A bag contains $4$ red and $3$ black balls.A second bag contains $2$ red and $4$ black balls.If one bag is selected at random and a ball is drawn from it,find probability that the ball is red.
My attempt:-
P(Red ball from bag 1)=4/7
P(Red ball from bag 2)=2/6
P(Selecting one bag from 2 bags)=1/2.
So,P(Red ball from bag 1 U Red ball from bag 2) =4/7+2/6=38/42.
So,our required probability from any one bag at random=19/42.
Am I correct?
Best Answer
Using the law of total probability and then Bayes rule, we can write the probability of picking a red ball as
\begin{align*} P(R) &= P(R \cap \text{picked bag 1}) + P(R \cap \text{picked bag 2}) \\ &= P(R \mid \text{picked bag 1})P(\text{picked bag 1}) + P(R \mid \text{picked bag 2})P(\text{picked bag 2}) \\ &= \frac{4}{7} \cdot \frac{1}{2} + \frac{2}{6} \cdot \frac{1}{2} = \frac{2}{7} + \frac{1}{6} = \frac{19}{42} \end{align*}