Let's say I have a bag with 3 coins. 1 coin is a fair coin and 2 coins have a bias (70% chance of heads). I choose a coin at random and flip it 6 times. I observe exactly 3 heads. What is the probability that the coin I chose is the fair coin?
My initial intuition is that my observation doesn't mean anything, and there is a 1/3 chance that it is the fair coin because I chose it at random from the bag with 3 coins.
Here's a similar problem you might use as a guide. Suppose the coin you chose got HHH in three tosses. Coin A is fair; Coins B and C both have P(H) = 0.7.
$$P(A\,|\,\text{HHH}) = \frac{P(A \cap \text{HHH})}{P(\text{HHH})} = \frac{\frac 1 3(0.5)^3}{\frac 1 3(0.5)^3 + \frac 2 3(0.7)^3} \approx .12 \ne \frac 1 3 $$
In Bayesian language, your prior probability of getting Coin A is $1/3.$ Your data is 'three H's in three tosses'. Your posterior probability for Coin A is around 0.12. The prior probability and the data are combined to give the posterior probability.