document.write( "Question 1163936: A box contains 6 balls, 2 are blue, 2 are red and 2 are green. Balls are drawn successively without replacement. If random variable X is number of draws until last red ball is obtained. Find the probability density function of random variable X? \n" ); document.write( "
Algebra.Com's Answer #788732 by math_helper(2461)\"\" \"About 
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document.write( "Probability / Calculation
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document.write( "All the fractions can be reduced, I didn't bother as the reduction is trivial.
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document.write( " P(X=2) =  (2/6)*(1/5) = 2/30 
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document.write( "         Explanation:  (R,R) is the only possible way for this to happen, 
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document.write( "                      The first R is drawn with prob. 2/6, the 2nd with 1/5.\r
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document.write( " P(X=3) = ((2/6)(4/5) + (4/6)(2/5)) * (1/4) = 4/30
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document.write( "         Explanation:    There are two ways this can happen: (R,*,R)  or
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document.write( "                  (*,R,R), where * is any non-red ball.  Notice how it must
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document.write( "                  end with R, and that last R has probability 1/4 since there
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document.write( "                  will be 4 balls left when that one is chosen.\r
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document.write( " P(X=4) = ((2/6)(4/5)(3/4) + (4/6)(2/5)(3/4) + (4/6)(3/5)(2/6)) * (1/3) = 6/30
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document.write( "         Explanation/Notes:  we could have assumed (R,*,*,R) and for the (R,*,*)
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document.write( "               portion you have 3 places to put the R:  3*(2/6)(4/5)(3/4) 
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document.write( "               = 3*(6/30)  (then *(1/3) for the 2nd R on the 4th selection 
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document.write( "               gives you 6/30). \r
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document.write( "By now you should see the pattern...
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document.write( " P(X=5) =  8/30\r
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document.write( " P(X=6) = 10/30\r
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document.write( "Let's re-write the PDF:
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document.write( " P(X=2)  =    2/30
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document.write( " P(X=3)  =    4/30
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document.write( " P(X=4)  =    6/30
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document.write( " P(X=5)  =    8/30
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document.write( " P(X=6)  =   10/30\r
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document.write( " and   (2+4+6+8+10)/30 = 30/30 = 1, as expected\r
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document.write( "There may be more elegant solutions, but this 'brute force' method also works.\r
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