Question

Each day, Robin commutes to work by bike with probability 0.4 and by walking with probability 0.6. When biking to work injuries occur according to a Poisson process with an average of 0.1 injuries per year and when walking, injuries occur according to a Poisson process with an average of 0.02 injuries per year. The two process are independent of each other.
What is the probability of at least one injury commuting to work in the next 20 years?

Answer 1

Answer:

64.65% probability of at least one injury commuting to work in the next 20 years

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

Each day:

Bikes to work with probability 0.4.

If he bikes to work, 0.1 injuries per year.

Walks to work with probability 0.6.

If he walks to work, 0.02 injuries per year.

20 years.

So

$\mu = 20*(0.4*0.1 + 0.6*0.02) = 1.04$

Either he suffers no injuries, or he suffer at least one injury. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X \geq 1) = 1$

We want $P(X \geq 1)$. Then

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-1.04}*1.04^{0}}{(0)!} = 0.3535$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.3535 = 0.6465$

64.65% probability of at least one injury commuting to work in the next 20 years