# Question 5a6e7

Jul 22, 2016

a) ~~74.06%
b) ~~24.94%
c) ~~1.94%
d) ~~0%

#### Explanation:

Let ${d}_{i} \in \left\{{d}_{1} , {d}_{2} , {d}_{3}\right\}$ be the number of $i$ dogs a person owns where ${d}_{3}$ denotes 3 or more.

A) probabilty of no dogs is $= 1 - p \left({d}_{1} \cup {d}_{2} \cup {d}_{3}\right)$
We use the inclusion exclusion principle because we don't want to double count the homes that have 1 dog with the homes who have 2 or 3 dogs. The same is true for 2 dog owner homes.

Firs we calculate  p(d_1uud_2uud_3) = .18+.04+.01 - (.18*.04) - (.18*.01) - (.04*.01) = 13.94%

We subtract 1 from above to find no dogs in just one of the homes. We need to have this happen in two homes so let $p \left(a \cap b\right)$ be the probability of event $a$ and $b$ where $a$ is the event in the first home and $b$ is the event in second. $p \left(a \cap b\right) = p \left(a\right) \cdot p \left(b\right)$ if the two events are mutually exclusive and indeed knowing more about one does not help in the second. The final answer is thus ${\left(.8606\right)}^{2} = .74063236$

B) Some dogs would be $p \left({d}_{1} \cup {d}_{2} \cup {d}_{3}\right)$ which is just 13.94%# but again this is in one of the homes. Using the events defined above we are now interested in $p \left(a \cup b\right)$ thus $p \left(a\right) + p \left(b\right) - p \left(a\right) \cdot p \left(b\right) = .1394 + .1394 - {\left(.1394\right)}^{2} = .25936764$

c) Dogs in each home is equivalent to finding $p \left(a \cap b\right) = {\left(.1394\right)}^{2} = .01943236$

d) More then 1 dog in each home would be the statement $p \left(a \cap b\right)$. However, this time we need to calculate a different probability then the one we have been using since we are no longer interested in homes with only one dog. The update is $p \left({d}_{2} \cup {d}_{3}\right)$

$p \left({d}_{2} \cup {d}_{3}\right) = .04 + .01 - \left(.04 \cdot .01\right) = .0496$

and $p \left(a \cap b\right) = {.0496}^{2} = .00256016$