Find the probability that during a certain seven-day week, more than 40 birds have been seen on the power line.

On Monday there were more than 10 birds seen on the power line. Show that the probability of there being more than 40 birds seen on the power line from that Monday to the following Sunday, inclusive, can be expressed as:

\(\frac{{{\text{P}}(X > 40) + \sum\limits_{r = 11}^{40} {{\text{P}}(X = r){\text{P}}(Y > 40 - r)} }}{{{\text{P}}(X > 10)}}\) where \(X \sim {\text{Po}}(5.84)\) and \(Y \sim {\text{Po}}(35.04)\).

mean for week is 40.88     (A1)

\({\text{P}}(S > 40) = 1 - {\text{P}}(S \leqslant 40) = 0.513\)     A1

[2 marks]

\(\frac{{{\text{probability there were more than 10 on Monday AND more than 40 over the week}}}}{{{\text{probability there were more than 10 on Monday}}}}\)     M1

possibilities for the numerator are:

there were more than 40 birds on the power line on Monday     R1

11 on Monday and more than 29 over the course of the next 6 days     R1

12 on Monday and more than 28 over the course of the next 6 days … until

40 on Monday and more than 0 over the course of the next 6 days     R1

hence if X is the number on the power line on Monday and Y, the number on the power line Tuesday – Sunday then the numerator is     M1

\({\text{P}}(X > 40) + {\text{P}}(X = 11) \times {\text{P}}(Y > 29) + {\text{P}}(X = 12) \times {\text{P}}(Y > 28) +  \ldots \)

\( + {\text{P}}(X = 40) \times {\text{P}}(Y > 0)\)

\( = {\text{P}}(X > 40) + \sum\limits_{r = 11}^{40} {{\text{P}}(X = r){\text{P}}(Y > 40 - r)} \)

hence solution is \(\frac{{{\text{P}}(X > 40) + \sum\limits_{r = 11}^{40} {{\text{P}}(X = r){\text{P}}(Y > 40 - r)} }}{{{\text{P}}(X > 10)}}\)     AG

[5 marks]