n.o 1. I made 10C5 - 8C3. However the answer is 8C4 :|. And of course the answers are not the same number form the calculator. Please check for me :(

3/ I do (4C2 x 5C2) / 9C4. And i know the answer is not correct as the book :(

Please help me :( The exam is coming !

In (1), why are you counting (A but no B)+(B but no A)+(no A and no B)? The question was about (B but no A). Same issue for (2). For (3), why are you using the formula for probabilities without replacement, when the problem stated very clearly that drawing was with replacement?
 
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1/ that is how the answer of the book gives and i understand it in thay way
3/ "replacement" part is which i do not know how to deal with
 
Sarina3003 said:
1/ that is how the answer of the book gives and i understand it in thay way
3/ "replacement" part is which i do not know how to deal with

If you copied out problem #1 correctly from the book, then the book's answer is wrong. However, if the actual problem given in the book is different from the one you wrote here, then the book's answer may be correct. The book's answer for #2 is correct, and the way you wrote #1, it is the same as #2 but with slightly different numbers.
 
#1 It is actually from the book but when i realize thay i can do like the 2nd way it's the same answer. It's not wrong when no A but B + B but no A +... because i have asked the teacher. And that question is exactly what is said in the book. And then when i see 2#, i see it almost the same but i cannot use the method of #1 to apply for 2#. Thats why I'm asking. May i miss out something? :(
 
For number 2# its not from the book I've been dealing but in the past papers exam of the college!
 
Sarina3003 said:
#1 It is actually from the book but when i realize thay i can do like the 2nd way it's the same answer. It's not wrong when no A but B + B but no A +... because i have asked the teacher. And that question is exactly what is said in the book. And then when i see 2#, i see it almost the same but i cannot use the method of #1 to apply for 2#. Thats why I'm asking. May i miss out something? :(

Your solution to #1 is NOT the solution to the problem you wrote, which says "How many committees can be form excluding A if B is included ?"---copied and pasted from your post. Your answer is for the related, but different problem: "How many committees can be form(ed) that includes at most one of the two students A or B?" Can't you see the difference? The second problem is for (A and not B)+(B and not A) + (not A and not B), but the question you wrote asks simply for (B and not A)---just that, and no more.
 
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