Probability on expected number of days

Question

It doesn't rain often in Tucson. Yet, when it does, I want to be prepared. I have 2 umbrellas at home and 1 umbrella in my office. Before I leave my house, I check if it is raining. If it is, I take one of the umbrellas with me to work, where I would leave it. When I go back home, I check if it is raining. If it is, I take one of the umbrellas with me home; therefore, the number of umbrellas at my house and in my office changes with time. The probability of rain is 0.1 every time I leave either my office or my house. The event of rain is independent of location and what happened in the past. Find the expected number of days before I run out of umbrellas where I am and it is raining outside. Also find the probability that I am home when that happens.

Answer 1

The person has 2 umbrellas at home and 1 in office. Also the probability of raining is independent of other factors and is equal to 0.1.

Now let us find the probability distribution of
X: Number of days before he running out of umbrellas.

Now X can take values from 0,1,2,3,..

Let us find the probability X=0, Now since he have 2 umbrellas at home and one at office, this probability will be zero.

Now let us find the probability x=1, now here we are interested in finding the probability that he runs without umbrella on second day, this will happen in following manner.

When he will go office there is no rain, so probability is 0.9, now on returning there is rain with prob 0.1 now on second day leaving office there is no raining with 0.9 and at the time of return it rains with 0.1

Hence total probability is .9*.1*.9*.1

Now let us find the probability x=2, now here we are interested in finding the probability that he runs without umbrella on second day, this will happen in following manner.

This probability will be .1*.9*.1*.9*.1  (The probabilities are arranged according to event)
Now let us find the probability x=3, now here we are interested in finding the probability that he runs without umbrella on second day, this will happen in following manner.

The probability is .9*.1*.1*.1*.9*.1

The probability that more x ≥ 4 will be 1 minus all these probabilities

Hence the expected number of days is 3.97,

Means on an average more than 3 days required to run without umbrella.