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Continuous Probabiltiy - Bus Arriving Watch

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    I am trying to do the following question:

    Number 24 and number 42 buses arrive independently at the corner of Mayeld Road at a random rate of 3 and 4 per hour respectively. You arrive at the bus stop at 10 a.m. What is the probability that it takes at least 20 minutes for a bus (either the 24 and 42 bus) to arrive?I am kinda of stuck on how to start it.

    So I've learned about Uniform distribution, normal distribution and exponential distribution. I actually not really sure which one to use here. Could you also explain what sort of problems to apply each distribution to?

    If I set up a random variable, what should it denote? The number of minutes pass 10 for a bus to arrive? Or that the 42 and 24 bus arrive?
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    (Original post by aspiring_doge)
    I am trying to do the following question:

    Number 24 and number 42 buses arrive independently at the corner of Mayeld Road at a random rate of 3 and 4 per hour respectively. You arrive at the bus stop at 10 a.m. What is the probability that it takes at least 20 minutes for a bus (either the 24 and 42 bus) to arrive?I am kinda of stuck on how to start it.

    So I've learned about Uniform distribution, normal distribution and exponential distribution. I actually not really sure which one to use here. Could you also explain what sort of problems to apply each distribution to?

    If I set up a random variable, what should it denote? The number of minutes pass 10 for a bus to arrive? Or that the 42 and 24 bus arrive?
    I'm not much of a probabilist, and it's too late to think about this in detail, but
    you have two Poisson processes, namely the arrival of no 24 buses, mean \lambda and no 42 buses, mean \mu, occurring in the same location. It can be shown that this gives you a new Poisson process with mean \lambda +\mu. This process describes the probability of arrival of either type of bus.

    The inter-arrival times of your Poisson process are random, with mean \frac{1}{\lambda +\mu}. The distribution of these times follow an exponential distribution, random variable X, with parameter \lambda +\mu. You can look up the formulae for P(X > a), P(b > X >a) and so on.

    Does that help? May be best to wait for confirmation from someone like Gregorius that I haven't led you astray though.
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    (Original post by atsruser)
    I'm not much of a probabilist, and it's too late to think about this in detail, but
    you have two Poisson processes, namely the arrival of no 24 buses, mean \lambda and no 42 buses, mean \mu, occurring in the same location. It can be shown that this gives you a new Poisson process with mean \lambda +\mu. This process describes the probability of arrival of either type of bus.

    The inter-arrival times of your Poisson process are random, with mean \frac{1}{\lambda +\mu}. The distribution of these times follow an exponential distribution, random variable X, with parameter \lambda +\mu. You can look up the formulae for P(X > a), P(b > X >a) and so on.

    Does that help? May be best to wait for confirmation from someone like Gregorius that I haven't led you astray though.
    Well I kind of understand about what you said in the first paragraph, but not in your second paragraph.Why is the mean \frac{1}{\lambda + \mu}? I thought you said that it was \lambda + \mu.
    I still cant figure how to express the probability of what I'm trying to find.
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    (Original post by atsruser)
    Does that help? May be best to wait for confirmation from someone like Gregorius that I haven't led you astray though.
    I couldn't have answered better myself.
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    (Original post by aspiring_doge)
    Well I kind of understand about what you said in the first paragraph, but not in your second paragraph.Why is the mean \frac{1}{\lambda + \mu}? I thought you said that it was \lambda + \mu.
    I still cant figure how to express the probability of what I'm trying to find.
    The interarrival time for the Poisson process is exponential with parameter \lambda + \mu. The mean of an exponential distribution is the reciprocal of its parameter.
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    (Original post by Gregorius)
    I couldn't have answered better myself.
    Then this is a red letter day. I've answered a question on probability without someone having to point out that I've misinterpreted the meaning of the word "either", or something.
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    (Original post by atsruser)
    Then this is a red letter day. I've answered a question on probability without someone having to point out that I've misinterpreted the meaning of the word "either", or something.

    Lol.
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    (Original post by aspiring_doge)
    Well I kind of understand about what you said in the first paragraph, but not in your second paragraph.Why is the mean \frac{1}{\lambda + \mu}? I thought you said that it was \lambda + \mu.
    I still cant figure how to express the probability of what I'm trying to find.
    OK, I'll try to fill out a few details.

