Question

Question 10 (1 point) Let V be the event that a computer contains a virus, and let W be the event that a computer contains a worm. Suppose PV)-0.10, PW) 0.05, and PV and W)-0.06. 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 Your Answer: Answer Question 11 (1 point) At a local college, the probability that there are no parking places available for an 11:00 am class is 0.53 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. Your Answer:

Answer 1

Solution :-

Question-10 :-

Given data :

P(V) = 0.10

P(W) = 0.05

P(V and W) = $P(V \cap W)$ = 1 -0.06

Here we need to calculate the probability that the computer contains neither a virus or a worm.

i.e, P(V' U W')

P(V' U W') = 1 - P(V U W)

$P(V' U W') = 1 - \left \{ P(V) + P(W) - P(V \cap W) \right \}$

P(V' U W') = 1 - { 0.10 + 0.05 - 0.06 }

P(V' U W') = 1 - 0.09

P(V' U W') = 0.91

The probability that the computer contains neither a virus or a worm is P(V' U W') = 0.91

Question-11 :-

Given data :

There are no parking places available for an 11:00 am class is 0.81. i.e, P( x=0 ) =0.81

Here we need to calculate the probability that there is at least one parking place available for the 11:00 class.

P( at least one ) = 1 - P( none )

$P(x\geq 1)=1-P(none)$

$P(x\geq 1)=1-P(x<1)$

$P(x\geq 1)=1-P(x=0)$

$P(x\geq 1)=1-0.81$

$P(x\geq 1)=0.19$

The probability that there is at least one parking place available for the 11:00 class is 0.19