# The probability of choosing a coin out of two different coins, one with tail on both faces

Sue has two coins. One is fair, with a head on one face and a tail on the other. The second is a trick coin and has a tail on both faces. Sue picks up one of the coins at random and flips it.

a) Find the probability that it lands heads up.

b) Given that it lands tails up , find the probability that she picked up the fair coin.

My turn:

a) We have one head out of three tails and one head, so the answer is $$\frac{1}{4}$$.

• Bayes' Theorem. Alternatively, as you noticed, if you were to have each of the different faces differently colored, each of the four faces are equally likely to have occurred. Three of the faces correspond to tails, and only one of the three tails faces came from the fair coin, so the answer is $\frac{1}{3}$. I strongly recommend doing this the pen-and-paper way using Bayes' theorem to confirm this number at least once, especially since generalizing the problem would make this second method not work, e.g. if the first coin was unbalanced. – JMoravitz Mar 12 at 18:42
• Before working on any problem like this you should have learned something about conditional probability. Have you ever heard of this? If not, you should edit the question to describe what kinds of probability you have heard about and also how it comes to be that you are looking at this problem (which is a problem that would be presented to someone studying conditional probability). If you have heard about conditional probability then it could help to edit the question to say what you know about it so people can answer in terms of things you know. – David K Mar 12 at 19:24

Another way: Among the four equally likely outcomes, three are favorable to "tails" and one of those three is favorable to "fair". Hence $$\Pr(\text{fair}\mid\text{tails}) = 1/3.$$

• I would have gone for 3 outcomes... – copper.hat Mar 12 at 19:35

Let A be the event that Tails comes up. You need to use Baye's theorem.

$$P(\text{fair} \mid A) = \frac{P(A \mid \text{fair})P(\text{fair})}{P(A)}$$ $$P(\text{fair} \mid A) = \frac{P(A \mid \text{fair})P(\text{fair})}{P(A\mid \text{fair})P(\text{fair}) + P(A\mid \text{biased})P(\text{biased})}$$

Is that clear?

(By "fair" I mean - the event that she picked up a fair coin.)

So we have, $$P(\text{fair} \mid A) = \frac{(1/2)(1/2)}{(1/2)(1/2) + (1)(1/2)} = \frac{1}{3}$$

• Please accept the answer if it is acceptable. Helps for new users like me :) – Vinayak Suresh Mar 12 at 18:47
• Excuse me , the second line is not clear . Would you illustrate it ?@Vinayak – Hussien Mohamed Mar 12 at 18:49
• @HussienMohamed The Law of Total Probability implies that $P(A) = P(A\cap B)+P(A\cap B^c) = P(B)P(A\mid B) + P(B^c)P(A\mid B^c)$ – JMoravitz Mar 12 at 18:51
• Sure, are you familiar with conditional probabilities? Do you see how I get the numerator? For the denominator, it follows from the so called Total probability law. (See en.wikipedia.org/wiki/Law_of_total_probability) for example. – Vinayak Suresh Mar 12 at 18:53
• See my edits for proper MathJax usage. – Michael Hardy Mar 12 at 19:15

The possible outcomes are $${(\text{FAIR}, H), (\text{FAIR}, T), (\text{TRICK}, T)}$$ with respective probabilities $${1 \over 4}, {1 \over 4}, {1 \over 2}$$.

So a) is $${ 1\over 4}$$ (only outcome with $$H$$) and b) $${ {1 \over 2} \over {1 \over 2}+ {1 \over 4}}$$.

To see b), note that the only outcomes of interest are $${ (\text{FAIR}, T), (\text{TRICK}, T)}$$ which have probabilities $${1 \over 4}, {1 \over 2}$$ of occurring. However, since you are given that one of these has occurred, we need to 'change' our probabilities in an appropriate way so that they sum to one. We do this by dividing by the sum of the probabilities of outcomes of interest, in this case $${1 \over 4}+ {1 \over 2}= {3 \over 4}$$. So the changed probabilities for the new experiment $${ (\text{FAIR}, T), (\text{TRICK}, T)}$$ are $${ { 1\over 4} \over {3 \over 4} }= {1\over 3}, { { 1\over 2} \over {3 \over 4} }= {2\over 3}$$, from which we can read off the probability of $$(\text{FAIR}, T)$$ as $${ 1\over 3}$$.

When there are only two hypotheses, this version of Bayes's formula can be used: \begin{align} & \frac{\Pr(\text{fair}\mid \text{tails})}{\Pr(\text{trick}\mid \text{tails})} = \frac{\Pr(\text{fair})}{\Pr(\text{trick})} \times \frac{\Pr(\text{tails}\mid \text{fair})}{\Pr(\text{tails}\mid \text{trick})} \\[12pt] = {} & \frac{1/2}{1/2} \times \frac{1/2}{1} = \frac 1 2. \end{align} So we have \begin{align} & \Pr(\text{fair}\mid\text{tails}) = \frac 1 2 \Pr(\text{trick}\mid\text{tails}) \\[8pt] & \Pr(\text{fair}\mid\text{tails}) + \Pr(\text{trick}\mid\text{tails}) = 1 \end{align} Finally we conclude: $$\Pr(\text{fair}\mid\text{tails}) = \frac 1 3.$$