Here is the question :If we are told that among the 10 coins, at least three have landed on heads. What is the probability that there are at most five heads?
In other words,
P( at most five head | at least three have landed).
The solution says, $P( X\leq 5 | X\geq 3) = P( 3\leq X\leq 5)/P( X\geq3) \approx 0.601$
Which completely makes sense. But initially, I got this question wrong because I assumed that
$P(\text{at most five head} | \text{at least three have landed})$ is equivalent to $P(\text{at most 2 heads out of 7 toss}) = \binom{7}{2}(\frac{1}{2})^2(\frac{1}{2})^5\approx 0.1641$
Why can't I make this assumption? "Given that three have already landed head," don't I only care about $2$ more heads out of $7$ tosses? and aren't tosses all independent events?
I understand the correct method but I cannot resolve as to why my original solution is incorrect.