# Two problems on Poisson distribution [closed]

1) A car hire firm has 2 cars which it hires out day by day. The number of demands for a car on each day is distributed as a Poisson distribution with mean of 1.5. Calculate the proportion of days
(i) on which there is no demand,
(ii) on which demand is refused.

2) A source of liquid is known to contain a mean number of bacteria per cubic centimetre equal to 3. Ten 1 cc test tubes are filled with liquid. Assume that Poisson distribution is applicable. Calculate the probability that all the test tubes will show growth, i.e. contain at least one bacterium each.

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## closed as too localized by BenjaLim, LVK, sdcvvc, William, Matt N.Sep 3 '12 at 10:26

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$1.$ Let the random variable $X$ be the demand. We are told that $X$ has Poisson distribution with mean (parameter) $\lambda=1.5$. Thus $$\Pr(X=k)=e^{-\lambda} \frac{\lambda^k}{k!}\tag{1}$$ for every non-negative integer $k$.

(i) We want $\Pr(X=0)$. By $(1)$, this is simply $e^{-\lambda}$.

(ii) One has to interpret the meaning of "demand is refused." This seems to be an unusual way of saying that at least one customer is turned away, meaning that $X \ge 3$. To calculate $\Pr(X\ge 3)$, we find $\Pr(X=0)+\Pr(X=1)+\Pr(X=2)$, and subtract the sum from $1$.

We have $\Pr(X=0)=e^{-\lambda}$. By $(1)$, $\Pr(X=1)=\lambda e^{-\lambda}$ and $\Pr(X=2)=\frac{\lambda^2}{2!}e^{-\lambda}$. Add up. My calculator gives approximately $0.8088468$. Subtract this from $1$.

$2.$ Let the random variable $X_i$ be the number of bacteria in the $i$-th test tube ($i=1,2,\dots, 10$). The $X_i$, we are told, have Poisson distribution with mean (parameter) $\lambda=3$. We will assume that the $X_i$ are independent. This assumption really cannot be scientifically justified, and should have been mentioned explicitly in the statement of the problem.

For any $i$, $\Pr(X_i=0)=e^{-\lambda}$. So the probability that the $i$-th test tube shows some growth, meaning that $X_i \ge 1$, is $1-e^{-\lambda}$.

The probability that all the $X_i$ are $\ge 1$ is equal to $(1-e^{-\lambda})^{10}$. As a check on your calculations, the number turns out to be roughly $0.6000803$.

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 I have edited part (ii) of question 1.Instead of demand is reduced I have replaced by demand is refused.please help me to answer this question corect answer of question 1 part (ii) is 0.1913 – prashant kumar pathak Jun 27 '12 at 18:14 @prashantkumarpathak: The solution of $1$(ii) has been changed to reflect the change in the question. I do not completely agree with the answer given, get $0.1911532$ roughly. Perhaps whoever computed the answer found the separate probabilities and rounded each time. Not that it matters, the Poisson model would only be roughly right anyway. – André Nicolas Jun 27 '12 at 18:33 answer to 2nd question correct but I havent understood that why have you raised to the power of 10 the following expression (1−e^−lambda). – prashant kumar pathak Jun 27 '12 at 18:33 If $A$ and $B$ are independent events, the probability that $A$ happens and $B$ happens is $\Pr(A)\Pr(B)$. If $A$, $B$, $C$ are independent events, the probability that $A$, $B$, $C$ all happen is $\Pr(A)\Pr(B)\Pr(C)$. And so on. For example, if toss a fair coin $5$ times, probability we get $5$ heads in a row is $(1/2)^5$. – André Nicolas Jun 27 '12 at 18:40 In question 1 I did not understood this line "This seems to be an unusual way of saying that at least one customer is turned away, meaning that X≥3".why is X≥3? instead it should be X≥1 – prashant kumar pathak Jun 27 '12 at 18:55
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