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

Question

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 $\frac{1}{4}$.

b) I do not understand how can i start with this?!

Answer 1

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

Answer 2

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

Answer 3

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 ${ 1\over 3}$.

Answer 4

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: $$ \Pr(\text{fair}\mid\text{tails}) = \frac 1 3. $$