What is the probability that randomly selected $2$ pets from this vet are cat or sick? Why doesnt addition rule for probability work?

Question

There are $7$ dogs and $6$ cats in a vet. $2$ of the dogs and $2$ of the cats are sick.What is the probability that randomly selected $2$ pets from this vet are cat or sick ?

My approach : Let the set of cats (sick and normal cats)called "A" and the set of sick pets (sick dogs and cats) called "B" ,then $$P(A \cup B)=P(A)+P(B)-P(A \cap B)=\frac{\binom{6}{2}}{\binom{13}{2}}+\frac{\binom{4}{2}}{\binom{13}{2}}-\frac{\binom{2}{2}}{\binom{13}{2}}=20/78$$

My friends'approach : The set of $A \cup B$ have $8$ elements in it , so the answer is $$\frac{\binom{8}{2}}{\binom{13}{2}}=28/78$$

Who is right here ? .. Can you help me ?

Addentum : By comments of @L.F. ,I saw that my calculation does not contain the probability of selecting a normal cat and a sick dog. However , i cannot understand why , i applied addition rule of probability. Moreover , related question in this page confirms me.

Answer 1

As the comments indicated, your analysis went astray.

$~\displaystyle \binom{6}{2}~$ equals the number of ways of selecting $2$ cats, out of $6$ cats.

$~\displaystyle \binom{4}{2}~$ equals the number of ways of selecting $2$ sick pets out of $4$ sick pets.

Your enumeration of $~\displaystyle \binom{6}{2} + \binom{4}{2} - \binom{2}{2}~$ represents

the union of selecting $2$ cats, and $2$ sick pets.

In other words, your enumeration represents either:

• selecting $2$ cats, that may or may not be sick
• selecting $2$ sick pets, that may or may not be cats
• or both.

However, the above options are not the only ways that you can select 2 pets, with neither pet being a non-sick dog. That is, you can select one dog that is sick, and one cat that is not sick. This specific option is not included in your enumeration of $~\displaystyle \binom{6}{2} + \binom{4}{2} - \binom{2}{2}.~$

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In case what I said doesn't clarify the situation,

let $A$ denote the set of all combinations of $2$ cats.

Let $B$ denote the set of all combinations of $2$ sick pets.

You enumerated $A \cup B$.

However, consider the case of $1$ well cat and $1$ sick dog. This case is not included in either set $A$ or set $B$.

Answer 2

Let us first go back to the very basic concept of probability by counting
$\dfrac{\text {favorable points in sample space}}{\text{total points in sample space}}$

Favorable points in sample space $= n(A\cup B) = n(A)+n(B) - n(A\cap B) = 6+4-2 = 8$
Total points in sample space $= 8+5 = 13$
Pr $=\dfrac{\binom82}{\binom{13}2} = \dfrac{28}{78}$

For the probability law to be valid, the union of the defined events must not miss any favorable sample point.

Here, as has already been pointed out, $A\cup B$ as defined by you misses out one ok cat + one sick dog, so your computation works out lower.

To correct for this, you will need to redefine/add events, but it is both safer and simpler to instead use

$Pr = \dfrac{\text {favorable points in sample space}}{\text{total points in sample space}}$