Question

Two coins A and B are kept in an urn. When coin A is flipped, it comes up heads with probability 1/4, when coin B is flipped, it comes up heads with probability 3/4. A coin is randomly chosen from the urn and flipped. Given that the flip landed on heads, what is the probability that it is coin A is?\[\]
(1) $\dfrac{9}{10}$\[\]
(2) $\dfrac{1}{4}$ \[\]
(3) $\dfrac{3}{4}$\[\]
(4) $\dfrac{1}{10}$\[\]

Answer 1

Hint: The choice of a coin out of two is $P\left( A \right)=P\left( B \right)=\dfrac{1}{2}$( prior probabilities). We see that the events the flipping of coin A for the first time and showing of head with probability $\left( P\left( H|A \right)=\dfrac{1}{4} \right)$ , coin B for the first time and showing of head with probability $\left( P\left( H|B \right)=\dfrac{1}{4} \right)$ happens(posterior probabilities ) before a new event that is a coin is tossed and has come up heads. We have to find the probability that it is a coin of type A $P\left( A|H \right)$. We use the Bayes’ theorem for two events here $P\left( A|H \right)=\dfrac{P\left( A \right)P(H|A)}{P\left( A \right)P(H|A)+P\left( B \right)P(H|B)}$\[\]

Complete step by step answer:
Dependent events are events whose occurrence of one affects the occurrence of other events when they occur successively. We measure it by conditional probability. If there are two events say $A$ and $B$where $B$ has already happened and we know its probability as $P\left( B \right)$then the probability of $A$ happening conditioned by (subjected to ) $B$ has happened is defined as
\[\begin{align}
  & P\left( A|B \right)=\dfrac{P\left( A\bigcap B \right)}{P\left( B \right)} \\
 & \Rightarrow P\left( A\bigcap B \right)=P\left( A|B \right).P\left( B \right) \\
\end{align}\]
Bayes’ theorem is used when there are more than two events dependent on each other. If there are $n$ events say ${{E}_{1}},{{E}_{2}},{{E}_{3}},...,{{E}_{n}}\left( {{E}_{i}}\ne \Phi \right)$ which are mutually exclusive (do not occur at the same time) and exhaustive (sum of events sets is the sample space). Let $X$ be a new event such that $X\subset \bigcup\limits_{i=1}^{n}{{{E}_{i}}},P\left( X \right)\ne 0$, then the probability of event ${{E}_{i}}$ has happened subjected to $X$ has happened is given by

\[P\left( {{E}_{i}}|X \right)=\dfrac{P\left( {{E}_{i}} \right)P\left( X|{{E}_{i}} \right)}{\sum\limits_{i=1}^{n}{P\left( {{E}_{i}} \right)P\left( X|{{E}_{i}} \right)}}\]
Where $P\left( {{E}_{i}} \right)$ are prior probabilities, $P\left( X|{{E}_{i}} \right)$ are posterior probabilities and $P\left( X \right)$ is the probability of a new event. \[\]

In the given question there are four events. Let the selection of coin A be denoted by event $A$ and its probability $P\left( A \right)$and the selection be coin B be denoted by and its probability is $P\left( B \right)$. As there are only two cases for the selection of coins , $P\left( A \right)=P\left( B \right)=\dfrac{1}{2}$ . After the flipping of coin let the showing of heads be denoted by $P\left( H \right)$. Here there two cases head or tail, $P\left( H \right)=\dfrac{1}{2}$\[\]
We are given in the question that the showing of head conditioned by the selection of coin A $P\left( H|A \right)=\dfrac{1}{4}$ and the showing of head conditioned by the selection of coin B $P\left( H|B \right)=\dfrac{3}{4}$ .
Now we select a coin randomly. The using Bayes’ theorem we calculate the probability of selection of coin A denoting as $P\left( A|H \right)$ . Here the prior probabilities are $P\left( A \right)=P\left( B \right)=\dfrac{1}{2}$ and the posterior probabilities are $P\left( H|A \right)=\dfrac{1}{4}$ and $P\left( H|A \right)=\dfrac{1}{4}$. The new event is showing off the head. Now,

\[ P\left( A|H \right)=\dfrac{P\left( A \right)P(H|A)}{P\left( A \right)P(H|A)+P\left( B \right)P(H|B)}=\dfrac{\dfrac{1}{2}\times \dfrac{1}{4}}{\dfrac{1}{2}\times \dfrac{1}{4}+\dfrac{1}{2}\times \dfrac{3}{4}}=\dfrac{1}{4}\]

So, the correct answer is “Option 2”.

Note: We need to be careful to distinguish between prior and posterior probabilities. We also need to take care of the fact that the events must not be dependent where the probabilities of two dependent events are related as $P\left( A\bigcap B \right)$=P(A).P(B). The question can also be framed with more than two types of coins.