Small combinatorics question

[Muffins]

Hi ya'll ! I was hoping you could help me get a 'second opinion' to a problem I ran into. It's most likely one of the most basic combinatorics question one can run into but I'm notoriously awful at it.
Here goes:

There are 8 tops in a closet. 3 red ones by the sizes of S,M and L; 3 blue ones, sizing S,M and L again, and 2 white ones, which are M and L.
2 tops are chosen randomly.
What's the probability that the chosen tops will not be white or/and small? *clarification: We need 2 shirts to be a combination of blue/red/medium/large.

I tried to solve this using combinatorics as well as conditional probability, both led me to different answers.

using combinatorics, I did-
[(2 out of 3)+(2 out of 3)+(1 out of 3)*(1 out of 3)]/(2 out of 8)
meaning we either select 2 shirts out of the 3 red OR 1 out of the red, 1 out of the blue, OR 2 out of the 3 blue ones.

My husband thought we should do it this way:
[(6 out of 8)*(4 out of 6)+(5 out of 7)*(3 out of 5)]\(2 out of 8)
so- we choose total 6 out of the 8 tops, (3 blue, 3 red), time it by the selection of sizes we want, and we repeat this picking the second top.
His end result was >1 however.

I tried using conditional probability too, and basically did (3/8*1/3)*2

Will gladly appreciate input about this as I'm a bit confused

P.S- Sorry for the headache!

 

[DrPhil]

Hi ya'll ! I was hoping you could help me get a 'second opinion' to a problem I ran into. It's most likely one of the most basic combinatorics question one can run into but I'm notoriously awful at it.
Here goes:

There are 8 tops in a closet. 3 red ones by the sizes of S,M and L; 3 blue ones, sizing S,M and L again, and 2 white ones, which are M and L.
2 tops are chosen randomly.
What's the probability that the chosen tops will not be white or/and small? *clarification: We need 2 shirts to be a combination of blue/red/medium/large.

I tried to solve this using combinatorics as well as conditional probability, both led me to different answers.

using combinatorics, I did-
[(2 out of 3)+(2 out of 3)+(1 out of 3)*(1 out of 3)]/(2 out of 8)
meaning we either select 2 shirts out of the 3 red OR 1 out of the red, 1 out of the blue, OR 2 out of the 3 blue ones.

My husband thought we should do it this way:
[(6 out of 8)*(4 out of 6)+(5 out of 7)*(3 out of 5)]\(2 out of 8)
so- we choose total 6 out of the 8 tops, (3 blue, 3 red), time it by the selection of sizes we want, and we repeat this picking the second top.
His end result was >1 however.

I tried using conditional probability too, and basically did (3/8*1/3)*2

Will gladly appreciate input about this as I'm a bit confused

P.S- Sorry for the headache!

"conditional probability" is a good approach, but lets make the arithmetic easier by doing it as a 2-dimensional joint distribution. Make a table:

\(\displaystyle \begin{array}{r|rrr|r} \\
& Red & Blue & White & Total \\ \hline \\ S & 1 & 1 & 0 & 2 \\ M & 1 & 1 & 1 & 3 \\ L & 1 & 1 & 1 & 3 \\ \hline \\ Total & 3 & 3 & 2 & 8 \end{array} \)

There are 4 in the boxes \(\displaystyle (M + L)\cap(Red + Blue)\), out of a total of 8 in all boxes. That is, 4 "true" out of 8 "total".

Do the probability of choosing 2 "true" without replacement:
P(2) = P(1st true)*P(2nd true)/(rearrangements) = (4/8)(3/7)/2

EDIT - can't convince myself that probabilities get divided by rearrangements! Counts do, so that only distinct pairs are counted, but then the total number of distinct pairs is (8*7)/2.

So I think the probability is (4/8)(3/7) = 12/56

2nd EDIT: made table prettier by using "array" in LaTeX

 

[Muffins]

Hat's off to you,

Thank you Dr Phil.