(a) (i) Bag I contains 5 red and 3 black balls, Bag II contains 6 red and 5 black balls. One bag is chosen at random and a ball is drawn which is found to be black. Find the probability that it was drawn from Bag I, (II).
(ii) Bag I contains 3 red and 5 white balls and bag II contains 4 red and 6 white balls. One of the bags is selected at random and a ball is drawn from it. The ball is found to be red. Find the probability that ball is drawn from Bag II.
(b) Bag I contains 4 black and 6 red balls, bag II contains 7 black and 3 red balls and bag III contains 5 black and 5 red balls. One bag is chosen at random and a ball is drawn from it which is found to be red. Find the probability that it was drawn from bag II.

Answer

To solve the given problem step by step, we will use Bayes' theorem, which helps us find the probability of an event based on prior knowledge of conditions related to the event.

### Part (a) (i)

**Step 1: Define the events**
- Let \( A_1 \) be the event that Bag I is chosen.
- Let \( A_2 \) be the event that Bag II is chosen.
- Let \( B \) be the event that a black ball is drawn.

**Step 2: Calculate the probabilities of choosing each bag**
- \( P(A_1) = P(A_2) = \frac{1}{2} \) (since one bag is chosen at random)

**Step 3: Calculate the probability of drawing a black ball from each bag**
- For Bag I: There are 3 black balls out of 8 total balls.
\[ P(B | A_1) = \frac{3}{8} \]
- For Bag II: There are 5 black balls out of 11 total balls.
\[ P(B | A_2) = \frac{5}{11} \]

**Step 4: Use the law of total probability to find \( P(B) \)**
\[ P(B) = P(B | A_1) P(A_1) + P(B | A_2) P(A_2) \]
\[ P(B) = \left(\frac{3}{8} \cdot \frac{1}{2}\right) + \left(\frac{5}{11} \cdot \frac{1}{2}\right) \]
\[ = \frac{3}{16} + \frac{5}{22} \]
To add these fractions, find a common denominator (which is 176):
\[ = \frac{3 \cdot 11}{176} + \frac{5 \cdot 8}{176} = \frac{33 + 40}{176} = \frac{73}{176} \]

**Step 5: Calculate \( P(A_1 | B) \) using Bayes' theorem**
\[ P(A_1 | B) = \frac{P(B | A_1) P(A_1)}{P(B)} \]
\[ = \frac{\left(\frac{3}{8}\right) \left(\frac{1}{2}\right)}{\frac{73}{176}} = \frac{\frac{3}{16}}{\frac{73}{176}} = \frac{3 \cdot 176}{16 \cdot 73} = \frac{528}{1168} = \frac{33}{73} \]

**Step 6: Calculate \( P(A_2 | B) \)**
\[ P(A_2 | B) = 1 - P(A_1 | B) = 1 - \frac{33}{73} = \frac{40}{73} \]

### Part (a) (ii)

**Step 1: Define the events**
- Let \( A_1 \) be the event that Bag I is chosen.
- Let \( A_2 \) be the event that Bag II is chosen.
- Let \( R \) be the event that a red ball is drawn.

**Step 2: Calculate the probabilities of choosing each bag**
- \( P(A_1) = P(A_2) = \frac{1}{2} \)

**Step 3: Calculate the probability of drawing a red ball from each bag**
- For Bag I: There are 3 red balls out of 8 total balls.
\[ P(R | A_1) = \frac{3}{8} \]
- For Bag II: There are 4 red balls out of 10 total balls.
\[ P(R | A_2) = \frac{4}{10} = \frac{2}{5} \]

**Step 4: Use the law of total probability to find \( P(R) \)**
\[ P(R) = P(R | A_1) P(A_1) + P(R | A_2) P(A_2) \]
\[ = \left(\frac{3}{8} \cdot \frac{1}{2}\right) + \left(\frac{2}{5} \cdot \frac{1}{2}\right) \]
\[ = \frac{3}{16} + \frac{1}{5} \]
To add these fractions, find a common denominator (which is 80):
\[ = \frac{3 \cdot 5}{80} + \frac{16}{80} = \frac{15 + 16}{80} = \frac{31}{80} \]

**Step 5: Calculate \( P(A_2 | R) \) using Bayes' theorem**
\[ P(A_2 | R) = \frac{P(R | A_2) P(A_2)}{P(R)} \]
\[ = \frac{\left(\frac{2}{5}\right) \left(\frac{1}{2}\right)}{\frac{31}{80}} = \frac{\frac{1}{5}}{\frac{31}{80}} = \frac{80}{155} = \frac{16}{31} \]

### Part (b)

**Step 1: Define the events**
- Let \( A_1 \) be the event that Bag I is chosen.
- Let \( A_2 \) be the event that Bag II is chosen.
- Let \( A_3 \) be the event that Bag III is chosen.
- Let \( R \) be the event that a red ball is drawn.

**Step 2: Calculate the probabilities of choosing each bag**
- \( P(A_1) = P(A_2) = P(A_3) = \frac{1}{3} \)

**Step 3: Calculate the probability of drawing a red ball from each bag**
- For Bag I: There are 6 red balls out of 10 total balls.
\[ P(R | A_1) = \frac{6}{10} = \frac{3}{5} \]
- For Bag II: There are 3 red balls out of 10 total balls.
\[ P(R | A_2) = \frac{3}{10} \]
- For Bag III: There are 5 red balls out of 10 total balls.
\[ P(R | A_3) = \frac{5}{10} = \frac{1}{2} \]

**Step 4: Use the law of total probability to find \( P(R) \)**
\[ P(R) = P(R | A_1) P(A_1) + P(R | A_2) P(A_2) + P(R | A_3) P(A_3) \]
\[ = \left(\frac{3}{5} \cdot \frac{1}{3}\right) + \left(\frac{3}{10} \cdot \frac{1}{3}\right) + \left(\frac{1}{2} \cdot \frac{1}{3}\right) \]
\[ = \frac{1}{5} + \frac{1}{10} + \frac{1}{6} \]
To add these fractions, find a common denominator (which is 30):
\[ = \frac{6}{30} + \frac{3}{30} + \frac{5}{30} = \frac{14}{30} = \frac{7}{15} \]

**Step 5: Calculate \( P(A_2 | R) \) using Bayes' theorem**
\[ P(A_2 | R) = \frac{P(R | A_2) P(A_2)}{P(R)} \]
\[ = \frac{\left(\frac{3}{10}\right) \left(\frac{1}{3}\right)}{\frac{7}{15}} = \frac{\frac{1}{10}}{\frac{7}{15}} = \frac{15}{70} = \frac{3}{14} \]

### Summary of Results
- Part (a)(i): Probability that the black ball was drawn from Bag I: \( \frac{33}{73} \)
- Part (a)(ii): Probability that the red ball was drawn from Bag II: \( \frac{16}{31} \)
- Part (b): Probability that the red ball was drawn from Bag II: \( \frac{3}{14} \)