I had responded to this earlier but it must not have gone through... First of all, Thanks for both of your inputs. I will have to look at Michael's as I have been looking for a reason to use Fold. For both answers though I don't have the combinedProbs[] function. Since it starts with a lowercase letter I am assuming it is a user defined function. Is this function available? Thank you Marco for giving me an example of how to generate random inputs. I will need them in the next chapter I am sure but the present chapter is looking for exact answers to the questions posed and so I am assuming that there is one number that will correctly answer the question. I know I could run the experiment several times and take the average, as you did above, but I would not necessarily get the same answer as what the book is looking for.

Hi,

yes there is indeed. This is the function:

combindedProbs[lis_List] := 1. - Product[1 - lis[[i]], {i, 1, Length[lis]}]

in your case that is 0.76. You can call this function like so:

combinedProbs[{0.4,0.5,0.2}]

With Mathematica you can of course also simulate that experiment. You can use RandomChoice with probabilities to simulate the "infections":

Table[Total[{RandomChoice[{0.6, 0.4} -> {0, 1}], RandomChoice[{0.5, 0.5} -> {0, 1}],  RandomChoice[{0.8, 0.2} -> {0, 1}]}], {10}]

You get a triple of three numbers for each simulation, which is made up of zeros and ones. Zero at position one means "no infection by virus one", a one at position two means "infection with virus two", etc. The 10 at the end simulates 10 computers.

Now we can run this 1000 times and check how many computers have no infection, i.e. the sum of the three entries is zero.

M = 1000; N[Length[Select[Table[Total[{RandomChoice[{0.6, 0.4} -> {0, 1}], RandomChoice[{0.5, 0.5} -> {0, 1}], RandomChoice[{0.8, 0.2} -> {0, 1}]}], {M}], # != 0 &]]/M]

This gives something close to 76%.

Cheers,

M.