This is a typical Bayes' Theorem problem. Recall that Bayes' Theorem is $$ P(A|B) = \frac{P(A)}{P(B)} P(B|A). $$ Let us denote the event of selecting the double-header as $\text{dh}$, and the event of throwing 10 $H$s as $10 H\text{s}$. We want to find the probability that the coin we selected was the double-header given that we threw 10 heads $$ P(\text{dh} | 10 H\text{s}) = \frac{P(\text{dh})}{P(10H\text{s})} P(10 H\text{s} | \text{dh}). $$ Clearly, the probability of throwing 10 heads given that we selected the double-header is $$ P(10 H\text{s} | \text{dh}) = 1 $$ and we infer from the question that the probability of selecting the double-header is $$ P(\text{dh}) = \frac{1}{1000}. $$ We use the Law of Total Probability to calculate the probability of throwing 10 heads, $$ P(10H\text{s}) = P(\text{dh})P(10 H\text{s} | \text{dh}) + P(\text{sh})P(10 H\text{s} | \text{sh}) $$ where $$ P(\text{sh}) = \frac{999}{1000} $$ is the probability of selecting a single-headed coin from the jar. The probability of flipping 10 heads with the single header is $$ P(10 H\text{s} | \text{sh}) = \frac{1}{2^{10}}. $$ Evaluating gives $$ P(10H\text{s}) = \frac{1}{1000} + \frac{1}{2^{10}} \frac{999}{1000} = \frac{1.976}{1000} $$ and plugging our results back into Bayes' Theorem gives $$ P(\text{dh} | 10 H\text{s}) = \frac{1}{1.976} \approx 50.6\%. $$