Solve the probability without replacement problem

[chwala]

Homework Statement

see attached.

Relevant Equations

understanding of probability concept.

My interest is solely on question $6.V$ only. This is the question paper;

This is my working to solution; I was able to use tree diagram in this question.

This is the mark scheme:

I would appreciate other ways of tackling the problem. Cheers!

 

[andrewkirk]

Rather than writing out the tree of all possible outcomes, use combinations.

Imagine the biscuits are numbered 1 to 30, with 1-12 wrapped and the others not.

The answer is N / M where
- N is the number of different possible unordered selections of four with two from 1-12 and two from 13-30
- M is the number of different possible unordered selections of four different biscuits from 1-30

Express M as a combination ${}_nC_k$ and get its value.

Express N as K x L where:
- K is the number of different possible unordered selections of two different biscuits from 1-12
- L is the number of different possible unordered selections of two different biscuits from 13-30

Express K and L as combinations and get their values.

Then you can calculate N, and the n get the answer.

That is the approach set out in your last photo, after the "OR"

 

[chwala]

Rather than writing out the tree of all possible outcomes, use combinations.

Imagine the biscuits are numbered 1 to 30, with 1-12 wrapped and the others not.

The answer is N / M where
- N is the number of different possible unordered selections of four with two from 1-12 and two from 13-30
- M is the number of different possible unordered selections of four different biscuits from 1-30

Express M as a combination ${}_nC_k$ and get its value.

Express N as K x L where:
- K is the number of different possible unordered selections of two different biscuits from 1-12
- L is the number of different possible unordered selections of two different biscuits from 13-30

Express K and L as combinations and get their values.

Then you can calculate N, and the n get the answer.

That is the approach set out in your last photo, after the "OR"

Yes, i am conversant with the combination approach... of course I already looked at it (the marks scheme) before posting this problem, ...i wanted to challenge myself and use a different approach and not just limit myself to the mark scheme. Cheers and thanks Andrewkirk...

the approach of,
$\frac {12C_2 ×18C_2}{30C_4},$ is more appropriate...