We flip $10$ coins that are all fair. After flipping all $10$ the first time, if the coin landed on tails we leave it alone. We then flip the remaining coins and try to get them all to be tails.
A) What is the probability that after the first round of flips, there are only $5$ coins on tails, conditioned on the event that after the second round all coins show tails?
B) What is the probability that after the second round of flips, all coins show tails?
For A, I'm thinking of using Bayes formula but I'm struggling on how to apply it.
For B, I know that the probability that after the first round, the probability that we have one tails is $ (\frac{1}{2})^9 $ so we would need $9$ tails in the second round and that would be $ (\frac{1}{2})^9 $ so this scenario would have probability $ \left(\frac{1}{2}\right)^{18} $. Do I just need to do this for each scenario and add all the probabilities? Doing so gives me a probability of
Best Answer
For B, the probability an individual coin is tails after one or two rounds is $\frac12+\frac12\times \frac12=\frac34$ so the probability they all are is $\left(\frac34\right)^{10}$. If you think the question does not include them all being tails after one round, then subtract the probability of that, i.e. $\left(\frac12\right)^{10}$.
For A, consider the probability there are five tails and five heads after one round and ten tails after two. This is ${10 \choose 5}\frac1{2^{10}} \times{5 \choose 5}\frac1{2^{5}} = {10 \choose 5}\frac1{2^{15}}$. This is the joint probability; for the conditional probability, divide by the answer to B.