# Conditional probability with 4 coins

The question: A box contains four coins, two of which are fair, one double-headed (i.e., heads on both sides), and the third is biased in such a way that it comes up heads with probability 1/4. A coin is drawn at random from the box and flipped twice. If both flips result in heads, what is the probability that the coin drawn was double-headed?

From what I understand, there are 2 fair coins with a 1/2 chance to get heads, 1 coin that has 100% chance of getting heads, and one coin with a 1/4 chance to get heads.

Does this mean, for example, that the chance of getting heads in both flips from a fair coin is 1/8? Since there's a 2/4 (1/2) chance to pick one? I just cannot understand the formulation of the question but I assume we have to use Baye's theorem here? I would appreciate any help here! Thanks

• Go ahead, try out your idea of Bayes' Theorem, but be careful re probabilities. Jul 8 at 18:04

Probability of choosing double-hedaded is $$\frac{1}{4}\times 1$$

Probability of choosing normal one is $$\frac{2}{4}\times \frac{1}{2}$$

Probability of choosing biased, heads with probability 1/4 is $$\frac{1}{4}\times \frac{1}{4}$$

If the result is H

$$P=\frac{\frac{1}{4}\times 1}{\frac{1}{4}\times 1 + \frac{2}{4}\times \frac{1}{2} + \frac{1}{4}\times \frac{1}{4}}=\frac{4}{9}$$

If the result is HH

$$P=\frac{\frac{1}{4}\times 1 \times 1}{\frac{1}{4}\times 1 \times 1 + \frac{2}{4}\times \frac{1}{2}\times \frac{1}{2} + \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}}=\frac{16}{25}$$

• Is this answer taking into account that the chosen coin was flipped twice (resulting in HH) rather than just flipped once (resulting in H)? Jul 9 at 2:15
• @paw88789 - I solved for H then added result for HH. Thanks for informing. Jul 9 at 7:05