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Does anyone know how to solve this question?

Traffic on a road follows a Poisson process with rate $\displaystyle \frac 23$'s of a car per minute. $10$% of vehicles are trucks and the other $90$% are cars.

a) What is the probability at least one truck passes in an hour?

b) Given that $10$ trucks have passed by in an hour, what is the expected number of total vehicles that have passed by?

c)Given that $50$ vehicles have passed by in an hour, what is the probability there were exactly $5$ trucks and $45$ cars?


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What do you know? What have you tried? Where are you stuck? – Did May 26 '11 at 5:12

The rate $2/3$ per minute means $2/3\times 60 = 40$ vehicles per hour. The random selection of the car types means that we may divide the Poisson process to two - with 10% = 4 trucks per hour and 90% = 36 cars per hour.

a) The probability that no truck passes in some time is exponentially decreasing with the rate indicated above, so after one hour it is $\exp(-4)$. The probability that at least one truck passes is the complement, $1-\exp(-4)=98.16$ percent.

b) The expected number of vehicles is the expected number of trucks plus the expected number of cars. The former is known to be 10. The latter is independent and it is 36, as explained in the first paragraph. So the total is 46.

c) This is harder. The probability that 50 vehicles passed - something we know that occurred - is $40^{50} \exp(-40)/50! = 1.77$ percent. The probability of a more refined event - subset of the previous one - that it was 5 trucks and 45 cars is $$4^{5} \exp(-4)/5! \times 36^{45} \exp(-36)/45! = 0.156 \times 0.021 = 0.00327 $$ Now just divide $0.327$ percent by $1.77$ percent and you get $18.5$ percent.

If you're cheating on a homework, you should write down in the homework that you got some help which also makes you less guilty if my solution is wrong.

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This solution for (a) corresponds to two thirds of a vehicle per minute. // The answer to (c) can be phrased in terms of a binomial distribution. – Did May 26 '11 at 5:24
I corrected the word "cars" to "vehicles" in the first sentence. Otherwise it was totally deliberate to avoid complicated and pre-cooked notions such as a binomial distribution - as well as Poisson distribution (mostly), in fact. The homework is also formulated in such an elementary language. – Luboš Motl May 26 '11 at 8:56
Re (a), the ambiguity lies in the OP's post. Strictly speaking, the post indicates 2/3s of cars plus 1/9 of 2/3s, that is 2/27s trucks. But I guess this is one of these situations where one should deliberately understand something different from what is written. // But this is the first time I see binomial distributions characterized as a complicated and pre-cooked notion... :-) – Did May 26 '11 at 12:17
hi, 2/3s cars is not a typo on my part that is how it is written in the book. – J.M May 26 '11 at 21:02
OK, if there are 2/3 of a car and not a vehicle, then you just change the numbers correspondingly. At least some work. ;-) – Luboš Motl May 29 '11 at 16:20

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