| Date | November 2014 | Marks available | 4 | Reference code | 14N.3dm.hl.TZ0.5 |
| Level | HL only | Paper | Paper 3 Discrete mathematics | Time zone | TZ0 |
| Command term | Show that | Question number | 5 | Adapted from | N/A |
Question
Andy and Roger are playing tennis with the rule that if one of them wins a game when serving then he carries on serving, and if he loses then the other player takes over the serve.
The probability Andy wins a game when serving is \(\frac{1}{2}\) and the probability he wins a game when not serving is \(\frac{1}{4}\). Andy serves in the first game. Let \({u_n}\) denote the probability that Andy wins the \({n^{{\text{th}}}}\) game.
State the value of \({u_1}\).
Show that \({u_n}\) satisfies the recurrence relation
\[{u_n} = \frac{1}{4}{u_{n - 1}} + \frac{1}{4}.\]
Solve this recurrence relation to find the probability that Andy wins the \({n^{{\text{th}}}}\) game.
After they have played many games, Pat comes to watch. Use your answer from part (c) to estimate the probability that Andy will win the game they are playing when Pat arrives.
Markscheme
\(\frac{1}{2}\) A1
[1 mark]
Andy could win the \({n^{{\text{th}}}}\) game by winning the \(n - {1^{{\text{th}}}}\) and then winning the \({n^{{\text{th}}}}\) game or by losing the \(n - {1^{{\text{th}}}}\) and then winning the \({n^{{\text{th}}}}\) (M1)
\({u_n} = \frac{1}{2}{u_{n - 1}} + \frac{1}{4}(1 - {u_{n - 1}})\) A1A1M1
Note: Award A1 for each term and M1 for addition of two probabilities.
\({u_n} = \frac{1}{4}{u_{n - 1}} + \frac{1}{4}\) AG
[4 marks]
general solution is \({u_n} = A{\left( {\frac{1}{4}} \right)^n} + p(n)\) (M1)
for a particular solution try \(p(n) = b\) (M1)
\(b = \frac{1}{4}b + \frac{1}{4}\) (A1)
\(b = \frac{1}{3}\)
hence \({u_n} = A{\left( {\frac{1}{4}} \right)^n} + \frac{1}{3}\) (A1)
using \({u_1} = \frac{1}{2}\) M1
\(\frac{1}{2} = A\left( {\frac{1}{4}} \right) + \frac{1}{3} \Rightarrow A = \frac{2}{3}\)
hence \({u_n} = \frac{2}{3}{\left( {\frac{1}{4}} \right)^n} + \frac{1}{3}\) A1
Note: Accept other valid methods.
[6 marks]
for large \(n{\text{ }}{u_n} \approx \frac{1}{3}\) (M1)A1
[2 marks]
Total [13 marks]
Examiners report
Not all candidates wrote this answer down correctly although it was essentially told you in the question.
Very badly answered. Candidates seemed to think that they were being told this relationship (so used it to find u(2)) rather than attempting to prove it.
This distinguished the better candidates. Some candidates though that they could use the method for homogeneous recurrence relations of second order and hence started solving a quadratic. Only the better candidates saw that it was a combined AP/GP.
The best candidates saw this but most had not done enough earlier to get to do this.
