Show that ${\text{P}}(X = x + 1) = \frac{\mu }{{x + 1}} \times {\text{P}}(X = x),{\text{ }}x \in \mathbb{N}$.

Given that ${\text{P}}(X = 2) = 0.241667$ and ${\text{P}}(X = 3) = 0.112777$, use part (a) to find the value of $\mu$.

METHOD 1

${\text{P}}(X = x + 1) = \frac{{{\mu ^{x + 1}}}}{{(x + 1)!}}{{\text{e}}^{ - \mu }}$    A1

$= \frac{\mu }{{x + 1}} \times \frac{{{\mu ^x}}}{{x!}}{{\text{e}}^{ - \mu }}$    M1A1

$= \frac{\mu }{{x + 1}} \times {\text{P}}(X = x)$    AG

METHOD 2

$\frac{\mu }{{x + 1}} \times {\text{P}}(X = x) = \frac{\mu }{{x + 1}} \times \frac{{{\mu ^x}}}{{x!}}{{\text{e}}^{ - \mu }}$    A1

$= \frac{{{\mu ^{x + 1}}}}{{(x + 1)!}}{{\text{e}}^{ - \mu }}$    M1A1

$= {\text{P}}(X = x + 1)$    AG

METHOD 3

$\frac{{{\text{P}}(X = x + 1)}}{{{\text{P}}(X = x)}} = \frac{{\frac{{{\mu ^{x + 1}}}}{{(x + 1)!}}{{\text{e}}^{ - \mu }}}}{{\frac{{{\mu ^x}}}{{x!}}{{\text{e}}^{ - \mu }}}}$    (M1)

$= \frac{{{\mu ^{x + 1}}}}{{{\mu ^x}}} \times \frac{{x!}}{{(x + 1)!}}$    A1

$= \frac{\mu }{{x + 1}}$    A1

and so ${\text{P}}(X = x + 1) = \frac{\mu }{{x + 1}} \times {\text{P}}(X = x)$     AG

[3 marks]

${\text{P}}(X = 3) = \frac{\mu }{3} \bullet {\text{P}}(X = 2){\text{ }}\left( {0.112777 = \frac{\mu }{3} \bullet 0.241667} \right)$    A1

attempting to solve for $\mu$     (M1)

$\mu  = 1.40$    A1

[3 marks]