Window probability...

There 5 panes on one diagonal.
choosing 3 out of 5 is 5C3 = 10
One the other diaginal = 10
Total desired events = 20

Choosing 3 out of 25 = 25C3 = 25 * 24 * 23 / ( 3 * 2 )

P = 5 * 4 / ( 25 * 4 * 23 ) = 1 /115

None of the answers match.

Ans = 2/115

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

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

Sorry I got it.
I considered only 2 biggest 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

independent events.

Prob = 21/575

Anupag,

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):

2-pane diagonal
3-pane diagonal
4-pane diagonal
5-pane diagonal
4-pane diagonal
3-pane diagonal
2-pane diagonal

= 7 diagonals.

Diagonals go in 2 directions = 2* 7 = 14 possible diagonals

Of these 14 diagonals, 4 do not qualify i.e. the 2-pane diagonals, because we're looking for cases where the ball hits 3 panes that lie on a diagonal.

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.

Yes Anupag, all 3 panes do lie on the same diagonal. I thought I made that clear but I see that I didn't.

Sorry for the confusion.