Complete Question
The complete question is shown on the first uploaded image
Answer:
The correct option is the second option
Step-by-step explanation:
Let D be the event that a resident has a dog
Let C be the event that a resident has a cat
Generally according to Bayes rule the probability that a resident has no dog give that he/she has one cat is mathematically represented as
[tex]P(D' | C) = \frac{P ( D' \ n \ C)}{ P( C )}[/tex]
Here [tex]P(D' \ n \ C )[/tex] is the probability that a resident has one cat and no dog and from the table the value is
[tex]P(D' \ n \ C ) = 0.2[/tex]
and [tex]P(C)[/tex] is the probability of having one cat which is mathematically evaluated as
[tex]P(C) = P(C \ n \ D') + P(C \ n \ D) + P(C \ n \ 2D) + P(C \ n \ 3D)[/tex]
From the table
P(C \ n \ D') = 0.2
P(C \ n \ D) = 0.05
P(C \ n \ 2D) = 0.04
P(C \ n \ 3D) = 0.06
[tex]P(C) = 0.2 + 0.05 + 0.04 + 0.06[/tex]
=> [tex]P(C) = 0.35[/tex]
So
[tex]P(D' | C) = \frac{0.2}{ 0.35}[/tex]
=> [tex]P(D' | C) = \frac{4}{7}[/tex]
