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Firstly a quick check,

With Poisson distribution; can you rewrite $\Pr(X\ =\ 2\ |\ X\ >\ 1)$ as $1\ -\ \Pr(X \geq 3)?$ or is it equal to $\frac{\Pr(X\ =\ 2)}{\Pr(X\ >\ 1)}?$

With my main question now. Could you check my answers and help with the last one please.

Every morning, I roll a die to decide how to travel to work. If I roll a $1$ or $2$, I take the train, if I roll a $3$, I catch a bus and otherwise I cycle. The probability that I am late for work is $\frac1{10}$ if I travel by train, $\frac15$ if I travel by bus and $\frac1{20}$ if I cycle. I work a 5-day week.

$(i)$ Show that the probability that I will be late for work tomorrow is $\frac{11}{120}$

$(ii)$ If I am late for work, what is the probability that I travelled by train.

$(iii)$ Calculate the probability that I am on time every day during a week.

$(iv)$ I work for 46 weeks per year. Let $Y$ denote the number of weeks in a year for which I am on time every day of the week. Find the mean and variance of $Y$

Answers ive got so far:

(i) $(\frac13\ \cdot\ \frac1{10})\ +\ (\frac16\ \cdot\ \frac15)\ +\ (\frac12\ \cdot\ \frac1{20})\ =\ \frac{11}{120}.$

(ii) Im not exactly sure but something like: $\frac{11}{120} \div \frac1{10}\ =\ \frac{11}{12}.$

(iii) $(\frac13\ \cdot\ \frac9{10})\ +\ (\frac16\ \cdot\ \frac45)\ +\ (\frac12\ \cdot\ \frac{19}{20})\ =\ \frac{109}{120}.$

$(\frac{109}{120})^5 = 0.618$

(iv) im not sure about this question

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For the Poisson, the first suggestion $1-\Pr(X\ge 3)$ is not correct, but the second suggestion is correct.

The setup in the calculation of (i) is of course fine.

For (ii), the answer is wrong and moreover numerically very implausible. By the standard formula for conditional probability, our answer is the probability that we travelled by train and were late, divided by the probability that we were late. The thing we need to divide by was computed in (i). The probability we travelled by train and were late is $(1/3)(1/10)$.

For (iii) we have to assume independence. On any day, the probability we are not late is $1$ minus the answer of (i). For the probability of not late $5$ times in a row, take the fifth power.

For (iv), let $X_i=1$ if we are on time every day in Week $i$, and $0$ otherwise. Then $Y=\sum X_i$. From (iii) we can calculate the mean and variance of $X_i$. From this we can calculate the variance of $Y$. To calculate the variance, assume independence.

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    Sorry i meant $1\ -\ Pr(X\ \geq\ 3)$ not just $Pr(X\ \geq\ 3)$, is that still wrong?2012-12-09
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    Still not correct. And the second one (division) is still correct.2012-12-09
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    Given $Var(x)\ =\ E(x^2)\ -\ (E(x))^2$ and Mean is $E(x)\ =\ \sum_xx Pr(x\ =\ x)$ how do you apply that to get mean and var of $X_i$? Do you use answer from (iii) as $X$ and use 46 somehow? im not exactly sure in this case.2012-12-09
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    Mean of any sum is sum of means. Variance of any sum of independent random variables is sum of the variances. Can be proved from definitions you give. It takes a while.2012-12-09
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    I understand the definitions but I still don't understand how to apply (iii) to get the mean and variance of $X_i$, and then $Y$?2012-12-11
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    The probability of "success" (no lates) in Week $i$ is $p=(109/120)^5$. Since $X_i=1$ with probability $p$, and $0$ with probability $1-p$, $E(X_i)=p$. ($X_i$ is a Bernoulli random variable). For the variance of $X_i$, use the formula $E(X_i^2)-(E(X_i))^2$. Since $X_i^2=X_i$, we get $p-p^2=p(1-p)$. For mean and variance of $Y$, multiply individual means, variances by $n$. Note that $Y$ has **binomial** distribution.2012-12-11