Question

A bag A contains 2 white and 2 red balls and another bag B contains 4 white and 5 red balls. A ball is drawn and is found to be red.The probability that it was drawn from bag B is:

(a) \(\frac{5}{19}\) 

(b) \(\frac{21}{52}\) 

(c) \(\frac{10}{19}\) 

(d) \(\frac{25}{52}\) 

Answer 1

Answer: (c)  \(\frac{10}{19}\)

Let the events E_1, E_2, and A be defined as follows: 

E₁ = Choosing bag A 

E₂ = Choosing bag B 

A = Choosing red ball. 

Then, P(E₁) = P(E₂) = \(\frac{1}{2}\) 

(∵ There are two bags that have an equally likely chance of being chosen) 

P(A/E₁) = P(Drawing red ball from bag A) = \(\frac{2}{4} =\frac{1}{2}\)  (∵ 2 red out of 4 balls) 

P(A/E₂) = P(Drawing red ball from bag B) = \(\frac{5}{9}\)        (5 red out of 9 balls)

∴ P(Red balls are drawn from bag B)

\(=P(E_2/A) = \frac{P(E_2)\times P(A/E_2)}{P(E_1)\times P(A/E_1)+P(E_2)\times P(A/E_2)}\)

\(=\frac{\frac{1}{2}\times \frac{5}{9}}{\frac{1}{2}\times \frac{1}{2}+ \frac{1}{2}\times \frac{5}{9}}\) = \(\frac{\frac{5}{18}}{\frac{1}{4}+\frac{5}{18}}\) = \(=\frac{5/18}{19/36}\) = \(\frac{10}{19}\)