The Jones family household includes Mr. and Mrs. Jones, four children, two cats,

Question

The Jones family household includes Mr. and Mrs. Jones, four children, two cats, and three dogs. Every six hours there is a Jones family stroll. The rules for a Jones family stroll are: Each scroll must include at least one parent and at least one pet. There can never be a dog and a cat on the same stroll unless both the parents go. All acceptable stroll groupings are equally likely. Given that exactly one parent went on the 6 P.M. stroll, what is the probability that Rover, the oldest dog, also went?

Explanation / Answer

Let us denote P the event that exactly one parent went

and R be the event that Rover, the oldest dog, went.

Pr(R|P) = Number of ways exactly one parent and rover can go/ Number of ways exactly one parent goes

, we can choose the parent in two ways and the remaining three positons can be filled by any of the remaining two dogs or the four kids.

so we have =2*6C3=40

Based on the rules, we have 2 possibilities regarding those four members:

(1) At least one dog and some children, no cat;

(2) At least one cat and some children, no dog.

7C4-1 =34

WE HAVE SIMILARLY FOR SECOND CASE  the pet is cat, the number of ways we can fill the remaining four positions is

6C4-1=14

the total number of acceptable groupings with exactly one parent is: 2 · (34 + 14) = 96

Hence, Pr(R|P) = 40/ 96 = 0.4167

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