Alpha particles are emitted by a radio active source at an average rate of $5$ in a $20$ minutes interval. Using Poisson distribution find the probability that there will be
- $2$ emissions;
- at least $2$ emissions
Average rate of emission in $20$ minute interval $= 5$ emissions
i.e. $\;$ $\lambda = 5$
- P(there will be $2$ emissions) $= P \left(X = 2\right)$
$\begin{aligned} P \left(X = 2\right) & = \dfrac{e^{- \lambda} \times \lambda^{2}}{2!} \\\\ & = \dfrac{e^{- 5} \times 5^2}{2} \\\\ & = \dfrac{0.0067 \times 25}{2} = 0.08375 \end{aligned}$
- P(at least $2$ emissions) $= P \left(\text{2 or more emissions}\right)$
$\begin{aligned} \text{Now, } P \left(\text{2 or more emissions}\right) & = 1 - P \left(\text{less than 2 emissions}\right) \\\\ & = 1 - P \left(X < 2\right) \end{aligned}$
$\begin{aligned} P \left(X < 2\right) & = \sum \limits_{x = 0}^{1} \dfrac{e^{- \lambda} \times \lambda^{x}}{x!} \\\\ & = e^{- \lambda} \times \left[\dfrac{\lambda^{0}}{0!} + \dfrac{\lambda^{1}}{1!} \right] \\\\ & = e^{- 5} \times \left[1 + \dfrac{5}{1} \right] \\\\ & = 0.0067 \times 6 = 0.0402 \end{aligned}$
$\therefore \;$ $P \left(\text{at least 2 emissions}\right) = 1 - 0.0402 = 0.9598$