Найти вероятности состояний системы после 1-го

1 шаг:
P1(1) = P1(0) P11 + P2(0) P21 + P3(0) P31 + P4(0) P41 = 0.25∙0 + 0∙0.2 + 0.75∙0 + 0∙0.8 = 0;
P2(1) = P1(0) P12 + P2(0) P22 + P3(0) P32 + P4(0) P42 = 0.25∙0.2 + 0∙0.3 + 0.75∙0 + 0∙0.1 =0.05 ;
P3(1) = P1(0) P13 + P2(0) P23 + P3(0) P33 + P4(0) P43 = 0.25∙0.3 + 0∙0.1 + 0.75∙0.5 + 0∙0 =0,45;
P4(1) = P1(0) P14 + P2(0) P24 + P3(0) P34 + P4(0) P44 = 0.25∙0.5 + 0∙0.4 + 0.75∙0.5 + 0∙0.1 =0.5;
P1(1)+ P2(1) +P3(1) +P4(1) =0 +0.05+ 0.45+ 0.5=1;
2 шаг:
P1(2) = P1(1) P11 + P2(1) P21 + P3(1) P31 + P4(1) P41 = 0∙0 + 0.05∙0.2 + 0.45∙0 + 0.5∙0.8 = 0.41
P2(2) = P1(1) P12 + P2(1) P22 + P3(1) P32 + P4(1) P42 = 0∙0.2 + 0.05∙0.3 + 0.45∙0 + 0.5∙0.1 =0.065;
P3(2) = P1(1) P13 + P2(1) P23 + P3(1) P33 + P4(1) P43 = 0∙0.3 + 0.05∙0.1 + 0.45∙0.5 + 0.5∙0 =0,23;
P4(2) = P1(1) P14 + P2(1) P24 + P3(1) P34 + P4(1) P44 = 0∙0.5 + 0.05∙0.4 + 0.45∙0.5 + 0.5∙0.1 =0.295;
P1(2)+ P2(2) +P3(2) +P4(2) =0.41+0.065+0.23 +0.295=1;
3 шаг:
P1(3) = P1(2) P11 + P2(2) P21 + P3(2) P31 + P4(2) P41 = 0,41∙0 + 0.065∙0.2 + 0.23∙0 + 0.295∙0.8 = 0.249
P2(3) = P1(2) P12 + P2(2) P22 + P3(2) P32 + P4(2) P42 = 0,41∙0.2 + 0.065∙0.3 + 0.23∙0 + 0.295∙0.1=0.131.;
P3(3) = P1(2) P13 + P2(2) P23 + P3(2) P33 + P4(2) P43 = 0,41∙0.3 + 0.065∙0.1 + 0.23∙0.5 + 0.295∙0 =0.2445;
P4(3) = P1(2) P14 + P2(2) P24 + P3(2) P34 + P4(2) P44 = 0,41∙0.5 + 0.065∙0.4 + 0.23∙0.5 + 0.295∙0.1=0.3755;
P1(3)+ P2(3) +P3(3)+P4(3) =0.249+ 0.131+ 0.2445 + 0.3755=1;
Ответ