Two players, A and B, alternately throw a fair six-sided dice, with A starting, until one of them obtains a six. Find the probability that A obtains the first six.

P(six in first throw) $= \frac{1}{6}$     (A1)

P(six in third throw) $= \frac{{25}}{{36}} \times \frac{1}{6}$     (M1)(A1)

P(six in fifth throw)$= {\left( {\frac{{25}}{{36}}} \right)^2} \times \frac{1}{6}$

P(A obtains first six) $= \frac{1}{6} + \frac{{25}}{{36}} \times \frac{1}{6} + {\left( {\frac{{25}}{{36}}} \right)^2} \times \frac{1}{6} +  \ldots$     (M1)

recognizing that the common ratio is ${\frac{{25}}{{36}}}$     (A1)

P(A obtains first six) $= \frac{{\frac{1}{6}}}{{1 - \frac{{25}}{{36}}}}\,\,\,\,\,$(by summing the infinite GP)     M1

$= \frac{6}{{11}}$     A1

[7 marks]

This question proved difficult to the majority of the candidates although a few interesting approaches to this problem have been seen. Candidates who started the question by drawing a tree diagram were more successful, although a number of these failed to identify the geometric series.