Question

Given three identical boxes I,II and III, each containing two coins. In box I, both coins are gold coins, in box II both are silver coins and in the box III, there is one gold and one 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 is also of gold?

Answer 1

Hint: Conditional probability is the probability of one event occurring with some relationship to one or more other events.

Complete step-by-step answer:
Let, P(A)= Probability of selecting box I.
 P(B)= Probability of selecting box II.
 P(C)= Probability of selecting box III.
 P(G|A)= Probability that the second coin is of gold in box I.
 P(G|B)= Probability that the second coin is of gold in box II.
 P(G|C)= Probability that the second coin is of gold in box III.
Now, P(A) =P(B) =P(C)= \[\dfrac{1}{3}\]; P(G|A)=1; P(G|B)=0; P(G|C)= \[\dfrac{1}{2}\]
The probability that the other coin in the box is also of gold, if the first coin is of gold i.e. P(A|G)
\[ \Rightarrow P(A|G) = \dfrac{{P(A).P(G|A)}}{{P(A).P(G|A) + P(B).P(G|B) + P(C).P(G|C)}}\]
Putting the values in above equation we get;
\[ \Rightarrow P(A|G) = \dfrac{{\dfrac{1}{3} \times (1)}}{{\dfrac{1}{3} \times (1) + \dfrac{1}{3} \times (0) + \dfrac{1}{3} \times \dfrac{1}{2}}} = \dfrac{{\dfrac{1}{3}}}{{\dfrac{1}{3} + \dfrac{1}{6}}} = \dfrac{1}{3} \times \dfrac{2}{1} = \dfrac{2}{3}\]
Required probability = \[\dfrac{2}{3}\]

Note: Student’s must be familiar with probability concepts and to solve such questions must know complete usage of conditional probability.