Lesson Probability for a computer to be damaged by viruses

Probability for a computer to be damaged by viruses

Problem

Since virus A damages the system with probability 0.4, the computer is not damaged by the virus A with probability 1-0.4.

Since virus B damages the system with probability 0.5, the computer is not damaged by the virus B with probability 1-0.5.

Since virus C damages the system with probability 0.2, the computer is not damaged by the virus C with probability 1-0.2.


Since viruses A, B, and C act independently, probability that the computer is not damaged by the viruses A, B and C is the product (1-0.4)*(1-0.5)*(1-0.2).


Hence, probability that the computer is not damaged by the viruses A, B and C is the complement


    1 - (1-0.4)*(1-0.5)*(1-0.2) = 1 - 0.6*0.5*0.8 = 1 - 0.24 = 0.76.


Answer.  The probability that the system gets damaged is 0.76.

Problem 2

Do it backwards, find the probability that none fail to work.

    P = 0.96*0.95*0.90,

    P = 0.8208.

The complement of this calculated probability is that at least one of the modules fails to work. So,

    P = 1-0.8208,

    P = 0.1792.


Answer.  the probability that at least one of these three modules fails to work properly is  0.1792.

Problem 3

(a)  At least one of these components will need repair within 1 year.

     According to the condition, the probability that

        V-component will not need a repair within 1 year is  (1-0.03),
        E-component will not need a repair within 1 year is  (1-0.007),
        A-component will not need a repair within 1 year is  (1-0.004).


     Hence, the probability that NO ONE of the three components will not need a repair is the product  (1-0.03)*(1-0.007)*(1-0.004).

     Then the probability under the question has the complementary value

         1 - (1-0.03)*(1-0.007)*(1-0.004) = 0.0406 = 4.06%.
  


(b)  Exactly one of these components will need repair within 1 year.

         (1-0.03)*(1-0.007)*0.004 + (1-0.03)*0.007*(1-0.004) + 0.03*(1-0.007)*(1-0.004) = 0.04030 = 4.030%.


     Explanation. 


         The probability that only A-component will need a repair within 1 year is  (1-0.03)*(1-0.007)*0.004.

         The probability that only E-component will need a repair within 1 year is  (1-0.03)*0.007*(1-0.004).

         The probability that only V-component will need a repair within 1 year is  0.03*(1-0.007)*(1-0.004).

         The probability that exactly one of these three independent events will happen within 1 year is the sum 
         of these particular probabilities, which coincides with my answer. 

Problem 4

The plane will crash if and only if one of the following events will happen


    - (a)  central and left engines fail

    - (b)  central and right engines fail

    - (c)  central and both left and right engines fail.



The probabilities for each of these events are


       P(a) = 0.005*0.008*(1-0.008) = 0.00003968

       P(b) = 0.005*0.008*(1-0.008) = 0.00003968

       P(c) = 0.005*0.008*0.008     = 0.00000032


The probability of the crash is the sum P = P(a) + P(b) + P(c) = 0.00007968 (approximately).    ANSWER