Question

Three bags contain 2 silver, 5 copper coins and 3 silver, 4 copper coins and 5 silver, 2 copper coins respectively. A bag is chosen at random and a coin is drawn from it which happens to be silver. What is the probability that it has come from the third bag?

Answer 1

To do this question, we will make use of bayes’ theorem. For that, we will first find the probability of choosing all the bags out of the three bags. Then we will find the probability of choosing a silver coin from that particular bag. Then we will use the bayes’ theorem given as:
If ${{E}_{1}},{{E}_{2}},...,{{E}_{n}}$ are events which constitute a partition of sample space S, i.e. ${{E}_{1}},{{E}_{2}},...,{{E}_{n}}$ are pairwise disjoint and ${{E}_{1}}\cup {{E}_{2}}\cup ...\cup {{E}_{n}}=S$ and A be any event with nonzero probability, then:
$P\left( {{E}_{i}}|A \right)=\dfrac{P\left( {{E}_{i}} \right)P\left( A|{{E}_{i}} \right)}{\sum\limits_{j=1}^{n}{P\left( {{E}_{j}} \right)}P\left( A|{{E}_{j}} \right)}$
Then we will assume the following:
Event A = the event of choosing a silver coin out of the three bags
Event ${{E}_{j}}=$ the event in which the silver coin chosen is from bag-j
Event ${{E}_{i}}=$ the event in which the silver coin chosen is from bag-i
Then we will find the probability for i=3 and hence we will have our answer.

Complete step by step answer:
Here, we have been given 3 bags and we have been told that one bag is chosen at random.
Since, choosing one bag out of the three bags is unbiased, the probability of choosing any one bag out of the 3 bags is equally likely for all three bags. Thus, let us first assume the probability of choosing bag-1 out of the three bags to be ‘x’. Since, all the bags have the same probability to be chosen, the probability of choosing bag-2 and bag-3 will also be equal to x.
Now, these are elementary events and we know that the sum of probabilities of all the elementary events is 1.
Thus, we can say that:
$x+x+x=1$
From this, we get the value of x as:
$\begin{aligned} & 3x=1 \\\ & \Rightarrow x=\dfrac{1}{3} \\\ \end{aligned}$
Now, if we consider bag-1, the probability of getting a silver coin out of it will be given as follows.
The bag-1 contains 2 silver and 5 copper coins. Thus, the total coins in bag-1 are $2+5=7$ out of which 2 are silver.
We know that the probability of any event is given as $\text{Probability=}\dfrac{\text{favourable outcomes}}{\text{total outcomes}}$. Hence, the probability of getting a silver coin out of bag-1 is given as:
$\begin{aligned} & \dfrac{\text{no}\text{. of silver coins in bag-1}}{\text{total no}\text{. of coins in bag-1}} \\\ & \Rightarrow \dfrac{2}{7} \\\ \end{aligned}$
Similarly, if we consider bag-2, it has 3 silver and 4 copper coins. Thus, the probability of getting a silver coin out of bag-2 is given as:
$\begin{aligned} & \dfrac{3}{3+4} \\\ & \Rightarrow \dfrac{3}{7} \\\ \end{aligned}$
Now similarly, if we consider bag-3 it has 5 silver and 2 copper coins. Thus, the probability of getting a silver coin out of bag-3 is given as:
$\begin{aligned} & \dfrac{5}{5+2} \\\ & \Rightarrow \dfrac{5}{7} \\\ \end{aligned}$
Now, we know that according to Bayes’ theorem, if ${{E}_{1}},{{E}_{2}},...,{{E}_{n}}$ are events which constitute a partition of sample space S, i.e. ${{E}_{1}},{{E}_{2}},...,{{E}_{n}}$ are pairwise disjoint and ${{E}_{1}}\cup {{E}_{2}}\cup ...\cup {{E}_{n}}=S$ and A be any event with nonzero probability, then:
$P\left( {{E}_{i}}|A \right)=\dfrac{P\left( {{E}_{i}} \right)P\left( A|{{E}_{i}} \right)}{\sum\limits_{j=1}^{n}{P\left( {{E}_{j}} \right)}P\left( A|{{E}_{j}} \right)}$
Here, let us assume the following:
Event A = the event of choosing a silver coin out of the three bags
Event ${{E}_{j}}=$ the event in which the silver coin chosen is from bag-j
Event ${{E}_{i}}=$ the event in which the silver coin chosen is from bag-i
Now, for different values of j, we will get:
$\begin{aligned} & P\left( {{E}_{1}} \right)=\dfrac{1}{3} \\\ & P\left( {{E}_{2}} \right)=\dfrac{1}{3} \\\ & P\left( {{E}_{3}} \right)=\dfrac{1}{3} \\\ \end{aligned}$
Similarly, as established above, we get for different values of j:
$\begin{aligned} & P\left( A|{{E}_{1}} \right)=\dfrac{2}{7} \\\ & P\left( A|{{E}_{2}} \right)=\dfrac{3}{7} \\\ & P\left( A|{{E}_{3}} \right)=\dfrac{5}{7} \\\ \end{aligned}$
Now, if we take i=3, we will get:
$\begin{aligned} & P\left( {{E}_{3}}|A \right)=\dfrac{P\left( {{E}_{3}} \right)P\left( A|{{E}_{3}} \right)}{\sum\limits_{j=1}^{3}{P\left( {{E}_{j}} \right)}P\left( A|{{E}_{j}} \right)} \\\ & \Rightarrow P\left( {{E}_{3}}|A \right)=\dfrac{\dfrac{1}{3}.\dfrac{5}{7}}{\dfrac{1}{3}.\dfrac{2}{7}+\dfrac{1}{3}.\dfrac{3}{7}+\dfrac{1}{3}.\dfrac{5}{7}} \\\ \end{aligned}$
Now, solving this we get:
$\begin{aligned} & P\left( {{E}_{3}}|A \right)=\dfrac{\dfrac{1}{3}.\dfrac{5}{7}}{\dfrac{1}{3}.\dfrac{2}{7}+\dfrac{1}{3}.\dfrac{3}{7}+\dfrac{1}{3}.\dfrac{5}{7}} \\\ & \Rightarrow P\left( {{E}_{3}}|A \right)=\dfrac{\dfrac{5}{21}}{\dfrac{2}{21}+\dfrac{3}{21}+\dfrac{5}{21}} \\\ & \Rightarrow P\left( {{E}_{3}}|A \right)=\dfrac{\dfrac{5}{21}}{\dfrac{10}{21}} \\\ & \therefore P\left( {{E}_{3}}|A \right)=\dfrac{1}{2} \\\ \end{aligned}$

Hence, the required probability is $\dfrac{1}{2}$.

Note: There are different kinds of events while doing probability we should know the meaning of. They are given as follows:
1. Independent events: An event A is said to be independent of another event B if the probability of occurrence of one of them is not affected by the occurrence of the other. This is characterized as:
$P\left( A|B \right)=P\left( A \right)$
2. Mutually exclusive events: Two events are said to be mutually exclusive if they cannot occur at the same time or simultaneously. They are also known as disjoint events. They are characterized as:
$P\left( A\cup B \right)=P\left( A \right)+P\left( B \right)$
3. Exhaustive events: When a sample space is distributed down into some mutually exclusive events such that their union forms the sample space itself, then such events are called exhaustive events.
4. Sure events: When an event is sure to happen it is known as a sure event. Its probability is always 1.
5. Impossible events: When an event is impossible to happen, it is known as an impossible event. Its probability is always 0.