Question

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

Answer 1

Hint :In mathematics probability is a term which is used to indicate the chances of occurrence of an event. Addition theorem of probability is used to solve this problem which state that if $ x $ and $ y $ are two events then probability is $ P\left( {XUB} \right) = $ $ P\left( X \right) + P\left( Y \right) - P\left( {X\bigcap Y } \right) $

Complete Step By Step Answer:
As we see in the question there are two purses which contain copper and silver coins.
The chances of choosing any one of the purses is always $ 50/50 $ .
Hence, probability of choosing purse is
Purse $ A $ $ = $ Purse $ B = \dfrac{1}{2} $
As we know purse $ A $ has total number of $ 4 $ copper coins and $ 3 $ silver coins, hence, probability of choosing copper coin out of total $ 7 $ coins present in purse will be
 $ = \dfrac{1}{2} \times \dfrac{4}{7} $
After solving this equation, we get
 $ = \dfrac{2}{7} $
Hence probability of choosing copper coin from Purse $ A $ will be $ \left( {\dfrac{2}{7}} \right) $
Similarly, purse $ B $ has $ 6 $ copper and $ 2 $ silver coins, therefore probability of choosing copper coin out of total $ 8 $ coins present in purse will be
 $ = \dfrac{1}{2} \times \dfrac{6}{8} $
After solving this equation, we get
 $ = \dfrac{3}{8} $
Hence probability of choosing copper coin from Purse $ B $ will be $ \left( {\dfrac{3}{8}} \right) $
on adding the probability of choosing copper from both the purses will give the final probability of choosing copper from any purse
 $ = \dfrac{2}{7} + \dfrac{3}{8} $
On taking the L.C.M
 $ \dfrac{{2 \times 8 + 3 \times 7}}{{56}} $
On solving the above equation, we get
 $ \dfrac{{16 + 21}}{{56}} $
On further solving we finally get
 $ \dfrac{{37}}{{56}} $
Hence, the probability of choosing copper coin from any purse will be $ \dfrac{{37}}{{56}} $ .Therefore, option $ \left( 4 \right) $ is the correct option.

Note :
Value of probability is usually ranging from 0 to 1 where o indicates absence of chances or impossibility of occurrences and 1 indicates certainty. Above condition is defined as a mutually exclusive event which states that two events cannot occur at same time.