The contents of three urns 1, 2, 3 are as follows: 1 has 2 black, 3 red balls; 2 has 1 black, 2 white, 1 red balls; 3 has 5 black, 4 white, 3 red balls. One urn is selected at rand... The contents of three urns 1, 2, 3 are as follows: 1 has 2 black, 3 red balls; 2 has 1 black, 2 white, 1 red balls; 3 has 5 black, 4 white, 3 red balls. One urn is selected at random and two balls are drawn. What is the probability that they come from urn 2? Read more

Steps to Solve

1. Identify the Urns and Their Contents

   Urn 1: 2 black, 3 red
   Urn 2: 1 black, 2 white, 1 red
   Urn 3: 5 black, 4 white, 3 red

2. Determine the Total Probability of Selecting Urn 2 and Drawing One White and One Red Ball

   Using the law of total probability:
   We need to find the probability of drawing one white ball and one red ball once Urn 2 is chosen.

   For Urn 2, the total number of balls is:
   $$ 1 + 2 + 1 = 4 $$
   To choose one white and one red:
   $$ P(\text{White, Red | Urn 2}) = \frac{{\text{Number of ways to pick 1 white}} \cdot {\text{Number of ways to pick 1 red}}}{\text{Total ways to pick 2 balls}} $$ The number of ways to choose 2 balls out of 4:
   $$ \binom{4}{2} = 6 $$
   The number of ways to choose 1 white and 1 red:
   $$ 2 \cdot 1 = 2 $$
   Hence,
   $$ P(\text{White, Red | Urn 2}) = \frac{2}{6} = \frac{1}{3} $$

3. Calculate the Total Probability of Drawing One White and One Red Ball from All Urns

   We need to find the probabilities for each urn:

   • Urn 1: No white balls, so $P = 0$
   • Urn 2: As calculated, $P(\text{White, Red | Urn 2}) = \frac{1}{3}$
   • Urn 3: Total number of balls = 12.
     Choosing one white and one red:
     Ways to choose 1 red: 3
     Ways to choose 1 white: 4
     Total ways to choose 2 balls:
     $$ \binom{12}{2} = 66 $$
     Hence,
     $$ P(\text{White, Red | Urn 3}) = \frac{3 \cdot 4}{66} = \frac{12}{66} = \frac{2}{11} $$

   Total probability of drawing one white and one red ball from all urns is:
   $$ P(\text{White, Red}) = \frac{1}{3} \cdot \frac{1}{3} + 0 + \frac{2}{11} \cdot \frac{1}{3} $$
   Calculate:
   $$ P(\text{White, Red}) = \frac{1}{9} + \frac{2}{33} $$
   Find a common denominator (99):
   $$ = \frac{11}{99} + \frac{6}{99} = \frac{17}{99} $$

4. Use Bayes' Theorem to Find the Conditional Probability

   We want ( P(\text{Urn 2 | White, Red}) ):
   $$ P(\text{Urn 2 | White, Red}) = \frac{P(\text{White, Red | Urn 2}) \cdot P(\text{Urn 2})}{P(\text{White, Red})} $$
   Since each urn is equally likely to be chosen, ( P(\text{Urn 2}) = \frac{1}{3} ):
   $$ P(\text{Urn 2 | White, Red}) = \frac{\left(\frac{1}{3}\right) \left(\frac{1}{3}\right)}{\frac{17}{99}} $$
   Simplifying:
   $$ = \frac{1}{9} \cdot \frac{99}{17} = \frac{11}{17} $$