Recent content by maria papadakh

yes,forget this idea.i will take the solution above with P(D)=...=0,325

or the possibility of 0,1 =P(W) and P(D/F)=O,25 and from the fact that P(D/W) and P(D/F) are independent we could calculate P(D/W)

it was corrected some hours ago.there are not other changes.But i think that i miss something.Why P(F1)=0,3 and P(F2)=0,05 are given?

As the problem is written i think that P(D/F) and P(D/W) are independent P(D/FUD/W)=P(D/F)+P(D/W)-P(D/F)*P(D/W)=O,1+0,25-O,1*0,25=0,35-O,O25=0,325 Now i think that it might be the solution.Is it right?Do you mean that the P(F1)=0,3 and P(F2) are not need to solve the problem?

it is the posibility of the disasters that happen in a forest.you can read the hole problem above.thank you.

I think that P(D)=P(D/W)*P(W)+P(D/F)*P(F) =P(D/W)0,1+0,25*0,35 P(F)=P(F1)+P(F2)=0,3+0,05=0,35 i don't know how to find P(D/W)

hi,finally it was wrong the propability of the wind.it is ten per cent.Can you help me now please?

Summary:: problem with independent events in possibilities A forest has a fire per year with probability 30% and it has 2 fires per year with possibility 5%.The probability of 3 or more fires is 0%.The forest might have disasters from the wind which blows with probability of 100% or has...

i will try with the jordan normal form.thank you!

i know that there is the Cayley -Hamilton theorem but i don't know if i can use it and how.Do you have any ideas about it?Please give me any help.