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Window probability...

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Senior Manager
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Window probability... [#permalink]  17 Mar 2004, 09:16
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?

a) 20/115
b) 15/115
c) 2/115
d) 15/92
E) 1/92
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Last edited by kpadma on 17 Mar 2004, 11:26, edited 2 times in total.
Director
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I would go with 2/115 or C)
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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

Senior Manager
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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

Last edited by Makky07 on 18 Mar 2004, 06:34, edited 1 time in total.
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ndidi204 wrote:
Ans = 2/115

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 =

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

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Paul

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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 ?
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anandnk wrote:
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
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Paul

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Sorry I got it.
I considered only 2 biggest diagonals.
Manager
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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
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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.
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hmmm.. [#permalink]  18 Mar 2004, 19:12
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.
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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.
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