An apartment complex allows residents to have up to 3 dogs and up to 2 cats. The joint probability model for the number of dogs (X) and number of cats (Y) that a resident has, is given below. Given that a resident has 1 cat, what is the probability that the resident does not have any dog

Complete Question

The complete question is shown on the first uploaded image

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

Here is the probability that a resident has one cat and no dog and from the table the value is

and   is the probability of having one cat which is mathematically evaluated as

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

=>  

So

=>  

x > 1.3

step by step:

8.5−0.9x> 9.8−1.9x

step 1: simplify both sides of the inequality.

−0.9x+8.5> −1.9x+9.8

step 2: add 1.9x to both sides.

−0.9x+8.5+1.9x> −1.9x+9.8+1.9x

1x+8.5> 9.8

step 3: subtract 8.5 from both sides.

1x+8.5−8.5> 9.8−8.5

1x> 1.3

step 4: divide both sides by 1.

1x/1 > 1.3/1

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step-by-step explanation: