Bayes' theorem

Step 1: Define the Events

Let's define the events for selecting the bags and drawing the ball:

• $$E_1$$: Event that the die shows 1 or 2 (Bag $$X$$ is chosen).
• $$E_2$$: Event that the die shows 3, 4, 5, or 6 (Bag $$Y$$ is chosen).
• $$W$$: Event that the drawn ball is white.

Step 2: Find individual probabilities

• Probabilities of choosing the bags:$$P(E_1) = \frac{2}{6} = \frac{1}{3}$$  $$P(E_2) = \frac{4}{6} = \frac{2}{3}$$
• Conditional probabilities of drawing a white ball:
  • From Bag $$X$$ (3 White, 2 Black $$\implies$$ Total = 5):$$P(W|E_1) = \frac{3}{5}$$
  • From Bag $$Y$$ (2 White, 4 Black $$\implies$$ Total = 6):$$P(W|E_2) = \frac{2}{6} = \frac{1}{3}$$

Step 3: Apply Bayes' Theorem

We need to find the conditional probability that the ball came from bag $$Y$$ given that it is white, which is $$P(E_2|W)$$:

$$P(E_2|W) = \frac{P(E_2) \cdot P(W|E_2)}{P(E_1) \cdot P(W|E_1) + P(E_2) \cdot P(W|E_2)}$$

Substitute the values into the formula:

$$P(E_2|W) = \frac{\frac{2}{3} \cdot \frac{1}{3}}{\left(\frac{1}{3} \cdot \frac{3}{5}\right) + \left(\frac{2}{3} \cdot \frac{1}{3}\right)}$$  $$P(E_2|W) = \frac{\frac{2}{9}}{\frac{1}{5} + \frac{2}{9}}$$

Find a common denominator for the bottom expression ($$45$$):

$$\frac{1}{5} + \frac{2}{9} = \frac{9 + 10}{45} = \frac{19}{45}$$

Now, simplify the main fraction:

$$P(E_2|W) = \frac{\frac{2}{9}}{\frac{19}{45}} = \frac{2}{9} \times \frac{45}{19} = \frac{10}{19}$$

Step 4: Calculate the Final Integer Value

The question defines this probability as $$P$$:

$$P = \frac{10}{19}$$

We need to find the value of $$19P$$:

$$19P = 19 \times \frac{10}{19} = 10$$

Final Answer

10