Mathematics, 05.05.2020 12:52 dessy6546

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?

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Answer from: solikhalifeoy3j1r

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.

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

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

Answer from: Quest

understand that i don't know the answers

will someone answer his question

Answer from: Quest

step-by-step explanation:

i found the life tables for the year 2012.

life table of males at age 30. probability of dying between ages x and x+1

total population:

29–30 . . . . . . . . . . . . 0.001030

30–31 . . . . . . . . . . . . 0.001061

males:

29–30 . . . . . . . . . . . . 0.001420

30–31 . . . . . . . . . . . . 0.001448

white population

29–30 . . . . . . . . . . . . 0.001005

30–31 . . . . . . . . . . . . 0.001037

white males

29–30 . . . . . . . . . . . . 0.001372

30–31 . . . . . . . . . . . . 0.001399

non hispanic white population

29–30 . . . . . . . . . . . . 0.001083

30–31 . . . . . . . . . . . . 0.001116

non hispanic white males

29–30 . . . . . . . . . . . . 0.001477

30–31 . . . . . . . . . . . . 0.001503

jacob, non hispanic white male. 30 years old.

probability that jacob lives to be 30 years old.

since the given probabilities are probabilities of dying, i think we simply subtract it from 1 to get the probability of living. we will use the probability of non hispanic white male.

probability of living to be 30 y.o.

29 - 30 : 1 - 0.001477 = 0.998523

30 - 31 : 1 - 0.001503 = 0.998497

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