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.

$X \sim {\text{Po(0.25T)}}$     (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]

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).