Question 1051994: Three computer viruses arrived as an e-mail attachment. Virus A damages the system
with probability 0.4. Independently of it, virus B damages the system with probability 0.5.
Independently of A and B, virus C damages the system with probability 0.2. What is the
probability that the system gets damaged?
Found 3 solutions by stanbon, ikleyn, MathTherapy:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Three computer viruses arrived as an e-mail attachment. Virus A damages the systemwith probability 0.4. Independently of it, virus B damages the system with probability 0.5.
Independently of A and B, virus C damages the system with probability 0.2. What is the
probability that the system gets damaged?
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Ans:: 0.4*0.5*0.2 = 0.04
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Cheers,
Stan H.
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Answer by ikleyn(52779)  (Show Source):
You can put this solution on YOUR website! .
Three computer viruses arrived as an e-mail attachment. Virus A damages the system
with probability 0.4. Independently of it, virus B damages the system with probability 0.5.
Independently of A and B, virus C damages the system with probability 0.2. What is the
probability that the system gets damaged?
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1 - (1-0.4)*(1-0.5)*(1-0.2) = 1 - 0.6*0.5*0.8 = 1 - 0.24 = 0.76.
Answer by MathTherapy(10552)  (Show Source):
You can put this solution on YOUR website! Three computer viruses arrived as an e-mail attachment. Virus A damages the system
with probability 0.4. Independently of it, virus B damages the system with probability 0.5.
Independently of A and B, virus C damages the system with probability 0.2. What is the
probability that the system gets damaged?
With all 3 being INDEPENDENT of each other, we get: P(Being damaged) = 
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