# Math Help - probability

1. ## probability

two computers are connected to the wireless network that is password protected. If the code is removed temporarily, the virus can attack the first computer with probability 0.5, a second computer with probability 0.7, and both simultaneously with probability 0.4

2. ## Re: probability

Originally Posted by krist
two computers are connected to the wireless network that is password protected. If the code is removed temporarily, the virus can attack the first computer with probability 0.5, a second computer with probability 0.7, and both simultaneously with probability 0.4
What is the question?

3. ## Re: probability

1.First computer is infected with a virus. What is the probability that the latter is infected with a virus?

2. if it is known that the first computer, was not attacked by the virus, which is the probability that the second computer was attacked?

4. ## Re: probability

Originally Posted by krist
1.First computer is infected with a virus. What is the probability that the latter is infected with a virus?

2. if it is known that the first computer, was not attacked by the virus, which is the probability that the second computer was attacked?
You should understand that this is not a homework service nor is it a tutorial service. Please either post some of your own work on these problems or explain what you do not understand about the question

5. ## Re: probability

Hello, krist!

Two computers are connected to the wireless network that is password protected.
If the code is removed temporarily,
. . the virus can attack the 1st computer with probability 0.5,
. . a 2nd computer with probability 0.7,
. . and both simultaneously with probability 0.4

Sure, I'd be delighted to help . . . What is the question?

. . $\begin{array}{ccccccc}P(A) &=& P(\text{1st attacked}) &=& 0.5 \\ P(B) &=& P(\text{2nd attacked}) &=& 0.7 \\P(A \cap B) &=& P(\text{both attacked}) &=& 0.4 \end{array}$

Probability that at least one computer is attacked.

. $P(A \cup B) \;=\;P(A) + P(B) - P(A\cap B) \;=\;0.5 + 0.7 - 0.4 \;=\;0.8$

Probabaility that neither computer is attacked.

. $P(\sim\!A\:\wedge \sim\!B) \;=\;1 - P(A\cup B) \;=\;1 - 0.8 \;=\;0.2$