Find the probability that exactly 4 taxis arrive during T.

Find the most likely number of taxis that would arrive during T.

Given that more than 5 taxis arrive during T, find the probability that exactly 7 taxis arrive during T.

During quiet periods of the day, taxis arrive at a mean rate of 1.3 taxis every 10 minutes.

Find the probability that during a period of 15 minutes, of which the first 10 minutes is busy and the next 5 minutes is quiet, that exactly 2 taxis arrive.

\(X \sim {\text{Po}}\left( {5.3} \right)\)

\({\text{P}}\left( {X = 4} \right) = {{\text{e}}^{ - 5.3}}\frac{{{{5.3}^4}}}{{4{\text{!}}}}\)     (M1)

= 0.164      A1

[2 marks]

METHOD 1

listing probabilities (table or graph)      M1

mode X = 5 (with probability 0.174)     A1

Note: Award M0A0 for 5 (taxis) or mode = 5 with no justification.

METHOD 2

mode is the integer part of mean      R1

E(X) = 5.3 ⇒ mode = 5      A1

Note: Do not allow R0A1.

[2 marks]

attempt at conditional probability       (M1)

\(\frac{{{\text{P}}\left( {X = 7} \right)}}{{{\text{P}}\left( {X \geqslant 6} \right)}}\) or equivalent \(\left( { = \frac{{0.1163 \ldots }}{{0.4365 \ldots }}} \right)\)      A1

= 0.267       A1

[3 marks]

METHOD 1

the possible arrivals are (2,0), (1,1), (0,2)       (A1)

\(Y \sim {\text{Po}}\left( {0.65} \right)\)     A1

attempt to compute, using sum and product rule,      (M1)

0.070106… × 0.52204… + 0.026455… × 0.33932… + 0.0049916… × 0.11028…      (A1)(A1)

Note: Award A1 for one correct product and A1 for two other correct products.

= 0.0461       A1

[6 marks]

METHOD 2

recognising a sum of 2 independent Poisson variables eg Z = X + Y      R1

\(\lambda  = 5.3 + \frac{{1.3}}{2}\)

P(Z = 2) = 0.0461     (M1)A3

[6 marks]