SOLVED Probability of books on a shelf

May 2010
1,260
410
Mauritius
Ok, this one is giving me some problems:

Qu. On a bookshelf there are works by three authors, three volumes by Gilmore, three
volumes by Lawson and one volume by Patterson. If the books are placed at
random on the shelf, the probability that the volumes of the same author are
together is

A. \(\displaystyle \frac{3}{70}\)

B. \(\displaystyle \frac{1}{140}\)

C. \(\displaystyle \frac{3}{140}\)

D. \(\displaystyle \frac{1}{70}\)

E. \(\displaystyle \frac{1}{35}\)

I could find the probability that three books from Gilmore are together, the separate probability that the books from Lawson being together, but I don't know how to find the intersection between those two sets of probabilities... (Thinking)

P(3 Gilmore together) =\(\displaystyle \frac{3!5!}{7!} = \frac{1}{7}\)

P(3 Lawson together) = \(\displaystyle \frac{1}{7}\)

P(3 Lawson together and 3 Gilmore together) = \(\displaystyle \frac{3!3!3!}{7!} = \frac{3}{70}\)

Total probability = \(\displaystyle \frac{1}{7} + \frac{1}{7} + \frac{3}{70} - P(intersections)\)

How can I solve this?
 
Last edited:

Plato

MHF Helper
Aug 2006
22,508
8,664
Look at \(\displaystyle \dfrac{(3!)^3}{7!}\). WHY?
 
May 2010
1,260
410
Mauritius
Um... I thought of it like this:

(G G G) (L L L) P

G permute among themselves, so 3!
L permute among themselves, so 3!
Then, there are three major items, so permuting again gives 3!
Finally, total different arrangements = 7!

So, I get ((3!)^3)/7!

Or... did I miss something?(Thinking)

EDIT: Ok, I'll see that tomorrow... bedtime here...
 
Last edited:

Plato

MHF Helper
Aug 2006
22,508
8,664
That is correct.
 
Dec 2009
3,120
1,342
Ok, this one is giving me some problems:

Qu. On a bookshelf there are works by three authors, three volumes by Gilmore, three
volumes by Lawson and one volume by Patterson. If the books are placed at
random on the shelf, the probability that the volumes of the same author are
together
is

A. \(\displaystyle \frac{3}{70}\)

B. \(\displaystyle \frac{1}{140}\)

C. \(\displaystyle \frac{3}{140}\)

D. \(\displaystyle \frac{1}{70}\)

E. \(\displaystyle \frac{1}{35}\)

I could find the probability that three books from Gilmore are together, the separate probability that the books from Lawson being together, but I don't know how to find the intersection between those two sets of probabilities... (Thinking)

P(3 Gilmore together) =\(\displaystyle \frac{3!5!}{7!} = \frac{1}{7}\)

P(3 Lawson together) = \(\displaystyle \frac{1}{7}\)

P(3 Lawson together and 3 Gilmore together) = \(\displaystyle \frac{3!3!3!}{7!} = \frac{3}{70}\)

Total probability = \(\displaystyle \frac{1}{7} + \frac{1}{7} + \frac{3}{70} - P(intersections)\)

How can I solve this?
The way I'm reading this is....

the volumes of any author who is the author of more than one book are together,
in other words, no author has volumes scattered on the shelf.
Hence, should not only one case be analysed??

no?
 
May 2010
1,260
410
Mauritius
Ok, I admit that I don't quite know what the question is really asking for.

If the volumes of each author needs to be grouped together, then the answer is only 3/70. So, this is the answer then?

It's so hard to understand what is required to find by the question... (Doh)

Thanks!
 
Dec 2009
3,120
1,342
Ok, I admit that I don't quite know what the question is really asking for.

If the volumes of each author needs to be grouped together, then the answer is only 3/70. So, this is the answer then?

It's so hard to understand what is required to find by the question... (Doh)

Thanks!
I would think so...

As if all of Gilmore's books are together,
but Patterson's book is between Lawson's books,
then it could be stated

"the volumes of the same author are not together".
 
May 2010
1,260
410
Mauritius
Ok, thanks. I'll be waiting for the solution and if I don't forget, I'll post it here (Happy).