Probability of getting either Black or tabby When randomly selecting a kitten for adoption, there is a $23 \%$ chance of getting a black kitten, a $50 \%$ chance of getting a tabby kitten, a $7 \%$ chance of getting a calico kitten, and a $20 \%$ chance of getting at ginger kitten.
Elisa asks the manager to randomly select two kittens. What is the probability that Elisa gets a black kitten or a tabby kitten?
My try:
The probability that she gets either black or tabby is one minus probability that she gets calico and ginger kitten, so the required answer is $1-0.07 \times 0.2=0.986$
But the answer is $0.73$?
 A: Also it seems like you were trying to calculate the probability of a black and a tabby kitten but the problem asks for the probability of a black or a tabby kitten.  That could be two black, one black and one tabby, one black and one calico, one black and one ginger, one tabby and one calico, or one tabby and one ginger.  You need to calculate the probability of each of those, then add.
A: In your final part of the question, you should have added the two probabilities together, rather than multiplied them, the chances of getting either calico or ginger are $0.2+0.07=0.27$. However, it seems that there is a possibility of getting, for example, one tabby and one black kitten.
Divide the kittens into two groups, A = tabby  or black ‍⬛ and B = calico or ginger. The chance of getting a cat from group A is 73% if she just gets one kitten, but if she gets two kittens the chance of getting at least one from group A is the chance of all of the outcomes {A,B}, {B,A}, and {A, A}, which is $2\times 0.73\times0.27+0.73^2$. That is not 0.73 but 0.9271, so the suggested answer to the question is definitely wrong and makes no sense either, since even if it is exactly one black or tabby kitten, and excludes the case of two black or tabby kittens, the answer is not 0.73 but $2\times0.73\times 0.27=0.3942$.
A: The answer is wrong. Let's visualize it:

As you may see, there are 16 possibilities, and only highlighted 4 of them have no “red” in them (i.e., neither black, nor tabby in them).
The probability to be in the highlighted part, i.e.

*

*in the last 2 rows, and at the same time

*in the last 2 columns

is $\ 27\% \times 27\%,\ $ i.e., $\ 0.27 \times 0.27 = 0.0729.$
So the probability to be in the “correct” part is $\ 1 - 0.0729 = \color{red}{0.9271}.$
