Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a probability question that reads:


A box has three coins. One has two heads, another two tails and the last is a fair coin. A coin is chosen at random, and comes up head. What is the probability that the coin chosen is a two headed coin.

My attempt:

P(two heads coin| given head) = P(two heads coin * given head)/P(given head)
= 1/3/2/3 = 1/2

Not sure whether this is correct?

share|improve this question

4 Answers 4

up vote 1 down vote accepted

Hint: No, it isn't. Let $H \equiv$ obtaining heads, $A \equiv$ picking a two-headed coin, $B \equiv$ picking a two-tailed coin, and $C \equiv$ picking a fair coin. Then observe that the probability of obtaining a head is: $$ \begin{align*} P(H) &= P(A)\cdot P(H \mid A) + P(B)\cdot P(H \mid B) + P(C)\cdot P(H \mid C) \\ &= \dfrac{1}{3} \cdot \dfrac{2}{2} + \dfrac{1}{3}\cdot \dfrac{0}{2} + \dfrac{1}{3}\cdot \dfrac{1}{2} \end{align*} $$

share|improve this answer
so my numerator is correct and my denominator is wrong. am i right? –  lakesh Aug 9 '13 at 10:34
No, the numerator is wrong as well. Using my notation, you have calculated $P(A)$ instead of $P(A \text{ and } H)$. –  Adriano Aug 9 '13 at 10:35
I don't think you're answering the question being asked. You correctly find $P(H)=1/2$, but the question was about $P(A\mid H)$ which your hint seems to say nothing about at all. –  Henning Makholm Aug 9 '13 at 12:17
@HenningMakholm I was hoping that the OP knew the formula: $$ P(A \mid H) = \dfrac{P(A) \cdot P(H \mid A)}{P(H)} $$ so I helped him out with the harder part, which was calculating the denominator. –  Adriano Aug 9 '13 at 16:32

For such a small number of options its easy to count them

The possible outcomes are:

heads or heads using the double head coin
tails or tails using the double tail coin
heads or tails using the fair coin

All these outcomes are equally likely. How many of these are heads and of those how many use the double headed coin?

$$Answer = \frac{2}{3}$$

share|improve this answer

The answer is 2/3.

P(A∣H)=P(A) * P(H∣A) / P(H) --> Bayes Theorem

P(A) = 1/3, because 1 out of 3 coins are 2-headed

P(H|A) = 1, because given you have the 2-headed coin, then it is for sure (100%) you will get a head

P(H) = 1/2, because there are 3 heads and 3 tails in total, so it's 50% chance to get heads

Therefore, P(A|H) = (1/3) * 1 / (1/2) = 2/3

share|improve this answer

1/2 is the only answer!! 50 50 chance, that's it period!!! The double tails is out for sure, so discard it. What is left besides the normal coin and 2 headed coin? Nothing! So, it's between two coins only. You can't rip the heads or tails from the coins and shake them up in the box. That would give you 3 heads and only 1 tail. That would be a 3 out of 4 chance! BUT THAT IS OUT! These are coins with no magical way of knowing if the coin chosen has a tail on the hidden side. So it's a 50 50 or 1/2 chance. Both the normal and 2 tail coins are out because it HAS to be a 2 headed coin chosen! So if you randomly draw out a coin with heads. You have a 50/50 or 1/2 chance. THAT'S IT, 1/2. No formula needed. I'm not great at math myself. This is common comprehension! Thanks Del

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.