Question

A bag X contains 2 white and 3 black balls and another bag Y contains 4 white and 2 black balls. One bag is selected at random and a ball is drawn from it. Then find the probability for the ball chosen to be a white ball.
A. \[\dfrac{2}{{15}}\]
B. \[\dfrac{7}{{15}}\]
C. \[\dfrac{8}{{15}}\]
D. \[\dfrac{{14}}{{15}}\]

Answer 1

Hint: First we will find the probability of choosing bag X and bag Y by using formula \[{\rm{probability = }}\dfrac{{{\rm{The}}\,{\rm{number}}\,{\rm{of}}\,{\rm{favorable}}\,{\rm{outcomes}}}}{{{\rm{The}}\,{\rm{number}}\,{\rm{of}}\,{\rm{total}}\,{\rm{outcomes}}}}\]. Then we will calculate the probability of getting white ball from bag X and bag Y. Then apply the formula law of total probability\[P\left( B \right) = P\left( {B|{A_1}} \right) \cdot P\left( {{A_1}} \right) + P\left( {B|{A_2}} \right) \cdot P\left( {{A_2}} \right) + \cdots + P\left( {B|{A_k}} \right) \cdot P\left( {{A_k}} \right)\].

Formula Used:
\[{\rm{Probability = }}\dfrac{{{\rm{The}}\,{\rm{number}}\,{\rm{of}}\,{\rm{favorable}}\,{\rm{outcomes}}}}{{{\rm{The}}\,{\rm{number}}\,{\rm{of}}\,{\rm{total}}\,{\rm{outcomes}}}}\]
Law of total probability:
\[P\left( B \right) = P\left( {B|{A_1}} \right) \cdot P\left( {{A_1}} \right) + P\left( {B|{A_2}} \right) \cdot P\left( {{A_2}} \right) + \cdots + P\left( {B|{A_k}} \right) \cdot P\left( {{A_k}} \right)\]

Complete step by step solution:
Let \[{A_1}\] be the event that a ball is drawn from bag X.
\[{A_2}\] be the event that a ball is drawn from bag Y.
\[B\] be the event to draw a white ball.
Given that bag X contains 2 white and 3 black balls and another bag Y contains 4 white and 2 black balls.
The total number of bags is 2.
The probability of choosing bag X is \[P\left( {{A_1}} \right) = \dfrac{{{\rm{The}}\,{\rm{number}}\,{\rm{of}}\,{\rm{bag}}\,{\rm{X}}}}{{{\rm{The}}\,{\rm{number}}\,{\rm{of}}\,{\rm{total}}\,{\rm{bags}}}}\]
                                                                             \[ = \dfrac{1}{2}\]
The probability of choosing bag Y is \[P\left( {{A_2}} \right) = \dfrac{{{\rm{The}}\,{\rm{number}}\,{\rm{of}}\,{\rm{bag}}\,{\rm{Y}}}}{{{\rm{The}}\,{\rm{number}}\,{\rm{of}}\,{\rm{total}}\,{\rm{bags}}}}\]
                                                                              \[ = \dfrac{1}{2}\]
Total number of balls in bag X is \[\left( {2 + 3} \right) = 5\]
The number of white balls in bag X is 2.
The probability that a ball is drawn from bag X is \[P\left( {B|{A_1}} \right) = \dfrac{2}{5}\]
Total number of balls in bag Y is \[\left( {4 + 2} \right) = 6\]
The number of white balls in bag Y is 4.
The probability that a ball is drawn from bag Y is \[P\left( {B|{A_2}} \right) \cdot = \dfrac{4}{6}\]
Now applying the law of total probability \[P\left( B \right) = P\left( {B|{A_1}} \right) \cdot P\left( {{A_1}} \right) + P\left( {B|{A_2}} \right) \cdot P\left( {{A_2}} \right) + \cdots + P\left( {B|{A_k}} \right) \cdot P\left( {{A_k}} \right)\]
\[P\left( B \right) = P\left( {B|{A_1}} \right) \cdot P\left( {{A_1}} \right) + P\left( {B|{A_2}} \right) \cdot P\left( {{A_2}} \right)\]
Now putting the values of \[P\left( {B|{A_1}} \right)\], \[P\left( {{A_1}} \right)\],\[P\left( {B|{A_2}} \right) \cdot \]and \[P\left( {{A_2}} \right)\]
\[ \Rightarrow P\left( B \right) = \dfrac{2}{5} \cdot \dfrac{1}{2} + \dfrac{4}{6} \cdot \dfrac{1}{2}\]
\[ \Rightarrow P\left( B \right) = \dfrac{1}{5} + \dfrac{1}{3}\]
\[ \Rightarrow P\left( B \right) = \dfrac{{3 + 5}}{{15}}\]
\[ \Rightarrow P\left( B \right) = \dfrac{8}{{15}}\]

Hence option C is the correct option.

Note:Students often get confused with bayes law and laws of total probability. The bayes law is \[P\left( {A \cap B} \right) = P\left( B \right) \cdot P\left( {A|B} \right)\] and the law of total probability is \[P\left( B \right) = P\left( {B|{A_1}} \right) \cdot P\left( {{A_1}} \right) + P\left( {B|{A_2}} \right) \cdot P\left( {{A_2}} \right) + \cdots + P\left( {B|{A_k}} \right) \cdot P\left( {{A_k}} \right)\].