Date | May 2013 | Marks available | 4 | Reference code | 13M.2.hl.TZ1.10 |
Level | HL only | Paper | 2 | Time zone | TZ1 |
Command term | Find | Question number | 10 | Adapted from | N/A |
Question
A ferry carries cars across a river. There is a fixed time of T minutes between crossings. The arrival of cars at the crossing can be assumed to follow a Poisson distribution with a mean of one car every four minutes. Let X denote the number of cars that arrive in T minutes.
Find T, to the nearest minute, if \({\text{P}}(X \leqslant 3) = 0.6\).
It is now decided that the time between crossings, T, will be 10 minutes. The ferry can carry a maximum of three cars on each trip.
One day all the cars waiting at 13:00 get on the ferry. Find the probability that all the cars that arrive in the next 20 minutes will get on either the 13:10 or the 13:20 ferry.
Markscheme
\(X \sim {\text{Po(0.25T)}}\) (A1)
Attempt to solve \({\text{P}}(X \leqslant 3) = 0.6\) (M1)
\(T = 12.8453 \ldots = 13{\text{ (minutes)}}\) A1
Note: Award A1M1A0 if T found correctly but not stated to the nearest minute.
[3 marks]
let \({X_1}\) be the number of cars that arrive during the first interval and \({X_2}\) be the number arriving during the second.
\({X_1}\) and \({X_2}\) are Po(2.5) (A1)
P (all get on) \( = {\text{P}}({X_1} \leqslant 3) \times {\text{P}}({X_2} \leqslant 3) + {\text{P}}({X_1} = 4) \times {\text{P}}({X_2} \leqslant 2)\)
\( + {\text{P}}({X_1} = 5) \times {\text{P}}({X_2} \leqslant 1) + {\text{P}}({X_1} = 6) \times {\text{P}}({X_2} = 0)\) (M1)
\( = 0.573922 \ldots + 0.072654 \ldots + 0.019192 \ldots + 0.002285 \ldots \) (M1)
\( = 0.668{\text{ }}(053 \ldots )\) A1
[4 marks]
Examiners report
There were some good answers to part (a), although poor calculator use frequently let down the candidates.
Very few candidates were able to access part (b).