Let V be the event that a computer contains a virus, and let W be the event that

Question

Let V be the event that a computer contains a virus, and let W be the event that a computer contains a worm. Suppose p(V) = 0.14 p(W) = 0.12 and p(V and W) = 0.08 What is the probability that the computer contains neither a virus nor a worm? Write only a number a your answer. Round to two decimal places (for example: 0.73). Do not write as a percentage. At a local college, the probability that there are no parking places available for an 11: 00 am class is 0.62. What is the probability that there is at least one parking place available for the 11: 00 am class? Write only a number a your answer. Round to two decimal places (for example: 0.73). Do not write as a percentage.

Explanation / Answer

Question 10:

By the addition rule, the probability that computer contains a virus or a worm is

P(V or W) = P(V) + P(W) - P(V and W) = 0.14+0.12- 0.08 = 0.18

The probability that computer neither contains virus nor worm is

P(not V and not W) = 1 -P(V or W) = 1 - 0.18 = 0.82

Question 11:

Given that :

P(no parking) = 0.62

So the probability that at least one parking is

P(at least one parking) = 1 - P(no parking) = 1 - 0.62 = 0.38

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