Question

A box contains 5 silver coins and 4 gold coins, two coins are stolen from the box. Now one coin is drawn at random from the box and is found to be of gold. Find the probability that the lost coins also were of gold. $$$$

Answer 1

We find the probability of both the stolen coins is silver coins as $P\left( {{E}_{1}} \right)$and the probability of both that both the stolen coins are gold coins as $P\left( {{E}_{2}} \right)$. We find the probability that the selected coin is gold subjected to condition that the stolen coins were silver $P\left( {{E}_{1}}/X \right)$ and the stolen coins were gold $P\left( {{E}_{2}}/X \right)$. We use the Bayes theorem and find the required probability of the stolen coins being gold subjected to the selected coin being gold as
$P\left( {{E}{2}}/X \right)=\dfrac{P\left( {{E}{2}} \right)P\left( X/{{E}{2}} \right)}{P\left( {{E}{1}} \right)P\left( X/{{E}{1}} \right)+P\left( {{E}{2}} \right)P\left( X/{{E}_{2}} \right)}$$$$$

Complete step-by-step answer:
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 and exhaustive and $X$ 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

$P\left( {{E}_{i}}|X \right)=\dfrac{P\left( {{E}_{i}} \right)P\left( X|{{E}_{i}} \right)}{\sum\limits_{i=1}^{n}{P\left( {{E}_{i}} \right)P\left( X|{{E}_{i}} \right)}}$

We are given the question that the box contains 5 silver coins and 4 gold coins, 2 coins are stolen from the box. Let ${{E}_{1}}$ be the event that both the stolen coins are silver coins and ${{E}_{2}}$ be the event that both the stolen coins are gold coins. Total number of coins is $5+4=9$.When the thief will draw two coins one after one the sample space will decrease by 1 from 9 to 8. So the probability for ${{E}_{1}},{{E}_{2}}$ is

& P\left( {{E}_{1}} \right)=\dfrac{5}{9}\times \dfrac{4}{8}=\dfrac{5}{18} \\\ & P\left( {{E}_{2}} \right)=\dfrac{4}{9}\times \dfrac{3}{8}=\dfrac{1}{6} \\\ \end{aligned}$$ We are further given in the question that one coin is drawn at random from the box and is found to be of gold. Let us denote that the new event of getting a gold coin after the stealing of coins be $X.$ We are going to get either a gold coin or silver coin which means the sample size is 2. So the probability of getting a gold coin after the loss of two silver coins $P\left( X/{{E}_{1}} \right)$ and probability of getting a gold coin after the loss of two gold coins $P\left( X/{{E}_{2}} \right)$ are equal and given by $$P\left( X/{{E}_{1}} \right)=P\left( X/{{E}_{2}} \right)=\dfrac{1}{2}$$ We are asked to find the probability that the lost coins are gold after the knowledge of the news of getting a gold coin that is $P\left( {{E}_{2}}/A \right)$. We use the Bayes’ theorem for two events and have, $$\begin{aligned} & P\left( {{E}_{2}}/X \right)=\dfrac{P\left( {{E}_{2}} \right)P\left( X/{{E}_{2}} \right)}{P\left( {{E}_{1}} \right)P\left( X/{{E}_{1}} \right)+P\left( {{E}_{2}} \right)P\left( X/{{E}_{2}} \right)} \\\ & \Rightarrow P\left( {{E}_{2}}/X \right)=\dfrac{\dfrac{1}{6}\times \dfrac{1}{2}}{\dfrac{5}{18}\times \dfrac{1}{2}+\dfrac{1}{6}\times \dfrac{1}{2}} \\\ & \Rightarrow P\left( {{E}_{2}}/X \right)=\dfrac{\dfrac{1}{12}}{\dfrac{5}{36}+\dfrac{1}{12}}=\dfrac{3}{8} \\\ \end{aligned}$$ **Note:** We note that the events ${{E}_{1}},{{E}_{2}}$ of stealing of gold and silver coins are mutually exclusive they cannot occur at the same time and also exhaustive means sum of the events is the sample size. The probabilities $P\left( {{E}_{1}} \right),P\left( {{E}_{2}} \right)$ are called prior probabilities and $P\left( {{E}_{1}}/X \right),P\left( {{E}_{2}}/X \right)$ are called posterior probabilities.