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Lots of questions on different subjects where you have to try to answer the latest question and if you are right then its your turn to ask the next question.

Moderators: playgirl, JUS, rodbox

#86692 by wildrover
Mon Sep 08, 2008 3:21 pm
:D If you draw the triangle described and use Pythagoras then you get:

(r-8)^2 + (r-4)^2 = r^2 which simplifies to
(r-20)(r-4) = 0
So the radius has to be 4cm or 20cm. Since if it were 4cm that would make one of the sides of the triangle -4cm long, it must be 20cm...
Well Done Buzzie!! 8)

#92575 by kizer
Mon Dec 08, 2008 6:56 pm
This needs a pick me up...

In the National Lottery, you pick 6 numbers out of 49. Then, 6 'Main' numbers are drawn on TV, and one 'Bonus' number.

Suppose you win a prize if:

a) All 6 main numbers match your numbers.
b) 5 main numbers match your numbers, and the bonus number.
c) 5 main numbers match your numbers, and you don't get the bonus ball.

Watching the Lottery one night, you realise you have matched all 4 of the main numbers that have been drawn so far. What is the probability that you win a prize, given that this has happened?

(Note - this is a real question from a finals paper on statistics!)

#92583 by wildrover
Mon Dec 08, 2008 9:06 pm
My take is that this can be simplified since the way the question is asked all that really matters is that at least one of the last two numbers drawn is one of yours - the bonus ball is irrelevant the way this is written:

a) When the 5th number is picked there is a 2/45 chance of you winning immediately and a 43/45 chance of you not winning.

b) On the 43/45 occasions you didn't win when ball 5 is drawn you have a 2/44 chance of winning when the 6th number is drawn.

Total chances of winning are 2/45 + (1/22*43/45) = 87/990 or a little worse than 1/11...

#92592 by kizer
Tue Dec 09, 2008 12:08 am
Yes you have it!

That is the crucial observation. From there I thought:

p(winning) = 1- p(not winning) = 1- (43/45 * 42/44) = your answer!

#92595 by wildrover
Tue Dec 09, 2008 12:52 am
Maths isn't a favourite subject for most on here - nevertheless I'll try one my daughter tried on me last week...

In the trapezium above, the top side and the bottom side are parallel and two diagonals are drawn as shown dividing the trapezium into 4 triangles. What is the area of the trapezium in terms of areas A and B?

#92616 by kizer
Wed Dec 10, 2008 1:32 am
I happen to know that the area of the two white triangles is the same - however, I don't know how they are related to A and B!

The answer is A+B+2W where W is the area of one of the white triangles.


#92630 by wildrover
Wed Dec 10, 2008 11:40 am
Consider the fact that B and the white triangle on the left share the same perpendicular height as do A and the white triangle on the right and work out the relationship between the areas of the 4 triangles......

#92920 by wildrover
Fri Dec 26, 2008 2:47 pm
I think this was a bit too Mathsy - anyway for anyone who's interested:

Firstly by examining the two larger triangles formed by triangle A and each of the white triangles in turn and using the formula Area = 1/2 base x perpendicular height we can see that the base is the same and the perpendicular height is the same so the area must be the same.

Therefore the two white triangles are equal in area.

Then taking triangle B and the left hand white triangle, and letting the part of the diagonal that forms the base of triangle B be y and the part of the diagonal that forms the base of the white triangle be x then:

Area triangle B/Area white triangle = y/x

Similarly for Triangle A and the other white triangle

Area White triangle/ Area triangle A = y/x

Therefore in terms of area since we know the two white triangles are equal in area: AxB = (white triangle)^2

Therefore the area of each white triangle is √(A+B)

Total area is A + B + 2√(A+B) or simplifying

Area = (√A + √B)^2

Phew - no more geometry problems....

#93136 by Buzzie
Tue Jan 06, 2009 12:26 pm

#93137 by wildrover
Tue Jan 06, 2009 12:44 pm

#93147 by Buzzie
Tue Jan 06, 2009 1:44 pm

#93939 by Buzzie
Mon Jan 26, 2009 8:57 am

#94006 by wildrover
Mon Jan 26, 2009 9:07 pm
I'm not surprised - it's a difficult concept to visualise and is counter-intuitive. The coin example is not a great one - at least it's not been explained well. The absolute key is that the two coin tosses have already both occurred and if you are told for example it's a tail, you don't know whether that is the first or the second toss otherwise the 50:50 advocates are correct.

#97090 by wildrover
Mon Mar 30, 2009 12:07 am
It's about time for the quarterly maths question:

A hoop made from metal tube 1cm thick is formed. It has an interior diameter of 24cm and is a perfect circle. It is held parallel to the ground and a ball 4cm in diameter is dropped randomly from above. Is the ball more likely to go through the hoop without touching it or is it more likely to hit the hoop? And how much more likely?

#97104 by pulse876
Mon Mar 30, 2009 8:48 am
The ball will only fall through the hoop without touching it if it is dropped from a position such that the center of the ball lies within a circle of radius 10cm that is centered on the center of the hoop. The area of that circle is pi*10^2 = 100pi sq cm.

If it is dropped from a position where the center of the ball lies outside that circle then it will either hit the hoop or, if it is far enough away, miss the hoop entirely (on the outside).

Assuming we are not interested in the cases where it misses the hoop entirely (on the outside) then it will hit the hoop if it is dropped from a position such that the center of the ball lies within a circular band of radius between 10cm and 15cm. The area of that band is pi*15^2-pi*10^2 = 125pi sq cm.

That area is 25% larger than the area of the first circle, ie there is a 25% greater chance that the ball will hit the hoop rather than fall through it (although the answer actually depends on the limits of the area from which the ball can be dropped).
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