Discussion
An unbiased die is tossed until 5 appears. If \(X\) denotes the number of tosses required, then \(\frac{P(X = 2)}{P(X = 5)}\) =
Answer: C
The ratio is \(\frac{216}{125}\).
The first 5 on toss \(k\) means \(k - 1\) failures, each with probability \(\frac{5}{6}\), then a success with probability \(\frac{1}{6}\): \(P(X = k) = \left(\frac{5}{6}\right)^{k - 1}\frac{1}{6}\).
\[ \begin{aligned} \frac{P(X = 2)}{P(X = 5)} &= \frac{(5/6)^{1}}{(5/6)^{4}} \\ &= \left(\frac{6}{5}\right)^{3} = \frac{216}{125} \end{aligned} \]
\(\frac{125}{216}\) is the reciprocal (the ratio taken the other way up), and \(\frac{36}{25}\) uses a difference of 2 failures instead of 3.
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