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

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

1 Answer 1


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

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

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

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