Question

There are 4 red and 4 black balls in a bag and another bag contains 2 red and 6 black balls. One bag is randomly selected and a red ball is drawn. Find the probability that the red ball is drawn from the first bag.

Hint:

Apply Bayes' theorem to calculate conditional probabilities by systematically identifying prior and likelihood terms.

Answer 1

Let\(A_1\)and\(A_2\)betheeventsofselectingthefirstandsecondbag,respectively.Let\(R\)betheeventofdrawingaredball.UsingBayes'theorem: \[ P(A_1\midR)=\frac{P(R\midA_1)P(A_1)}{P(R)}. \] Here: \[ P(R\midA_1)=\frac{4}{8},\quadP(R\midA_2)=\frac{2}{8},\quadP(A_1)=P(A_2)=\frac{1}{2}. \] \[ P(R)=P(R\midA_1)P(A_1)+P(R\midA_2)P(A_2)=\frac{4}{8}\cdot\frac{1}{2}+\frac{2}{8}\cdot\frac{1}{2}=\frac{3}{8}. \] Thus: \[ P(A_1\midR)=\frac{\frac{4}{8}\cdot\frac{1}{2}}{\frac{3}{8}}=\frac{2}{3}. \]