The lifts in the office buildings of a small city have occasional breakdowns. The breakdowns at any given time are independent of one another and can be modelled using a Poisson Distribution with mean 0.2 per day.

(a)     Determine the probability that there will be exactly four breakdowns during the month of June (June has 30 days).

(b)     Determine the probability that there are more than 3 breakdowns during the month of June.

(c)     Determine the probability that there are no breakdowns during the first five days of June.

(d)     Find the probability that the first breakdown in June occurs on June \({3^{{\text{rd}}}}\).

(e)     It costs 1850 Euros to service the lifts when they have breakdowns. Find the expected cost of servicing lifts for the month of June.

(f)     Determine the probability that there will be no breakdowns in exactly 4 out of the first 5 days in June.

(a)     mean for 30 days: \(30 \times 0.2 = 6\) .     (A1)

\({\text{P}}(X = 4) = \frac{{{6^4}}}{{4!}}{{\text{e}}^{ - 6}} = 0.134\)     (M1)A1     N3

[3 marks]

(b)     \({\text{P}}(X > 3) = 1 - {\text{P}}(X \leqslant 3) = 1 - {{\text{e}}^{ - 6}}(1 + 6 + 18 + 36) = 0.849\)     (M1)A1     N2

[2 marks]

(c)     EITHER

mean for five days: \(5 \times 0.2 = 1\)     (A1)

\({\text{P}}(X = 0) = {{\text{e}}^{ - 1}}\,\,\,\,\,( = 0.368)\)     A1     N2

OR

mean for one day: 0.2     (A1)

\({\text{P}}(X = 0) = {({{\text{e}}^{ - 0.2}})^5} = {{\text{e}}^{ - 1}}\,\,\,\,\,( = 0.368)\)     A1     N2

[2 marks]

(d)     Required probability \( = {{\text{e}}^{ - 0.2}} \times {{\text{e}}^{ - 0.2}} \times (1 - {{\text{e}}^{ - 0.2}})\)     M1A1

= 0.122     A1     N3

[3 marks]

(e)     Expected cost is \(1850 \times 6 = {\text{11}}\,{\text{100 Euros}}\)     A1

[1 mark]

(f)     On any one day \({\text{P}}(X = 0) = {{\text{e}}^{ - 0.2}}\)

Therefore, \(\left( {\begin{array}{*{20}{c}}
  5 \\
  1
\end{array}} \right){({{\text{e}}^{ - 0.2}})^4}(1 - {{\text{e}}^{ - 0.2}}) = 0.407\)     M1A1     N2

[2 marks]

Total [13 marks]

Many candidates showed familiarity with the Poisson Distribution. Parts (a), (b), and (c) were straightforward, as long as candidates multiplied 0.2 by 30 to get the mean. Part (e) was answered successfully by most candidates. Parts (d) and (f) were done very poorly. In part (d), most candidates calculated \({\text{P}}(X = 1)\) rather than \({\text{P}}(X \leqslant 1)\). Although some candidates realized the need for the Binomial in part (e), some incorrectly used 0.8 and 0.2.