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Hi, 


I'm doing question 3c and am struggling to work out the tP45^(HS) value for the integral of A^(HSD).


I have attached my workings for the probability. Can you please help me to see where I am wrong or explain how they got tP45^(HS) to equal 3/5(e^(-0.005t) - e^(-0.001t))Test 2 Quedtion 3c.pdf (0,3 MB)

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Hi, just a note: the answer should be 3/5(e^(-0.005t)-e^(-0.01t)).


For your tP45^HS calculation you need to take into account staying in the S state until time t. So basically what this probability is doing is saying that you start off in H state, you make a transition to the S state at time s and then stay in S state until time t. So you just need to include the holding time in the Sick state until time t. Attached are the workings. 


Scanned Documents (2).pdf (91 kb)

by (350 points)
Thanks so much!
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