# Poisson Distrobution-HELP

## Main Question or Discussion Point

Poisson Distrobution-HELP!! URGENT!!!

the problem:

The Breakdowns of a robot follow a Poisson Dist. with an avg of .5 breakdowns per 8-hour workday. If this robot is placed in service at the beginning of the day, find the probability that:

a. It will break down durring the day.
b. It will work for at least 4 hours without breakdown
c. Does what happend the day before have any effect on your answers? why?

I'm a newb to probability and am I'm haveing trouble figuring out where to begin. I know that the rate(lamda) has something to do with the percent of break downs and the work day. Is it .5*workday or just .5. I just don't know where to begin with this problem. Related Set Theory, Logic, Probability, Statistics News on Phys.org
i know that P{X >= a } = e^(-גa) where ג is lamda, would P{ X < a } = -e^(-גa)?

Well the answer to a. is

P{ X < a } = e^(-גa)
= e^(-0.5)
= 0.6065

where ג = 0.5 <- the rate equals lamda

Since P{ X >= a } = 1 - ( 1 - e^( גa ) ) = e^(-גa)

right?

i need help on b

anybody.......

akito458 said:
the problem:

The Breakdowns of a robot follow a Poisson Dist. with an avg of .5 breakdowns per 8-hour workday. If this robot is placed in service at the beginning of the day, find the probability that:

a. It will break down durring the day.
b. It will work for at least 4 hours without breakdown
c. Does what happend the day before have any effect on your answers? why?

I'm a newb to probability and am I'm haveing trouble figuring out where to begin. I know that the rate(lamda) has something to do with the percent of break downs and the work day. Is it .5*workday or just .5. I just don't know where to begin with this problem. A Poisson Process is a counting process; in your case define a success or occurrence when a breakdown happends. The probability of having n breakdowns within a period of time t is: P{N(t)=n}=exp(-lamda*t)*(lamda*t)^n/n!
In this case lamda = 0.5 breakdowns/day (since a day has 8 hours of work. Also, we can express lamda per hour as lamda = 0.5/8 = 1/16 breakdowns/hour.

a. It will break down during the day.
Here we can look for the probability of having a breakdown in a day (8 hours period), then P{N(1) >= 1} = 1 - P{N(1) < 1} = 1 - P{N(1) = 0} = 1 - exp(-0.5)
Or if we look for the probability that the first breakdown occurs within 8 hours; let say X1 is a random variable representing the interarrival time or interarrival occurrence time of breakdowns. By definition the interarrival times are ~Exponential, f(t) = lamda*exp(-lamda*t), 0 <= t < infinity
Then P{X1 <= 8} = 1 - exp[-(0.5/8)*8] = 1 - exp(-0.5)

b. It will work for at least 4 hours without breakdown.
Here we look for the probability of no breakdown in at least 4 hours, or a breakdown in at most 4 hours.
P{N(4) = 0} = exp[-(0.5/8)*4]
P{X1 > 4} = 1 - P{X1 <= 4} = exp[-(0.5/8)*4]

I hope it helps.

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