Acertain box contains 200 particles of an ideal gas. how many times more likely is it to find the particles evenly split between the left and right halves of the box, than to find 160 particles on one side and 40 on the other?

 Each particle has two possibilities: either it is on the left hand side or it is on the right hand side.  

The probability of having them equally split is  

P(100,100) = 200C100 / 2^200  

where 200C100 is the binomial coefficient.  

For the uneven distribution we have  

P= 2 * 200C160 / 2^200  

where the factor 2 comes in because we could have 160/40 or 40/160 as a division.  

So the ratio of probabilities is  

200C100 /(2 * 200C160 )  

= 2.2 * 10^16

Hope this helps!

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a pan containing water is placed in a positio n where it is exposed to a 10 mile per hour wind. the pan contains a uniform depth of 1.27 cm of water, is quite wide, and has a length of 2.44 m in the wind direction. the water in the pan is at a constant temperature of 15.5 °c.