I was trying to solve a probability problem as follows:
Suppose you are flipping an unfair coin. It flips 70% heads and 30% tails, and each flip is independent from the previous.
Now you flip this coin three times. What is the probability you get 3 heads given that you flipped at least 1 head?
My first guess is one of the heads fixed, therefore it boils down to what is the probability that two coins are heads:
$0.7 * 0.7 = 0.49$
However I also solved it another way using Bayes Theorem:
Let probability of flipping 3 heads be $P(H3)$
Let the probability of at least 1 head be denoted as $P(A1)$
Using the Bayes theorem $$P(H3|A1) = P(A1|H3) * P(H3) / P(A1) = 1 * 0.7^3 / (1 - 0.3^3) = 1 * 0.343 / 0.973 = 0.353 $$
Can someone explain which way is right and which way is wrong and also the intuition behind it? I'm think the first way is wrong, but I don't see where the fallacy is.