If the probability of clearing a virus is 80% per sweep....?

Question

and halves on each subsequent sweep, what's the probability of the virus being gone after 5 sweeps.

My tutorial sheet says P = 0.082. I say it's more like 0.91792, or 1 minus the sheet's answer.

For clarity, the wording is "What is the probability that the virus will be eradicated with 5 applications of the antiviral software", just to confirm I haven't misread.

Answer 1

Yes, as everyone has said, you are correct.

After first sweep = 80%
Second sweep = 8%
Third sweep = 2.4%
Fourth sweep = 0.96%
Fifth sweep = 0.432%

Total = 91.792%

So the probability of the virus being gone after 5 sweeps is 0.91792.

And the probability of your tutorial sheet being correct is 0.

Answer 2

I agree with you. I think your tutorial sheet is incorrect.

We know the probability of eradicating the virus in the first sweep is 0.8; in the second sweep, 0.4; then 0.2, 0.1, 0.05.

The virus will only still be there if it's NOT eradicated during ANY of the sweeps. The probability of the virus NOT being eradicated during the first sweep is 0.2; not being eradicated during the second sweep is 0.6; then 0.8, 0.9, 0.95.

So the probability the virus will still be there after five sweeps is 0.08208.

Therefore, the probability it will be gone is 0.91792.

So I'd say your answer is correct (and the sheet's is wrong).

Answer 3

You're right.
The answer in the tutorial sheet is the prob of the virus NOT being cleared in 5 sweeps. To get the probability that it IS cleared you have to minus this from 1 as you have done.