Moderate -0.8 This is a straightforward Poisson distribution problem requiring students to equate two probability expressions, simplify using the Poisson formula, and solve for λ. The algebra is simple (consecutive Poisson probabilities have a clean ratio), and the calculation is routine. Easier than average as it's purely mechanical application of a standard formula with no problem-solving insight required.
It is given that \(Y \sim\) Po\((\lambda)\), where \(\lambda \neq 0\), and that P\((Y = 4) =\) P\((Y = 5)\). Write down an equation for \(\lambda\). Hence find the value of \(\lambda\) and the corresponding value of P\((Y = 5)\). [5]
Poisson formula used [not just quoted] correctly once
\(\frac{\lambda^4}{4!} = \frac{\lambda^5}{5!}\)
A1
This equation or exact equivalent, needs \(e^{-\lambda}\) seen somewhere
M1
Correct method for cancelling \(e^{-\lambda}\)
\(\Rightarrow \lambda = 5\)
A1
Solve to get \(\lambda = 5\) only, www
\(\mathbf{0.175(46)}\)
B1
Probability, in range \([0.175, 0.176]\), allow from \(\lambda = 5\) from wrong working (5 marks)
$\frac{\lambda^4}{4!}e^{-\lambda} = \frac{\lambda^5}{5!}e^{-\lambda}$ | M1 | Poisson formula used [not just quoted] correctly once
$\frac{\lambda^4}{4!} = \frac{\lambda^5}{5!}$ | A1 | This equation or exact equivalent, needs $e^{-\lambda}$ seen somewhere
| M1 | Correct method for cancelling $e^{-\lambda}$
$\Rightarrow \lambda = 5$ | A1 | Solve to get $\lambda = 5$ only, www
$\mathbf{0.175(46)}$ | B1 | Probability, in range $[0.175, 0.176]$, allow from $\lambda = 5$ from wrong working (5 marks)
It is given that $Y \sim$ Po$(\lambda)$, where $\lambda \neq 0$, and that P$(Y = 4) =$ P$(Y = 5)$. Write down an equation for $\lambda$. Hence find the value of $\lambda$ and the corresponding value of P$(Y = 5)$. [5]
\hfill \mbox{\textit{OCR S2 2016 Q4 [5]}}