1. ## Independence Question - Can anyone check my solution please?

Your computer has been acting very strangely lately, and you suspect that it might have a virus on it. Unfortunately, all 15 of the different virus detection programs you own are outdated. You know that if your computer does have a virus, each of the programs, independently of the others, has a 0.77 chance of believing that your computer as infected, and a 0.23 chance of thinking your computer is fine. On the other hand, if your computer does not have a virus, each program has a 0.93 chance of believing that your computer is fine, and a 0.07 chance of wrongly thinking your computer is infected. Given that your computer has a 0.64 chance of being infected with some virus, and given that you will believe your virus protection programs only if 11 or more of them agree, find the probability that your detection programs will lead you to the right answer.

2. I did the following calculation for this question:

P(more than 10 program says infectious/infectious): i= from 11 to 15
Comb(15,i)*0.77^i*0.23^(15-i)
P(more than 10 program says not infectious/not infectious):
i= from 11 to 15
Comb(15,i)*0.93^i*0.07^(15-i)

so;

1365 * (0.77^11)*(0.23^4) = 0.215497369

455 * (0.77^12)*(0.23^3) = 0.240482571

105 * (0.77^13)*(0.23^2) = 0.185790883

15 * (0.77^14)*(0.23^1) = 0.0888565092

1* (0.77^15) * 1 = 0.0198317426

and

1365 * (0.93^11) * (0.07^4) = 0.0147515361

455 * (0.93^12) * (0.07^3) = 0.0653282312

105 * (0.93^13) * (0.07^2) = 0.200292049

15 * (0.93^14) * 0.07 = 0.380146135

1* (0.93^15) * 1 = 0.336700862

Then, I found out;

P(more than 10 program says infectious/infectious):
0.215497369 + 0.240482571 + 0.185790883 + 0.0888565092 + 0.0198317426 =

0.750459075

P(more than 10 program says not infectious/not infectious):
0.0147515361 + 0.0653282312 + 0.200292049 + 0.380146135 + 0.336700862 =

0.997218813

Finally, I multiplied these 2 probabilities and get the result 0.750459075 but it says "incorrect". What did I do wrong...?

3. Originally Posted by essedra
Your computer has been acting very strangely lately, and you suspect that it might have a virus on it. Unfortunately, all 15 of the different virus detection programs you own are outdated. You know that if your computer does have a virus, each of the programs, independently of the others, has a 0.77 chance of believing that your computer as infected, and a 0.23 chance of thinking your computer is fine. On the other hand, if your computer does not have a virus, each program has a 0.93 chance of believing that your computer is fine, and a 0.07 chance of wrongly thinking your computer is infected. Given that your computer has a 0.64 chance of being infected with some virus, and given that you will believe your virus protection programs only if 11 or more of them agree, find the probability that your detection programs will lead you to the right answer.
Let X be the random variable 'number of virus detection programs that say that the computer has a virus' when there is a virus.
X ~ Binomial(n = 15, p = 0.77).

Let Y be the random variable 'number of virus detection programs that say that the computer has no virus' when there is no virus.
Y ~ Binomial(n = 15, p = 0.93).

Pr(11 or more detection programs say there is a virus | there is a virus) = $\Pr( X \geq 11)$.

Pr(11 or more detection programs say there is no virus | there is no virus) = $\Pr(Y \geq 11)$.

4. Originally Posted by mr fantastic
Let X be the random variable 'number of virus detection programs that say that the computer has a virus' when there is a virus.
X ~ Binomial(n = 15, p = 0.77).

Let Y be the random variable 'number of virus detection programs that say that the computer has no virus' when there is no virus.
Y ~ Binomial(n = 15, p = 0.93).

Pr(11 or more detection programs say there is a virus | there is a virus) = $\Pr( X \geq 11)$.

Pr(11 or more detection programs say there is no virus | there is no virus) = $\Pr(Y \geq 11)$.
Yes, I did exactly the same, as you can see above but the answers were incorrect it said.

5. Originally Posted by essedra
Yes, I did exactly the same, as you can see above but the answers were incorrect it said.
Well then, it will depend on what is meant by "find the probability that your detection programs will lead you to the right answer" .... Perhaps whoever wrote the question has a different interpretation than the one in this thread.