Let X be a random variable with probability distribution function pX(x)=P(X=x)=211,x{10,9 Find the probability distribution function pY(y) of Y=2X+1, and 1. evaluate pY(11)= 2. evaluate pY(1)= Instructions on how to input the answer: Round your answer to exactly 4 decimal places. Input the string of the answers separated by a comma. Do not include any space in the text. (pY(11),pY(1))=1)0.0476,0.0476 Since the range of X is {10,9,,1,0,1,,9,10}, the range of Y=2X+1 is {19,17,,1,1,3,,19,21}. Then we can work out the probabilities of P(Y=k),kP(Y=19)P(Y=17)P(Y=19)P(Y=21)=P(X=10)=211,=P(X=9)=211,==P(X=9)=211,=P(X =10)=211. Therefore pY(y)=P(Y=y)=211,y{19,17,,1,1,3,,19,21}.