Dividing into even groups

Printable View

September 8th 2009, 08:29 PM
jfz23

Dividing into even groups

I took an intro to probability class last semester and the first week of the intermediate class is a little review. I don't remember too much so can someone confirm to see if this is right?

There are nine indistinguishable items and they are to be divided evenly among three assembly lines.

(A) How many ways can this be done?

My answer is (9 C 3) * (6 C 3) * (3 C 3). There are nine items for the first assembly line, for which you choose 3 for. Then there are six items left for the second line, for which you choose 3. Then three are left for the third.

(B) Two of the items are used and seven are new. What is the probability that a particular line gets both used items?

This one I'm a little unsure about. Any hints?

Thank you.
September 9th 2009, 09:34 AM
Soroban

Hello, jfz23!

Quote:

There are nine indistinguishable items
and they are to be divided evenly among three assembly lines.

(A) How many ways can this be done?

I believe this is a trick question . . .

The key word is "indistinguishable".
The nine items are identical: . $\{\star\;\star\;\star\;\star\;\star\;\star\;\star\ ;\star\;\star \}$

I see only one way to distribute the items:

. . $\begin{array}{ccc} \text{Line 1} & \text{Line 2} & \text{Line 3} \\<br /> <br /> \{\star\;\star\;\star\} & \{\star\;\star\;\star\} & \{\star\;\star\;\star\} \end{array}$

Quote:

(B) Two of the items are used and seven are new.
What is the probability that a particular line gets both used items?

Another trick question . . .

If they had said "both used items are in the same line" (any of the lines),
. . we would use a certain approach.

Since they said "a particular line", the approach is different.

For simplicity, assume that the nine items are distinguishable.
They would be: . $\{A,B,c,d,e,f,g,h,i\}$ . where $A$ and $B$ are used.

There are: . ${9\choose3,3,3} \:=\:{\color{blue}1680}$ possible arrangements.

Suppose the particular line is Line 1.

There is 1 way for A and B to be in Line 1.
Then there are: . ${7\choose1,3,3} \:=\:{\color{red}140}$ ways to distribute the other seven items.

Hence, there are: . $1\cdot140 \:=\:{\color{blue}140}$ ways for $A$ and $B$ to be in Line 1.

Therefore: . $P(A\text{ and }B\text{ in Line 1}) \:=\:\frac{140}{1680} \:=\:\boxed{\frac{1}{12}}$
September 11th 2009, 08:39 AM
jfz23

Thanks for your help! But what if I were to assume that the wrenches are identical?

Would the answer to part A be (9 C 3) * (6 C 3)?

Also, is (7 C 1, 3, 3) = (7 C 1) * (7 C 3) * (7 C 3)?