There are two coins each having probability of heads as $0.9$ and $0.1$ respectively. And probability of choosing a coin for experiment is $0.5$ ( a coin once picked at beginning is used for every toss ).
P(Heads in $11$th toss) = $0.5$
P(Heads in $11$th toss | Heads in first $10$ tosses ) = $0.9$ (approx) as we are almost certain that $1$st coin is used therefore value obtained is $0.9$.
I tried to prove it mathematically
P(Heads in $11$th toss | Heads in first $10$ tosses ) = $$\frac{P(Heads\ in \ 11th \ toss \ ∩ Heads \ in \ first \ 10 \ tosses )}{ P(Heads \ in \ first \ 10 \ tosses)}$$
P(Heads in $11$th toss ∩ Heads in first $10$ tosses ) = $0.5 * ( (0.9)^{11}) + 0.5 * ( (0.1)^{11}) $
P(Heads in first $10$ tosses) = $0.5 * ( (0.9)^{10}) + 0.5 * ( (0.1)^{10})$
P(Heads in $11$th toss | Heads in first $10$ tosses ) = $( 0.9^{11} + 0.1^{11})/(0.9^{10}+0.1^{10}) = 0.8999$
But there is another way to write P(Heads in $11$th toss ∩ Heads in first $10$ tosses ) as P(Heads in first $10$ tosses|Heads in $11$th toss ) * P(Heads in $11$th toss)
I am not able to figure out what would be the value of P(Heads in first $10$ tosses|Heads on $11$th toss ) and how to find it?
If there were no biases in the coins would that make the above events independent?
If I understand your question correctly, I think that you should also consider the chance that a certain coin was chosen if you got 10 heads in a row.
P(0.9 coin | first 10 heads) = P(0.9 coin and first 10 heads)/P(first 10 heads) = $\frac{0.5*0.9^{10}}{0.5*0.9^{10} + 0.5*0.1^{10}} = 0.9999999997132$ (according to Wolfram Alpha). The chance of 11th coin heads is $0.9999999997132*0.9 + (1 - 0.9999999997132)*0.1$ which is super close to 0.9. Hope this helps!