Independent solution

How to solve this Joint Probability Mass Functions question

Setup

Setup

Sum over the first coordinate to obtain the marginal distribution of Y. The factor depending on Y can be taken outside that finite sum.

pY(y)=∑x=03(4−x)(3−y)60p_Y(y)=\sum_{x=0}^{3}\frac{(4-x)(3-y)}{60}
∑x=03(4−x)=4+3+2+1=10\sum_{x=0}^{3}(4-x)=4+3+2+1=10

Model

Model

The marginal masses decrease linearly across the three possible values.

pY(y)=3−y6,y=0,1,2p_Y(y)=\frac{3-y}{6},\qquad y=0,1,2
pY(0)=12,pY(1)=13,pY(2)=16p_Y(0)=\frac12,\quad p_Y(1)=\frac13,\quad p_Y(2)=\frac16

Compute

Compute

Compute the first two moments and apply the variance identity.

E⁡[Y]=0(12)+1(13)+2(16)=23\operatorname{E}[Y]=0\left(\frac12\right)+1\left(\frac13\right)+2\left(\frac16\right)=\frac23
E⁡[Y2]=1(13)+4(16)=1\operatorname{E}[Y^2]=1\left(\frac13\right)+4\left(\frac16\right)=1
Var⁡(Y)=1−(23)2=59=0.5555…\operatorname{Var}(Y)=1-\left(\frac23\right)^2=\frac59=0.5555\ldots

Answer

Answer

The variance rounds to 0.56.

0.56(A)\boxed{0.56\quad\text{(A)}}