Question

A purse contains $4$ copper coins and $3$ silver coins, the second purse contains $6$ copper coins and $2$ silver coins. If a coin is drawn out of any purse, then the probability that it is a copper coin is?
A) $\dfrac{4}{7}$
B) $\dfrac{3}{4}$
C) $\dfrac{{37}}{{56}}$
D) $\dfrac{3}{7}$

Answer 1

Hint:
We are given here the situation of two purses where there are some coins of different types. We are asked to calculate the probability of getting a particular type of coin if we pick it totally random from any of the purses. Thus, we will take into account that any of the purses can be chosen at random and out of that the copper coin can also be selected at random.

Complete step by step solution:
Here,
Clearly,
The probability of choosing the first purse=the probability of choosing the second purse.
Thus,
$P(1) = P(2) = \dfrac{1}{2}$
Now,
Number of copper coins in the first purse$ = 4$
Total Number of coins in the first purse $ = 4 + 3 = 7$
Thus,
Probability of choosing a copper coin out of the first purse, $P(C) = \dfrac{4}{7}$
Further,
Probability of choosing a copper coin if purse one is selected, $P(1 \cap C) = P(1).P(C) = \dfrac{1}{2} \times \dfrac{4}{7} = \dfrac{2}{7}$
Again,
Number of copper coins in the second purse $ = 6$
Total Number of coins in the second purse $ = 6 + 2 = 8$
Thus,
Probability of choosing a copper coin, $P({C_2}) = \dfrac{6}{8} = \dfrac{3}{4}$
Further,
Probability of choosing a copper coin if the second purse is chosen at random, $P(2 \cap {C_2}) = P(2).P({C_2}) = \dfrac{1}{2} \times \dfrac{3}{4} = \dfrac{3}{8}$
Thus,
Probability of choosing a copper coin at random out of the two purse,
$P(Cc) = P(1 \cap C) + P(2 \cap {C_2})$
Thus, plugging in the values, we get
$P(Cc) = \dfrac{2}{7} + \dfrac{3}{8} = \dfrac{{16 + 21}}{{56}} = \dfrac{{37}}{{56}}$

Hence, the correct option is (C).

Note:
We have simply added the individual possibilities without considering the intersection of both the events. This is because the events are clearly mutually exclusive with each other which means the intersection set of the events is a null set. In other words, there is not a single common element of the two events in accordance to each other.