I am a bit confused with the following problem of probability: "We select 3 phones of a sample in which we know that the 60% of them contain one mistake and the 10% contain more than one mistake. We define the variables $X$="number of phones with some mistake" and $Y$="number of products with less than two mistakes". Obtain the joint distribution table".
I have conluded from the information that $X$ is given by a binomial distribution of probability $0.7$ and $Y$ also by a binomial but with probability $0.9$. But I think I am not considering cases that are trivial and have probability zero. Maybe because the problem seems a bit ambiguous to me. Any help? Thanks.