MATH 111 April Exam Solutions 2016page 9 of 11 pagesXYE0.90.10.60.4(d) [3 marks] Using the node diagram and elementary reasoning (as ifyou were in high school but understood the diagram), work outdirectly (without any linear algebra) the probabilitiesxnandynthat acounter starting on node X will be on nodes X and Y after exactlynsteps.[For example, you should be able to see exactly what has tohappen for a counter on X to find itself on Y after exactly 5 moves.]Starting on X, a counter which does not get to E will simply cyclebetween X and Y.It will be on Y after all odd numbers of moves andon X after all even numbers of moves.The probability of a cycle is(0.9) (0.4) = 0.36.Thusodd0even)36.0(2/nnxnneven0odd)36.0)(9.0(2/)1(nnynn(e) [3 marks] Now simplify your answers to parts (c) and check that they agree with your answerto (d).nnnx6.0216.021neven:2/2/236.06.06.06.0216.0216.0)1(216.021nnnnnnnnnxnodd:06.0216.0216.0)1(216.021nnnnnnxnnny6.0436.043.nodd:16.0)6.0(236.0236.0436.0436.0)1(436.043nnnnnnnny2/)1(2/)1(236.0)9.0(6.0)9.0(nnneven:06.0436.0436.0)1(436.043nnnnnny.page68