#### Answer

Yes.

#### Work Step by Step

Evaluating the expressions at x=0, 0.5, 1:
$P(0)=\frac{1}{2(2-2\cdot0)!(2\cdot0)!}=\frac{1}{2\cdot2!\cdot0!}=\frac{1}{4}$
$P(0.5)=\frac{1}{2(2-2\cdot0.5)!(2\cdot0.5)!}=\frac{1}{2\cdot1!\cdot1!}=\frac{1}{2}=\frac{2}{4}$
$P(1)=\frac{1}{2(2-2\cdot1)!(2\cdot1)!}=\frac{1}{2\cdot0!\cdot2!}=\frac{1}{4}$
The probabilities are equal to the ones in exercise 15, therefore they describe the sampling distribution found there.