Melody

No you have to add the indices.   m^(-4+-2) =m^-6

That is what I would have though. They should have used large roast and small roast,  OR

large potatoes and small potatoes,  Something that is more spherical.

4/23 = 0.1739130434782609

Mmm there is a lot of digits repeating here.  You may need first of all to work out how many repeat 

I'd probably try a calculator that shows more  places or do it by hand.

Yes my first answer is wrong because of a careless error   4!/(2!2!) = 6   [not 3]

so     1+6+1=8   the answer is 8/16 = 1/2

The only ways you can get an even sum is if 

exactly 4 or 2 or 0 dice are even.

4 even     1/(2^4) = 1/16

2even      [1/(2^4) ]* 4!/(2!2!)   = 3/16

0even      1/(2^4) = 1/16

1+3+1=5

So I get a total probability of   5/16

There are 10C5 / 2 ways to split the kids up = 126

If Jenny and Kenny are togther and Lenny is not then there are   7C3 ways to choose the other three in L and K's group.

7C3=35

So I get     35/126

50% chance of getting an even on one die

4 even     1/(2^4)

2even      [1/(2^4) ]* 4!/(2!2!)

0even      1/(2^4)

Now add them together.  That's what I think.

Thanks very much for clarifying Qube73  :)

Drop a vertial line from A to DC and call the point of intesection E

Now you have a rectangle AECB and a triangle ADE

You can find the lengths of AE and DE using trigonometry

You can find the length of EC useing subtraction

So now you can find the  the 2 areas.

Add them and you have the area of the whole lot.

The area of the sector is    (50/360)* pi*r^2

Find that and then subtract to get the shaded area.