| Date | November 2015 | Marks available | 4 | Reference code | 15N.1.hl.TZ0.6 |
| Level | HL only | Paper | 1 | Time zone | TZ0 |
| Command term | Find | Question number | 6 | Adapted from | N/A |
Question
A box contains four red balls and two white balls. Darren and Marty play a game by each taking it in turn to take a ball from the box, without replacement. The first player to take a white ball is the winner.
Darren plays first, find the probability that he wins.
The game is now changed so that the ball chosen is replaced after each turn.
Darren still plays first.
Show that the probability of Darren winning has not changed.
Markscheme
probability that Darren wins \({\text{P}}(W) + {\text{P}}(RRW) + {\text{P}}(RRRRW)\) (M1)
Note: Only award M1 if three terms are seen or are implied by the following numerical equivalent.
Note: Accept equivalent tree diagram for method mark.
\( = \frac{2}{6} + \frac{4}{6} \bullet \frac{3}{5} \bullet \frac{2}{4} + \frac{4}{6} \bullet \frac{3}{5} \bullet \frac{2}{4} \bullet \frac{1}{3} \bullet \frac{2}{2}\;\;\;\left( { = \frac{1}{3} + \frac{1}{5} + \frac{1}{{15}}} \right)\) A2
Note: A1 for two correct.
\( = \frac{3}{5}\) A1
[4 marks]
METHOD 1
the probability that Darren wins is given by
\({\text{P}}(W) + {\text{P}}(RRW) + {\text{P}}(RRRRW) + \ldots \) (M1)
Note: Accept equivalent tree diagram with correctly indicated path for method mark.
\({\text{P (Darren Win)}} = \frac{1}{3} + \frac{2}{3} \bullet \frac{2}{3} \bullet \frac{1}{3} + \frac{2}{3} \bullet \frac{2}{3} \bullet \frac{2}{3} \bullet \frac{2}{3} \bullet \frac{1}{3} + \ldots \)
or \( = \frac{1}{3}\left( {1 + \frac{4}{9} + {{\left( {\frac{4}{9}} \right)}^2} + \ldots } \right)\) A1
\( = \frac{1}{3}\left( {\frac{1}{{1 - \frac{4}{9}}}} \right)\) A1
\( = \frac{3}{5}\) AG
METHOD 2
\({\text{P (Darren wins)}} = {\text{P}}\)
\({\text{P}} = \frac{1}{3} + \frac{4}{9}{\text{P}}\) M1A2
\(\frac{5}{9}{\text{P}} = \frac{1}{3}\)
\({\text{P}} = \frac{3}{5}\) AG
[3 marks]
Total [7 marks]
