Question

A purse contains 2 silver and 4 copper coins. A second purse contains 4 silver and 3 copper cons. If a coin at random from one of the two purses, what is the probability that it is a silver coin?

Answer 1

Hint: Let \[{{E}_{1}},{{E}_{2}}\] be the event of selecting coins from purse 1 and 2. Let A be the coin drawn as silver. Thus find the required probability by using the theorem of the total probability on Baye’s theorem.

Complete step-by-step answer:
It is said that the 1\[^{st}\] purse contains 2 silver and 4 copper coins.
Thus, the total number of coins in purse 1 = 2 + 4 = 6 coins
Now let us look into purse 2, it contains 4 silver and 3 copper coins.
Thus, the total number of coins in purse 2 = 4 + 3 = 7 coins
Now, one coin is taken from any one of the 2 purses randomly. We are asked to find the probability that the coin chosen among both the purses is a silver coin.
Let \[{{E}_{1}}\] be the event of selecting a coin from purse 1.
Let \[{{E}_{2}}\] be the event of selecting a coin from purse 2.
Let A be the coin drawn which is a silver coin.
The probability of selecting a coin from purse 1, \[P\left( {{E}_{1}} \right)=\dfrac{1}{2}\].
Similarly, the probability of selecting a coin from purse 2, \[P\left( {{E}_{2}} \right)=\dfrac{1}{2}\].
Now, the probability of getting a silver coin from purse 1 = \[P\left( \dfrac{A}{{{E}_{1}}} \right)\].
= Number of silver coins / Total number of coins = \[\dfrac{2}{6}\].
Now, the probability of getting a silver coin from purse 2 = \[P\left( \dfrac{A}{{{E}_{2}}} \right)\].
= Number of silver coins / Total number of coins = \[\dfrac{4}{7}\].
Thus we got \[P\left( {{E}_{1}} \right)=\dfrac{1}{2}\], \[P\left( {{E}_{2}} \right)=\dfrac{1}{2}\], \[P\left( \dfrac{A}{{{E}_{1}}} \right)=\dfrac{2}{6}\], \[P\left( \dfrac{A}{{{E}_{2}}} \right)=\dfrac{4}{7}\].
Now, by using Baye’s theorem, in a case where we consider A to be an event in a sample space. Thus, we can state simplified various of the theorem of total probability on baye’s theorem.
\[\therefore \] Required probability = \[P\left( A \right)=P\left( {{E}_{1}} \right)P\left( \dfrac{A}{{{E}_{1}}} \right)+P\left( {{E}_{2}} \right)P\left( \dfrac{A}{{{E}_{2}}} \right)\].
Let us substitute the values in the above expression.
\[\therefore \] Required probability = \[P\left( A \right)=\]\[\dfrac{1}{2}\times \dfrac{2}{6}+\dfrac{1}{2}\times \dfrac{4}{7}\]
                                                           \[\begin{align}
  & =\dfrac{1}{6}+\dfrac{2}{7} \\
 & =\dfrac{7+12}{42}=\dfrac{19}{42} \\
\end{align}\]
Hence, the probability of selecting a silver coin = \[\dfrac{19}{42}\].
Thus we got the required probability.

Note: In general Baye’s rule is used to flip a conditional probability, while the law of total probability is used when you don’t know the probability of an event, but you know its occurrence under several disjoint scenarios and the probability of each scenario.