The content of Urn I, II, III are as follows : Urn I : 1 white, 2 black and 3 - Maths - Probability

Question

The content of Urn I, II, III are as follows :
Urn I : 1 white, 2 black and 3 red balls
Urn II : 2 white, 1 black and 1 red balls
Urn II : 4 white, 5 black and 3 red balls
one urn is choosen and two balls are drawn They happen to be
white and red What is the probability that they come from urn III
?

Ajanta Trivedi · Accepted Answer

let E1 ,E2 , E3 and A denote the following events E1 = urn I is chosen, E2 = urn II is chosen, E3 = urn III is chosen, and A = two balls are drawn at random are white and red since one of the urn is chosen at random therefore $P (E_{1}) = \frac{1}{3} = P (E_{2}) = P (E_{3})$ if E1 has already occurred, then urn I has been chosen the urn I contains 1 white , 2 black and 3 red balls therefore, the probability of drawing 1 white and 1 red ball is $\frac{{}^{1}{C_{1}} \cdot {}^{3}{C_{1}}}{{}^{6}{C_{2}}}$ so $P (A / E_{1}) = \frac{{}^{1}{C_{1}} \cdot {}^{3}{C_{1}}}{{}^{6}{C_{2}}} = \frac{1 * 3}{\frac{6 * 5}{2}} = \frac{3}{3 * 5} = \frac{1}{5}$ similarly $P (A / E_{2}) = \frac{{}^{2}{C_{1}} \cdot {}^{1}{C_{1}}}{{}^{4}{C_{2}}} = \frac{2 * 1}{\frac{4 * 3}{2}} = \frac{2}{2 * 3} = \frac{1}{3}$ and $P (A / E_{3}) = \frac{{}^{4}{C_{1}} \cdot {}^{3}{C_{1}}}{{}^{12}{C_{2}}} = \frac{4 * 3}{\frac{12 * 11}{2}} = \frac{12}{6 * 11} = \frac{2}{11}$ we are required to find P(E3/A) by Baye's theorem, we have $P (E_{3} / A) = \frac{P (E_{3}) P (A / E_{3})}{P (E_{1}) P (A / E_{1}) + P (E_{2}) P (A / E_{2}) + P (E_{3}) P (A / E_{3})} = \frac{\frac{1}{3} \times \frac{2}{11}}{\frac{1}{3} \times \frac{1}{5} + \frac{1}{3} \times \frac{1}{3} + \frac{1}{3} \times \frac{2}{11}} = \frac{\frac{2}{11}}{\frac{1}{5} + \frac{1}{3} + \frac{2}{11}} = \frac{\frac{2}{11}}{\frac{118}{165}} = \frac{2}{11} \times \frac{165}{118} = \frac{15}{59}$