I got this question wrong on my midterm, can you please solve and show steps/ work on how to get the correct solution. Thank you.

7. (5 points) The probability mass function (pmf) of random variable X is given below. X 1 2 3 P(X=X) PI P2 P3 Find p1, P2 and p3 given that (E(X) = 1.7)and V(X) = 0.61. V ( x ) = E (x 2 ) - E (x ) 2 E (x ) = p, 1 + p, 2 + p, 3 0.61 = (P. (X1 ) + P2 ( X2 )2 + P's (x3)= ) - (1. 7) 1.7 = p 1+ 2 + 3 0. 61 = Cp ( 1 ) 2 + Pz ( 2 ) ? + p. ( 3 ) ? ) - 2. 89/ 1 . 7 - p2 2 - p, 3 = py 3. 5 = p + 4 p2 + 9 ps 1. 7-7 p2 - 3 ps = 2 pz 1.7- P2 - 2 pz = 3 p 3 3. 5 = ( 1.7 - P, 2 - p; 3 ) + 4 p2 + 9 pg 1.8 = 2 p,+ 6p3 0. 61 : 0.45 ( 1) + 0.4 ( 2) + 0.75 (3) 2 - 2.89 = 0.61 3.5 = P2 + 2 ( 1.7 - P2 - 3p3 ) + 9p3 0.1 = - P2+ 3 p3 3. 5 = Py + 4 p, + 3 ( 7-7 - P2 - 2 pz) - 1. 6 = -2p2 - 2 p2 Incorrect. -2.5 1. 6 = 2 p2 + 2pz P3 = 0. 2 1. 7 - 8, + Pz 0.4 0.2 0. 6 1.7 0. 7 0 . 1 Page 8 P. = 0.45 P2 = 0.4 P3= 0.15 X