A bag contains 1000 coins. 999 are normal, fair coins; 1 coin has 2 heads.
You pick a coin at random from the bag. You flip it 10 times. You get 10 heads.
What is the probability that the coin is the 2-headed coin?
A bag contains 1000 coins. 999 are normal, fair coins; 1 coin has 2 heads.
You pick a coin at random from the bag. You flip it 10 times. You get 10 heads.
What is the probability that the coin is the 2-headed coin?
On
You are right about using Bayes' Theorem. Let's systematically assemble the ingredients you need to apply that theorem.
Let $T$ be the event that you pick a two-headed coin; $P(T) = 1/1000.$ Let $F$ be the event that you pick a fair coin; $P(F) = 999/1000.$ Let $E$ be the event of getting ten heads in a row; $P(E|T) = 1$ and $P(E|F) = (1/2)^{10}.$
Bayes' Theorem: $$P(T|E) = \frac{P(T\cap E)}{P(E)} = \frac{P(T\cap E)}{P(T\cap E)+P(F\cap E)}.$$
Now, $P(T \cap E) = P(T)P(E|T) = ?$ And $P(F \cap E) = P(F)P(E|F) = ?$
I'll let you put the pieces together.
Option 1: you picked a fair coin and it came up heads 10 times in a row.
$P = 0.999\cdot \frac {1}{2^{10}}$
Or you picked the fake coin
$P = 0.001$
$\frac {0.001}{0.999\cdot \frac {1}{2^{10}} + 0.001} = \frac {1.024}{0.999 + 1.024} = \frac {1.024}{2.023}$