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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 14,850

PS # 1

Ten fair coins are tossed simultaneously. Find the probability of getting atleast seven heads.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

PS # 1

This is a binomial distribution, with 10 trials and a probability of success of 0.5.

The probability of at least 7 heads is P(7) + P(8) + P(9) + P(10).

This is worked out by 0.5*10 (10C7 + 10C8 + 10C9 + 10C10) = 11/64.

Why did the vector cross the road?

It wanted to be normal.

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**ganesh****Moderator**- Registered: 2005-06-28
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Very good, mathsyperson. You're correct.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 14,850

PS # 2

Two squares are chosen at random on a chessboard. What is the probability that they have a side in common?

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Ooh, interesting. Let's see... a chessboard has 4 corner squares, 24 edge squares and 36 centre squares.

On the corner squares, the chance of the other square being next to it is 2/63.

On the edge squares, the chance is 3/63 and on the centres, the chance is 4/63.

Combine this with the chances of the first square being each of the types and we get (4/64*2/63) + (24/64*3/63) + (36/64*4/63) = 224/4032 = 1/18.

Why did the vector cross the road?

It wanted to be normal.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 14,850

Excellent! Well done, mathsyperson!

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 14,850

PS # 3

The probabitlity that a student will pass in Statistics examination is 2/3 and the probability that he will not pass in mathematics is 5/9. The probability that he will pass in atleast one of the examinations is 4/5. Find the probability of his passing in both the examinations.

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**mathsyperson****Moderator**- Registered: 2005-06-22
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P(AnB) = P(A) + P(B) - P(AuB).

Therefore, P(Passing both) = 2/3 + 5/9 - 4/5 = 19/45.

Why did the vector cross the road?

It wanted to be normal.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 14,850

ganesh wrote:

and the probability that he will not pass in mathematics is 5/9.

mathsyperson, did you notice that?

The probability that he will pass in mathematics is, therefore, 4/9.

Hence, the probability of passing both the subjects would be

2/3 + 4/9 - 4/5 = 14/45.

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**ganesh****Moderator**- Registered: 2005-06-28
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PS # 4

The probability of rain was 40%. If it rained, the Redskins had a 30% chance of winning, if it did not rain, they had a 55% chance of winning. Given that the Redskins won, what is the probability that it rained?

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Oh, silly me. I hate it when I make stupid mistakes from not reading the question properly. Ah well.

The probability of it raining and of them winning is 0.4*0.3 = 0.12.

The probability of it not raining and of them winning is 0.6*0.55 = 0.33.

Therefore, the probability of rain if they won is 0.12/(0.12+0.33) = 0.26666... = 27% (nearest %)

Why did the vector cross the road?

It wanted to be normal.

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**ganesh****Moderator**- Registered: 2005-06-28
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** Excellent, mathsyperson!**

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**maudish****Member**- Registered: 2009-10-11
- Posts: 1

Hi, could you please answer this problem?

In how many ways you can choose 2 white squares on a chessboard such that they are either in the same row or same column?

Thanks in advance.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 88,953

Hi maudish;

I believe the sequence looks like this:

For various n x n boards:

2 x 2 =0,

3 x 3 =4

4 x 4 =8,

5 x 5 =22

6 x 6 =36

7 x 7 = ?,

8 x 8 = 96

All by actual count. If i find a formula for n x n will post it.

*Last edited by bobbym (2009-10-11 10:50:38)*

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 88,953

Hi maudish;

Using the ordinary generating function for the rook polynomials on a r x c board.

The series can be eliminated because we only seek the coefficient of x^2 ( 2 rooks). Also since the board is square r = c. So.

The above is the number of 2 non attacking rooks on a c x c chessboard.

From a combinatorical argument and playing much spot the pattern.

We can solve for n and clean up:

Where n is the number of ways 2 rooks can be positioned on the white squares of a c x c chessboard when c is even. The above will generate the table given in the previous post, i.e.

c = 2 then n = 0

c = 4 then n = 8

c = 6 then n = 36

c = 8 then n = 96

c = 10 then n = 200

.

.

.

This is not a proof, just a conjecture. I have tested it for c = 16 by direct count. I suppose it might be proven by induction but the correct method is by partitioning the chessboard with it's forbidden black squares into disjoint boards and then using the rook polynomials to prove it. When I do that I will post it.

*Last edited by bobbym (2009-10-12 15:17:55)*

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**G-man****Member**- Registered: 2011-02-28
- Posts: 16

If I'm not wrong.

Answer to question in first post.

*Last edited by G-man (2011-02-28 18:22:56)*

Maths!......

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 88,953

Hi G-man;

Welcome to the forum. Which question are you answering?

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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