Question

I have 4 umbrellas, some at home, some in the office. I keep moving between home and office.
I take an umbrella with me only if it rains. If it does not rain I leave the umbrella behind (at
home or in the office). It may happen that all umbrellas are in one place, I am at the other, it
starts raining and I must leave, so I get wet.
(a) If the probability of rain is p, what is the probability that I get wet? [Ans pq/q+4 where q = 1-p]
(b) Current estimates show that p = 0.6 in Guwahati. How many umbrellas should I have so
that, if I follow the strategy above, the probability I get wet is less than 0.01? Ans [24]

Answer 1

The problem can be modeled using markov chain.

We take the possible number of umbrellas that we can have at any point of time.

S = {0, 1, 2, 3, 4}

For the given scenario, suppose we have one umbrella at home/office and rest three at other location.

Then, if it rains, we will have to take the one and only umbrella from one place to other.

In that case,

Probability (1 umbrella at one side to umbrellas at other side) = Probability that it rains = p

Similarly, if it does not rain, we would not take umbrella from one location to other.

So, Probability (1 umbrella at one side to 3 umbrellas at other side) = Probability that it does not rain =1 - p = q

Similarly, for finding the probability of number of umbrellas from one place to other, in terms of p and q, we can create a transition probability table.

If we draw a transition probability table, it would look like -

	0	1	2	3	4
0					1
1				q	p
2				p
3		q	p
4	q	p

Rest numbers do not add up to 4 or 5 given certain condition of rain/ no rain.

All of these probabilities would add up to 1.

sum over i = 0 to 4 * π(i) = 1

stationary distribution is a probability distribution represented as a row vector π with probabilities summing to 1, and given transition matrix P, satisfying -

π = πP

So, we have -

π(0) = π(4) * q

π(1) = π(2) = π(3) = π(4)

Simplifying the equation we can write -

π(4) * q + 4 * π(4) = 1,

=> π(4) = 1 / q + 4

and π(0) = q / q + 4

Given,

probability of raining in Guwahati is 0.6,

Then, the probability of me getting wet is -

probability of rain * probability that i am having 0 umbrellas at one place (π(0))

= pq / q + 4

= 0.6 * 0.4 / 0.4 + 4

= 0.054 = 5.4%

The chance, as seen in formula, depends on the number of umbrellas (4 in this case)

If we want chance to be less than 1%,

Let the number of umbrellas we need be x,

pq / q + x < 0.01

or

x > 100pq - q

=> x > 23.6

To reduce chances of getting wet to less than 1%. we would need minimum 24 umbrellas.