Question

Box-I contains 2 gold coins, while another Box-II contains 1 gold and 1 silver coin. A person chooses a box at random and takes out a coin. If the coin is of gold, what is the probability that the other coin in the box is also of gold?

Answer 1

Now, we will first analyze the situation and take two events that is ${{E}_{1}}$ and ${{E}_{2}}$ which will be the events when boxes are chosen and then we will find the probability of events to happen by using the formula: $\text{Probability of an event to happen}=\dfrac{\text{Number of favourable outcome}}{\text{Total number of outcomes}}$ , and then we will use Bayes’ theorem $P\left( {{E}_{i}}|A \right)=\dfrac{P\left( {{E}_{i}} \right)P\left( A|{{E}_{i}} \right)}{\sum\limits_{k=1}^{n}{P\left( {{E}_{k}} \right)P\left( A|{{E}_{k}} \right)}}$ , where $k=1,2,3.....n$ . Here, we have A as the event of drawing the gold coins from the boxes, $P\left( A|{{E}_{1}} \right)$ which is the probability of drawing a gold coin from Box-I and $P\left( A|{{E}_{2}} \right)$ which is the probability of drawing a gold coin from Box-II. We will find each of the probabilities using the formula for probability defined above and then substitute them in the Bayes’ theorem to get the answer.

Complete step by step answer:
Now, it is given that the Box-I contains 2 gold coins and Box-II contains 1 gold and 1 silver coin.
Now, let ${{E}_{1}}$ and ${{E}_{2}}$ be the events that the boxes I and II are chosen, respectively.
Now, we know that the probability of an event to happen is:
$\text{Probability of an event to happen}=\dfrac{\text{Number of favourable outcome}}{\text{Total number of outcomes}}$
Therefore, $P\left( {{E}_{1}} \right)=P\left( {{E}_{2}} \right)=\dfrac{1}{2}\text{ }.......\left( 1 \right)$
Now, let A be the event that the coin drawn is of gold. Now we will find the probability $P\left( A|{{E}_{1}} \right)$ which is the probability of drawing a gold coin from Box-I. Now it is given that the Box-I contains 2 gold coins. Therefore, $\Rightarrow P\left( A|{{E}_{1}} \right)=\dfrac{2}{2}=1\text{ }.......\left( 2 \right)$
Now, we will find the probability $P\left( A|{{E}_{2}} \right)$ which is the probability of drawing a gold coin from Box-II. Now it is given that the Box-II contains 1 gold coins. Therefore, $\Rightarrow P\left( A|{{E}_{2}} \right)=\dfrac{1}{2}\text{ }.......\left( 3 \right)$
Now, we will use Bayes’ theorem, which is as following:
Let ${{E}_{1}},{{E}_{2}},........{{E}_{n}}$ be a set of events associated with a sample space S, where all the events ${{E}_{1}},{{E}_{2}},........{{E}_{n}}$ have nonzero probability of occurrence and they form a partition of S. Let A be any event associated with S, then according to Bayes theorem:
$P\left( {{E}_{i}}|A \right)=\dfrac{P\left( {{E}_{i}} \right)P\left( A|{{E}_{i}} \right)}{\sum\limits_{k=1}^{n}{P\left( {{E}_{k}} \right)P\left( A|{{E}_{k}} \right)}}$ , where $k=1,2,3.....n$
Now, we have to find out the probability that the other coin in the box is also of gold which means that the gold coin must be drawn from Box-1 as it has 2 gold coins: $P\left( {{E}_{1}}|A \right)$
So, using Bayes’ theorem,
$P\left( {{E}_{1}}|A \right)=\dfrac{P\left( {{E}_{1}} \right)P\left( A|{{E}_{1}} \right)}{P\left( {{E}_{1}} \right)P\left( A|{{E}_{1}} \right)+P\left( {{E}_{2}} \right)P\left( A|{{E}_{2}} \right)}$
Now, we will put the values in the given equation from equation 1, 2 and 3:
$P\left( {{E}_{1}}|A \right)=\dfrac{\dfrac{1}{2}\times 1}{\dfrac{1}{2}\times 1+\dfrac{1}{2}\times \dfrac{1}{2}}=\dfrac{2}{3}$
Hence, $P\left( {{E}_{1}}|A \right)=\dfrac{2}{3}$

Therefore, the probability that the other coin in the box is also of gold is: $\dfrac{2}{3}$

Note: A common mistake is made by the students while applying the Bayes’ Theorem, student must avoid writing the formula for Bayes’ Theorem as: $P\left( {{E}_{1}}|A \right)=\dfrac{P\left( {{E}_{1}} \right)}{P\left( A \right)}$. Remember that $P\left( A|B \right)$ is the probability of condition when event A is occurring while event B has already occurred. For these types of questions we will have to define each and every step in detail so that it will be better for the understanding of the examiner.