    1. Suppose you have a random process, such as the arrival of a bus, where you always get the same mean number of events per unit time. Call this mean \lambda. So we might have \lambda = 3 s^{-1} i.e. we have 3 events per second, on average, and this is true for every second. Note that usually we don't get *exactly* 3 events per second though - we can get any number, due to the randomness.

    Since we have 3 events per second, on average, we have \frac{1}{\lambda} = \frac{1}{3} s between events on average.

    2. Suppose further that the causes of the events are unrelated to each other - for example, if we are measuring radioactive decays, one atom doesn't "know about" or "influence" any other atom's decay. Then, together with fact 1 above, we are dealing with a random process called a Poisson process, random variable X with Poisson distribution, and you can analyse it to show that the probability of n events occuring per unit time is

    P(X=n)=\frac{\lambda^n e^{-\lambda}}{n!}

    Here \lambda is the mean number as described above, but when we write down this formula, we often refer to it as the "parameter" of the distribution. (It's quite easy to do the analysis if you're dead, French, and called Poisson, but otherwise you can think of a Poisson process as the limit of a set of binomial choices; imagine your unit time period chopped up into tiny pieces, and compute the probability of an event occurring or not occuring in each tiny piece)

    3. We can measure two interesting things about a Poisson process, namely the number of events per unit time, or the lengths of the gaps between events. We saw above that on average this is the inverse of the mean of the Poisson distribution, but of course, the gaps vary in length due to the randomness. We call the gaps the "inter-arrival times", and it can be shown that, for a Poisson process, these have an exponential distribution, random variable T, where

    P(T >t)= e^{-\lambda t}

    where \lambda is the same value as we use above in the Poisson distribution - bear in mind that we are merely measuring two different aspects of the same underlying random process, so we would expect only to have to deal with one characteristic number for that process.

    4. In your problem, you have two separate Poisson processes occuring simultaneously, so we need to know how that changes the results above. Well, the answer is "not much". That's because if we have, say, bus A giving us 4 mean arrivals per hour, and bus B with 3 mean arrivals per hour, then clearly we get 7 mean bus arrivals per hour, where we ignore the type of bus which is arriving. So we have a mean number of events for the combined distribution of 3 / hour + 4 / hour=7 / hour.

    In addition, the A buses don't influence the B buses (we assume, for this problem). A buses don't know anything about the arrival of B buses, or vice versa. So we have the conditions for a Poisson process, as I described them in points 1 and 2 above. That's cool, since we already figured out that the mean for this process is 3+4 =7; in general, of course, the combination of two Poisson processes, means \lambda and \mu, gives us a new Poisson process mean \lambda + \mu.

    That's nice, since we already have the formulae for the probability of number of events and inter-arrival times for this process, and we can therefore merely replace the \lambda parameter with \lambda + \mu to calculate the required probabilities.
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    (Original post by atsruser)
    ...
    This is great! Many thanks. :-)
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    (Original post by Gregorius)
    Lol.
    Probability theory: not enough mathematics, too much close textual analysis.

    (Unless it's in a course called "Probability and Measure", then it's the other way round).
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    (Original post by atsruser)
    Probability theory: not enough mathematics, too much close textual analysis.

    (Unless it's in a course called "Probability and Measure", then it's the other way round).
    Funnily enough, I hated probability and stats when I was at university. Now I earn my living from it!
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    (Original post by atsruser)
    OK, I'll try to fill out a few details.

    ....
    Thanks for that explanation! It was really helpful
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    Quick question,

    so I need to compute

    P\{X \geq 20\} = \int_{20}^{\infty} \frac{7}{60}e^{-\frac{7}{60}x} dx where \lambda = \frac{7}{60}

    right?


    Is it also possible to do it this way:

    P\{X \geq 20\} = 1 - F(20) where F(t) = 1 - e^{-\lambda t}
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    (Original post by aspiring_doge)
    Quick question,

    so I need to compute

    P\{X \geq 20\} = \int_{20}^{\infty} \frac{7}{60}e^{-\frac{7}{60}x} dx where \lambda = \frac{7}{60}

    right?
    Well, your value for lambda looks fine, and you're trying to work with the pdf, which is OK, but note that

    \int_t^\infty \lambda e^{-\lambda x} \ dx = - [e^{-\lambda x}]_t^\infty = -(0-e^{-\lambda t}) = e^{-\lambda t}

    which is the result that I quoted above. And the answer to your second question is then "yes", of course.
 
 
 
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