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