I came across this question where I was asked to calculate the probability of drawing 2 identical so

juniorychichoa70 2022-04-30 Answered
I came across this question where I was asked to calculate the probability of drawing 2 identical socks from a drawer containing 24 socks (7 black, 8 blue, and 9 green). When solving this question, I tried tackling it using the counting principle directly as opposed to simply applying the combinatorics formula in hopes to get comfortable with that principle. Though, when I looked up the solution at the end of the book, the answer was completely different, and I am not sure why that is.

What I did was as follows: First, I noticed that when we draw the first sock, we have 24 different options to choose from and then only 1 in order to match it with the first one we chose. Then, I divided it by the total number of ways we can draw 2 socks out of 24 and got the expression:
24 1 24 23 = 1 23
The solution at the back of the book took care of each case separately (choosing 2 out of 7 blue socks, OR choosing 2 out of 8 blue socks OR choosing 2 out of 9 green socks), which makes sense, but what is wrong with my approach?
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Answers (1)

alastrimsmnr
Answered 2022-05-01 Author has 18 answers
Say you take a green sock out of the drawer. Then for the socks to be identical, you need another green sock, but not all the socks are green. Furthermore, each colour has a different number of socks, so the probability of the second sock matching are all different and so you cannot group different colours of socks.
The same argument applies to any colour.
Not exactly what you’re looking for?
Ask My Question

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Relevant Questions

asked 2022-05-03
A stained glass window consists of nine squares of glass in a 3x3 array. Of the nine squares, k are red, the rest blue. A set of windows is produced such that any possible window can be formed in just one way by rotating and/or turning over one of the windows in the set. Altogether there are more than 100 red squares in the set. Find k.
first, there are 8 Isometries of a square.
Identity, three rotations (90,-90,180) four reflections (vertical, horizontal, two diagonal axis). let G be the permutation group, then |G|=8, and I can find fix(g) for every g.
can someone give me a hint of how to proceed from there.
asked 2022-05-09
Let A and B be subsets of the finite set S with S = A B and A B = . Denote by P ( X ) the power set of X and denote by | Y | the number of elements in the set Y.
Given a statement | P ( A ) | + | P ( B ) | = | P ( A ) P ( B ) |
Use the Addition Counting Principle to prove or disprove the statement.

I understand that its asking me to find the elements of P ( A ) and P ( B ), but where does P ( X ) and | Y | fit in to solve this question?
asked 2022-05-10
A catering service offers 12 appetizers, 9 main courses, and 6 desserts. A customer is to select 5 appetizers, 3 main courses, and 4 desserts for a banquet. In how many ways can this be done?
asked 2022-05-03
I am to create a six character password that consists of 2 lowercase letters and 4 numbers. The letters and numbers can be mixed up in any order and I can also repeat the same number and letter as well. How many possible passwords are there?
What I have pieced together so far:
Well, from the fundamental counting principle, we would definitely need 26 2 × 10 4 but obviously this is not all the possibilities since I can rearrange letters and numbers. Since it is a password the order matters so would I try and do a permuation of some sort like 6 P 2 since there are 6 slots to try to rearrange 2 objects (letters)?
asked 2022-04-30
I'm missing something here. Let X = { ( 123 ) , ( 132 ) , ( 124 ) , ( 142 ) , ( 134 ) , ( 143 ) , ( 234 ) , ( 243 ) }, A 4 act on X by conjugation (inner automorphisms) and x = ( 123 ), then 4 = | O ( x ) | = | G | / | G x | = 12 / | G x | . However, G x = { 1 }
What's wrong here?
asked 2022-05-02
Ministry of Education are inviting tender for four categories promoting the use of IT in education. Each category consists of 5, 4, 3, and 7 projects, respectively. Each project appears on exactly one category. How many possible projects are there to choose from? Explain your answer.
My Answer: ( 5 + 4 ) + ( 4 + 4 ) + ( 3 + 3 ) + ( 7 + 4 ) = 9 + 8 + 6 + 11 = 34 possible projects to choose from. I used the sum rule here.
Is this correct?
asked 2022-05-09
Let n be a positive integer. Find the number of permutations of ( 1 , 2 , . . . , n ) such that no number remains in its original place.

Solution: To do this, first we are going to count the number of permutations where at least one number remains in its place, according to the inclusion-exclusion principle, we must first add the permutations where a given number is fixed, then subtract the permutations where 2 given numbers are fixed and so on.

To find a permutation that fixes k given elements, we only have to arrange the rest, which can be done in ( n k ) ! ways. However, if we do this for every choice of k elements, we are counting ( n k ) ( n k ) ! = n ! k ! permutations. Since in total there are n ! permutations we get as our result:
n ! ( n ! 1 ! + n ! 2 ! n ! 3 ! + . . . + ( 1 ) n + 1 n ! n ! )
I'm a bit confused about a point of the solution, when we fix k elements and rearrange the other n k, some of the remaining elements will stay fixed in their position right? so why does this work?