SOLUTION: Suppose that a die is weighted such that prime numbers are twice as likely to appear as non-primes. What is the probability of rolling 5 at most 2 times in 15 tries? (recall: 1 i

The primes are 2,3,5, the non-primes are 1,4,6

Let x = probability of a non-prime, then
2x = probability of a prime

P(1)=P(4)=P(6)=x, P(2)=P(3)=P(5)=2x

P(1 or 4 or 6) = x+x+x = 3x
P(2 or 3 or 5) = 2x+2x+2x = 6x 

   3x + 6x = 1
        9x = 1
         x = 1/9

P(1 or 4 or 6) = 3x = 3(1/9) = 1/3
P(2 or 3 or 5) = 6x = 6(1/9) = 6/9 = 2/3 


At most 2 times means 0 times, 1 time, or 2 times. 
P(5) = 2x = 2(1/9) = 2/9
P(not 5) = 1-2/9 = 9/9-2/9 = 7/9 

P(no times) = (15C0)(2/9)^0(7/9)^15 = (1)(1)(0.0230586012) = 0.0230586012
P(1 time) = (15C1)(2/9)^1(7/9)^14 = (15)(2/9)(0.029646773) = 0.0988225767
P(2 times) = (15C2)(2/9)^2(7/9)^13 = (105)(4/81)(0.0381172796) = 0.1976461533
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P(rolling five 0,1, or 2 times) = sum of those = 0.3195263312

Or you can do it on a TI-83 or 84 this way:

Press 2ND VARS then scroll down to binomcdf( 
press ENTER        

         if you have wizard  [trials:15, p:1/9, x value: 2, 
         scroll to Paste]

either way you should have this:  

binomcdf(15,1/9,2)

press ENTER

see answer .3195263312

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Expected value = 1(1/9)+2(2/9)+3(2/9)+4(1/9)+5(2/9)+6(1/9) =
               = 1/9+4/9+6/9+4/9+10/9+6/9
               = 31/4 = 7 3/4 = 7.75

Now even though you'll never expect a value of 7 3/4 ever :), 
if you rolled the die 15 times every day for a year, and averaged 
up the value you got each day, you would get an average
very close to 7.75.  So "expected value" really means "expected
AVERAGE value if the experiment were repeated many times".

Edwin