Box I contains two white and three black balls. Box II contains four white and one black balls and box III contains three white and four black balls. A dice having three red, two yellow and one green face, is thrown to select the box. If red face turns up, we pick up box I, if a yellow face turns up we pick up box II, otherwise, we pick up box III. Then, we draw a ball from the selected box. If the ball drawn is white, what is the probability that the dice had turned up with a red face?

Answer

To solve the problem step by step, we will use the concept of conditional probability and Bayes' theorem.

### Step 1: Define the Events
Let:
- \( E_1 \): Event that the red face of the die shows up (Box I is chosen).
- \( E_2 \): Event that the yellow face of the die shows up (Box II is chosen).
- \( E_3 \): Event that the green face of the die shows up (Box III is chosen).
- \( W \): Event that a white ball is drawn.

### Step 2: Calculate the Probabilities of Selecting Each Box
The probabilities of selecting each box based on the die roll are:
- \( P(E_1) = \frac{3}{6} = \frac{1}{2} \) (3 red faces)
- \( P(E_2) = \frac{2}{6} = \frac{1}{3} \) (2 yellow faces)
- \( P(E_3) = \frac{1}{6} \) (1 green face)

### Step 3: Calculate the Probability of Drawing a White Ball from Each Box
Now, we calculate the probability of drawing a white ball from each box:
- For Box I (2 white, 3 black):
\[ P(W | E_1) = \frac{2}{2 + 3} = \frac{2}{5} \]
- For Box II (4 white, 1 black):
\[ P(W | E_2) = \frac{4}{4 + 1} = \frac{4}{5} \]
- For Box III (3 white, 4 black):
\[ P(W | E_3) = \frac{3}{3 + 4} = \frac{3}{7} \]

### Step 4: Apply Bayes' Theorem
We want to find the probability that the red face of the die was rolled given that a white ball was drawn, i.e., \( P(E_1 | W) \). By Bayes' theorem:
\[ P(E_1 | W) = \frac{P(W | E_1) \cdot P(E_1)}{P(W)} \]
Where \( P(W) \) can be calculated using the law of total probability:
\[ P(W) = P(W | E_1) \cdot P(E_1) + P(W | E_2) \cdot P(E_2) + P(W | E_3) \cdot P(E_3) \]

### Step 5: Calculate \( P(W) \)
Substituting the values:
\[ P(W) = \left(\frac{2}{5} \cdot \frac{1}{2}\right) + \left(\frac{4}{5} \cdot \frac{1}{3}\right) + \left(\frac{3}{7} \cdot \frac{1}{6}\right) \]
Calculating each term:
- First term: \( \frac{2}{5} \cdot \frac{1}{2} = \frac{1}{5} \)
- Second term: \( \frac{4}{5} \cdot \frac{1}{3} = \frac{4}{15} \)
- Third term: \( \frac{3}{7} \cdot \frac{1}{6} = \frac{1}{14} \)

Now, we need a common denominator to sum these fractions. The least common multiple of 5, 15, and 14 is 210.

Converting each term:
- \( \frac{1}{5} = \frac{42}{210} \)
- \( \frac{4}{15} = \frac{56}{210} \)
- \( \frac{1}{14} = \frac{15}{210} \)

Adding these:
\[ P(W) = \frac{42 + 56 + 15}{210} = \frac{113}{210} \]

### Step 6: Calculate \( P(E_1 | W) \)
Now substituting back into Bayes' theorem:
\[ P(E_1 | W) = \frac{P(W | E_1) \cdot P(E_1)}{P(W)} = \frac{\left(\frac{2}{5}\right) \cdot \left(\frac{1}{2}\right)}{\frac{113}{210}} = \frac{\frac{1}{5}}{\frac{113}{210}} = \frac{210}{565} = \frac{42}{113} \]

### Final Answer
Thus, the probability that the die had turned up with a red face given that a white ball was drawn is:
\[ \boxed{\frac{42}{113}} \]