Question

Question #f0d61

Answer 1

The Poisson distribution is discrete with probability:

`p(x) = (e^(-lambda))(lambda^x)/(x!)`

Explanation:

First, simplify the conditional probability:

P(X = 1 | X <= 1) = P(X=1) / [P(X=0) + P(X=1)]

Now, insert the Poisson probability for each expression above:

P(X=1) / [P(X=0) + P(X=1)] =

`[(e^(-lambda))(lambda^1)/(1!)} / [(e^(-lambda))(lambda^0)/(0!) + (e^(-lambda))(lambda^1)/(1!)`

Next, simplify:

`[(e^(-lambda))(lambda)} / [(e^(-lambda)) + (e^(-lambda))(lambda)] = lamda / (1+lamda)`

Finally, set equal to 0.8 and solve for `lambda`:

`lamda / (1+lamda)` = `0.8`

`lambda` = `4`

Hope that helped