If an event occurs $0$ times (out of $50$, in this case) then it does not occur at least once. So we can find the probability of it not occurring and then subtract that value from $1$.
So, what are the chances of it not occurring on $1$ trial? $1-.116 = .884$
What about not occurring on $2$ trials? $(1-.116)\cdot (1-.116)=.884 \cdot .884 = .781456$
Now what about not occurring on $50$ trials? $.884^{50} \approx .002102$
We must subtract this value from $1$ (recall that what we just calculated is the probability of it not occurring, so the probability of it occuring at least once is: $$1-.002102 \approx .9979 $$
Follow the hint. In the starting case, with probability $1/3$ it rains, the professor takes the umbrella, and with probability $2/3$, it does not rain when it is time for the professor to return home. So with probability $2/9$ the professor has not walked in the rain but the umbrella is at the office.
Similarly, with probability $1/9$, it has rained both to and from work and the umbrella has made a round trip.
With probability $2/9$, it did not rain on the way to work but it rained on the way back from work, meaning the professor got wet.
With probability $4/9$, it did not rain either way, and the professor is back home.
We can summarize this in a table for the round trip:
$$\begin{array}{ccccc}
\text{Umbrella} & \text{Rain} & \text{Got wet} & \text{Probability} \\
\hline
\text{Office} & \text{Yes, No} & \text{No} & 2/9 \\
\text{Home} & \text{Yes, Yes} & \text{No} & 1/9 \\
\text{Home} & \text{No, Yes} & \text{Yes} & 2/9 \\
\text{Home} & \text{No, No} & \text{No} & 4/9 \\
\end{array}$$
Therefore, with probability $5/9$, we have returned to the initial state (not wet, umbrella home), except a day has passed. Thus the expected number of additional days until getting wet is still $\mu$. With probability $2/9$, the professor got wet that day. With probability $2/9$, the professor has survived a day but now the umbrella is at the office. Since $v$ represents the expected number of days until getting wet when the professor is home but the umbrella is not, we summarize the expected number of days until getting wet is $$\mu = \frac{5}{9}(1 + \mu) + \frac{2}{9}(1) + \frac{2}{9}(1 + v).$$
Now for $v$, we suppose the professor begins the day at home but the umbrella is at the office. Then with probability $1/3$, the professor must walk in the rain to work. With probability $2/9$, the professor makes it to the office, and takes the umbrella home because it rains when it is time to leave. With probability $4/9$, it does not rain at all and the professor survives a day but returns to the state where the umbrella is not at home. So the expected number of days until getting wet in this case is...? I have not given the formula so that you have a chance to do the rest.
Best Answer
It says "at least once," which means it could be in the morning or evening or both. There are $5$ choices that have at least one teleport in the day - rows $1$, $2$, $3$, $4$, and $7$ in the list given - and there are nine total outcomes, so the answer is $5/9$.