If bag A contains 4 red and 4 black balls while another bag B contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag and it is found to be red. Find the probability that it is drawn from bag A ?

Concept:

Bayes' Theorem:

Let E₁, E₂, ….., E_n be n mutually exclusive and exhaustive events associated with a random experiment and let S be the sample space. Let A be any event which occurs together with any one of E₁ or E₂ or … or E_n such that P(A) ≠ 0. Then

\(P\left( {{E_i}\;|\;A} \right) = \frac{{P\left( {{E_i}} \right)\; \times \;P\left( {A\;|\;{E_i}} \right)}}{{\mathop \sum \nolimits_{i\; = 1}^n P\left( {{E_i}} \right) \times P\left( {A\;|\;{E_i}} \right)}}\)

i = 1, 2, ..., n

Calculation:

Let E₁ be the event of choosing bag A, E₂ be the event of choosing the bag B and X be the event of drawing a red ball.

⇒ P (E₁) = P (E₂) = 1/2

⇒ P (X | E₁) = P (Drawing a red ball from bag A) = 4/8 = 1/2

Similarly, P (X | E₂) = P (Drawing a red ball from bag B) = 2/8 = 1/4

Here, we have to find the probability of drawing a ball from bag A given that the ball is red in color i.e P (E₁ | X)

Applying Bayes Theorem, we get:

\(P{\rm{(}}{E_1}\;{\rm{|}}X) = \frac{{P\left( {{E_1}} \right) \cdot P\;\left( {X\;|\;{E_1}} \right)}}{{\left[ {P\left( {{E_1}} \right) \cdot P\;\left( {X\;|\;{E_1}} \right) + \;P\left( {{E_2}} \right) \cdot P\;\left( {X\;|\;{E_2}} \right)} \right]}}\;\)

\(P{\rm{(}}{E_1}\;{\rm{|}}X) = \frac{{\frac{1}{2} \cdot \frac{1}{{2}}}}{{\left[ {\frac{1}{2} \cdot \frac{1}{2} + \;\frac{1}{2} \cdot \frac{1}{{4}}} \right]}} = \frac{{2}}{{3}}\;\)