On a randomly chosen day, find the probability that

(i)     there are no complaints;

(ii)     there are at least three complaints.

In a randomly chosen five-day week, find the probability that there are no complaints.

On a randomly chosen day, find the most likely number of complaints received.

Justify your answer.

The department store introduces a new policy to improve customer service. The number of complaints received per day now follows a Poisson distribution with mean $\lambda$.

On a randomly chosen day, the probability that there are no complaints is now $0.8$.

Find the value of $\lambda$.

(i)     $P(X = 0) = 0.549{\text{ }}( = {{\text{e}}^{ - 0.6}})$     A1

(ii)     $P(X \ge 3) = 1 - P(X \le 2)$     (M1)

$P(X \ge 3) = 0.0231$     A1

[3 marks]

EITHER

using $Y \sim {\text{Po(3)}}$     (M1)

OR

using ${(0.549)^5}$     (M1)

THEN

${\text{P}}(Y = 0) = 0.0498{\text{ }}\left( { = {{\text{e}}^{ - 3}}} \right)$     A1

[2 marks]

${\text{P}}(X = 0)$ (most likely number of complaints received is zero)     A1

EITHER

calculating ${\text{P}}(X = 0) = 0.549$ and ${\text{P}}(X = 1) = 0.329$     M1A1

OR

sketching an appropriate (discrete) graph of ${\text{P}}(X = x)$ against $x$     M1A1

OR

finding ${\text{P}}(X = 0) = {e^{ - 0.6}}$ and stating that ${\text{P}}(X = 0) > 0.5$     M1A1

OR

using ${\text{P}}(X = x) = {\text{P}}(X = x - 1) \times \frac{\mu }{x}$ where $\mu  < 1$     M1A1

[3 marks]

${\text{P}}(X = 0) = 0.8{\text{ (}} \Rightarrow {e^{ - \lambda }} = 0.8)$     (A1)

$\lambda  = 0.223\left( { = \ln \frac{5}{4}, =  - \ln \frac{4}{5}} \right)$     A1

[2 marks]

Total [10 marks]

Parts (a), (b) and (d) were generally well done. In (a) (ii), some candidates calculated $1 - {\text{P}}(X \le 3)$.

Parts (a), (b) and (d) were generally well done.

A number of candidates offered clear and well-reasoned solutions to part (c). The two most common successful approaches used to justify that the most likely number of complaints received is zero were either to calculate ${\text{P}}(X = x)$ for $x = 0,{\text{ }}1,{\text{ }} \ldots$ or find that ${\text{P}}(X = 0) = 0.549{\text{ }}( > 0.5)$. A number of candidates stated that the most number of complaints received was the mean of the distribution $(\lambda  = 0.6)$.

Parts (a), (b) and (d) were generally well done.