Question

4. (10 pts.) An exterminator is trying to find a mouse in a 3-room house. It is equally likely that the mouse is in any of the three rooms. Since the mouse is asleep, it is not moving. The likelihood of finding the mouse in room k in one search is 1 - ak, where α1, α2, α3 are some constants between 0 and 1 Find the probability of the mouse being found in room k (for k-1,2,3), given that a search in room 1 was unsuccessful.

Answer 1

The probability that the mouse being found in room $k\left ( k=1,2,3 \right )$ is $1-\alpha _k$. The probability of not finding is $\alpha _k$.

a)The probability that the mouse not being found in room 1 and finding in room 2 is

$\left (1- \alpha _2 \right )\alpha _1$

b)The probability that the mouse not being found in room 1 and finding in room 3 is

$\left (1- \alpha _3 \right )\alpha _1$

The probability of finding the mouse in rooms 2 or 3 is ( cases a, b are mutually exclusive)

$P\left ( \textup{2 or 3 and }\overline{1} \right )=\left (2- \alpha _2- \alpha _3 \right )\alpha _1$

We need to find the probability ($\overline{1}$ means not finding in the first room) (use conditional probability)

$P\left ( \textup{2 or 3 }| \overline{1} \right )=\frac{P\left ( \textup{2 or 3 and }\overline{1} \right )}{P\left ( \overline{1} \right )}\\$

$P\left ( \textup{2 or 3 }| \overline{1} \right )=\frac{\left (2- \alpha _2- \alpha _3 \right )\alpha _1}{ \alpha _1}\\ {\color{Blue} P\left ( \textup{2 or 3 }| \overline{1} \right )=2- \alpha _2- \alpha_3}$