1. ## another urn probability

An urn contains 2 fair coins and one 2-headed coin. A coin is drawn randomly and tossed twice. Given that both tosses resulted in H, find the conditional probability that the 2-headed coin was chosen (as a fraction in lowest terms). HINT: use Baye's Formula

Next: Repeat problem, assuming the urn contains 8 fair coins and one 2-headed coin.

2. Originally Posted by mhitch03
An urn contains 2 fair coins and one 2-headed coin. A coin is drawn randomly and tossed twice. Given that both tosses resulted in H, find the conditional probability that the 2-headed coin was chosen (as a fraction in lowest terms). HINT: use Baye's Formula

Next: Repeat problem, assuming the urn contains 8 fair coins and one 2-headed coin.
Define the event A=Draw the 2-headed coin. B=get 2 heads in 2 tosses.

Bayes formula tells us that

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

Now we can figure out all of the probabilties on the right hand side

$P(B|A)P(A)=1\cdot \frac{1}{3}$

$P(B|A^c)P(A^c)=\frac{1}{4}\cdot \frac{2}{3}$

From here just plug into the formula