Answer
A fraction of $0.014$ of the air molecules are pushed outside the house.
Work Step by Step
We can find an expression for the original volume:
$P~V_1 = nRT_1$
$V_1 = \frac{nRT_1}{P}$
We can find an expression for the final volume:
$P~V_2 = nRT_2$
$V_2 = \frac{nRT_2}{P}$
We can divide $V_1$ by $V_2$:
$\frac{V_1}{V_2} = \frac{\frac{nRT_1}{P}}{\frac{nRT_2}{P}}$
$V_1 = \frac{T_1}{T_2}~V_2$
$V_1 = \left(\frac{289~K}{293~K}\right)~V_2$
$V_1 = 0.986~V_2$
Since a fraction of 0.986 of the air molecules remain in the house, the fraction of air molecules that are pushed outside is $1-0.986$ which is $0.014$
