Probability of getting 3 on loaded die

[jdstokes]

A six-sided die is loaded in such a way that an odd number is twice as likely to occur as an even number. We throw the die until a 3 is observed. What is the probability that the number of throws required is (strictly) more than 5 and (strictly) less than 10?

Answer: 0.1805.

My working.

P(odd) = 2P(even)
P(odd) + P(even) = 1
Therefore
P(odd) = 2/9.

X = "number of throws before success"

X is geometric with p = 2/9.

Therefore

$P( 6\leq X \leq 9 ) = \sum_{i=6}^9 (2/9)(1-2/9)^i = 0.1403642$.

Any help would be appreciated.

Thanks.

James

 

[jdstokes]

I finally figured it out. For anyone interested, the solution involves a change of variables from $X$ (the number of failures before success) to $Y = X + 1$ (the number of attempts required for success).
So $P( 6\leq X \leq 9 ) = \sum_{i=6}^9 (2/9)(1-2/9)^{i-1} = 0.1805$.