Unfair Coin Toss probability, Independence

Question

We have two coins. The first one is a fair coin and has $50\%$ chance of landing on head and $50\%$ of landing on tails. for the other one it is $60\%$ for head and $40\%$ for tails. We choose one of them randomly (with equal chance) and toss it three times.The result is Tails-Head-Head. If we toss it again for the 4th time, what is the probability of the coin landing on tails.

At first, it seems like a bayes theorem question, but after some thought I think because the result of every toss is independent of others, the answer is just:

$$\frac{1}{2} . \frac{1}{2} + \frac{1}{2} . \frac{4}{10} = \frac{9}{20} $$

The first $\frac{1}{2}$ is because we choose the coins randomly.

I want to know whether my argument is correct for this probability question or not.

Answer 1

Call the fair coin $A$ and the biased coin $B$.

$$P(THT|A) = (.5)^3$$

$$P(THT|B) = (.4)(.6)(.4)$$

$$P(A| THT) = \frac{P(THT|A) P(A)}{P(THT|A) + P(THT|B)}$$

$$P(B| THT) = \frac{P(THT|B) P(B)}{P(THT|A) + P(THT|B)}$$

(Note that $P(A|THT) + P(B|THT) = 1$, as it must)

Once you have the probability it is $A$, you can easily calculate the probability the 4th coin is $H$ ($0.5$). And likewise if the coin is $B$ ($0.6$).

You can plug in the numbers.

Answer 2

The answer, giving two terms corresponding to the actual coin being fair, then the actual coin being biased, will be $$P(T) = \frac{\left(\frac12\right)^3}{\left(\frac12\right)^3+\frac25\left(\frac35\right)^2} \left(\frac12\right) + \frac{\frac25\left(\frac35\right)^2}{\left(\frac12\right)^3+\frac25\left(\frac35\right)^2}\left(\frac25\right)\\ $$

Doing the arithmetic, $$ P(T) = \frac{1201}{2690} \approx 44.65\% $$ The probability that your coin is biased is about 53.5%, which is why the tails on the next throw is less likely than you might think.