I understand that this question has been asked before on this forum but, my question is not on how to do the question but on why my approach using reduced sample space fails.
Consider this question, also asked here:
When coin 1 is flipped, it lands on heads with probability .4; when coin 2 is flipped, it lands on heads with probability .7. One of these coins is randomly chosen and flipped 10 times.
(a) What is the probability that the coin lands on heads on exactly 7 of the 10 flips?
(b) Given that the first of these 10 flips lands heads, what is the conditional probability that exactly 7 of the 10 flips land on heads?
For part (b), my initial approach was just to disregard the biasness of the coins, and just calculate the probability that 6 of the remaining 9 coins are heads, so the conditional probability is just $\frac{9\choose{6}}{2^9}$ because in the reduced sample space of 9 tosses, we want 6 of the tosses to be heads, the remaining 3 to be tails. Is there a reason why does this way of thinking is wrong?
Your approach fails because you are not using the information that the first toss landed heads (except to count 1 heads). Here's why that is a problem.
Coin 2 is significantly more likely to land heads compared to coin 1. So if the first toss landed heads, it's more likely (not certain though) that coin 2 is being used. You are not using this information at all.
Instead you are using the probability of landing heads for each of the remaining tosses to be $0.5$, which is simply not true.