Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

To solve this problem, imagine that the window that will be used is a 5 x 5 i.e. with 25 panes.

The neighborhood children who live on Main street play softball outside every evening. During a 3 month period, 3 out of the 25 window-panes on 123 Main Street get broken by their ball. What is the probability that the broken panes all lie on one of the diagonals of the window?

There are 14 diagonals in total but only 10 (5 in each direction) have more than 3 or more panes.

Of these five, you can have (in order):

3C3 combinations for the diagonal that has 3 panes.
4C3 combinations for the diagonal that has 4 panes.
5C3 combinations for the diagonal that has 5 panes.
4C3 combinations for the diagonal that has 4 panes.
3C3 combinations for the diagonal that has 3 panes.

So the probability that 3 broken panes lie on a diagonal =

2(3C3 + 4C3 + 5C3 + 4C3 + 3C3)/25C3 = 2/115

Last edited by Makky07 on 18 Mar 2004, 07:34, edited 1 time in total.

There are 14 diagonals in total but only 10 (5 in each direction) have more than 3 panes.

Of these five, you can have:

3C3 combinations for the diagonal that has 3 panes. 4C3 combinations for the diagonal that has 4 panes. 5C3 combinations for the diagonal that has 5 panes. 4C3 combinations for the diagonal that has 4 panes. 3C3 combinations for the diagonal that has 3 panes.

So the probability that 3 broken panes lie on a diagonal =

I did not understand the explanation provided by ndidi

I assumed that the window panes are arranged as follows x - normal window pane d - diaginal window pane

d x x x d x d x d x x x d x x x d x d x d x x x d

Where are the 14 diagonals ?

There are more than the 2 diagonals that you just mentioned Anandnk. Everytime that you have the possibility of having 3 windows arranged diagonally, you will have a possibility. Thus, there are much more possiblities than you simple 2 large diagonals
_________________

The phrase "Diagonals of the window" will mean the diagonals as understood in common parlance. 14 diagonals is nice thinking but is stretching logic a wee bit IMHO.

I will look at it this way.

Diagonals defined as Anand says.

9 panes on "diagonals" say dp=diagonal pane, np=pane not on diagonal

First hit - one dp breaks, prob = 9/25
second hit - another dp breaks prob = 8/24
third hit - another dp breaks prob = 7/23

I think you're either misunderstanding the question, or you're not understanding the diagram. The question is asking for the probability that all 3 window panes that the ball hits lie on a diagonal.

If you draw the 5 by 5 with 25 squares, you'll get (in order):

Thanks ndidi. Yes indeed, the question is unclear to me !!. Diagonals of the window does not - IMHO- include the possibilities that you have enumerated.

Now if we assume that your definition of diagonals is the one to be taken, the solution given above by you is incorrect.

According to this definition, ALL panes are on one or the other diagonals (even when you take care to include only those diagonals which have at least three panes) - Check it out. Therefore, the required probability is 1.

However, your solution approach is ok if the question is reworded as "all three panes lie on the same diagonal". May be I am talking semantics here, but the question if reworded will be clearer.