I got this puzzle that I need help on. I hope its not too easy because I really don't understand it.
A bag contains 100 coins. 99 of them are fair and will give heads or tails with an equal probability. 1 biased coin will always yield heads. I pull a coin out randomly from the bag and toss it up 3 times. In each of the 3 times it lands on heads.
What is the probability that I pulled out the biased coin?
Let $A$ be the situation of picking the loaded coin and let $B$ the situation of getting heads thrice.
By Bayes' Theorem $$P(A|B) = \frac{P(B|A)P(A)}{P(B)} = \frac{1\cdot\frac{1}{100}}{1\cdot\frac{1}{100}+\frac{1}{8}\cdot\frac{99}{100}} = \frac{8}{107}.$$
This is a standard conditional probability problem. We can solve it using the formal machinery of conditional probability. Instead, we give an argument which is informal, but we hope persuasive.
Imagine repeating the experiment $100000$ times. Then "about" $1000$ times, we will pull out the funny coin, and of course get $3$ heads. About $99000$ times, we will pull out an ordinary coin. About $\frac{1}{8}$ of these times we will get $3$ heads, so it will happen about $12375$ times.
So we get three heads a total of about $13375$ times. In about $1000$ cases, it was the biased coin. So the required probability is $\dfrac{1000}{13375}$.
