Total probability 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

We use the theorem of total probability by conditioning on the outcome of the die roll, calculating the conditional probabilities using combinations, and evaluating the final sum.

1. Let $$E_n$$ be the event that the die shows the number $$n$$. Since the die is fair, the probability of rolling any specific face is constant:
$$P(E_n) = \frac{1}{6} \quad \text{for each } n \in \{1, 2, 3, 4, 5, 6\}$$

2. Let $$W$$ be the event that both drawn balls are white. Analyze the composition of the bag after rolling a number $$n$$:
* Initial balls: 2 white, 2 black
* Added balls: $$n$$ black
* Total balls in the bag: $$2 + 2 + n = 4 + n$$
* Total white balls: 2

3. Calculate the conditional probability $$P(W|E_n)$$ of drawing 2 white balls simultaneously from this composition:
The number of ways to choose 2 white balls out of 2 is $$\binom{2}{2} = 1$$.
The total number of ways to choose any 2 balls out of $$(4+n)$$ is $$\binom{4+n}{2}$$.
$$P(W|E_n) = \frac{1}{\binom{4+n}{2}} = \frac{1}{\frac{(4+n)(3+n)}{2}} = \frac{2}{(n+3)(n+4)}$$

4. Use partial fraction decomposition to simplify the conditional probability expression for easier summation:
$$\frac{2}{(n+3)(n+4)} = 2 \left( \frac{1}{n+3} - \frac{1}{n+4} \right)$$

5. Apply the Theorem of Total Probability to find the absolute probability $$P(W)$$:
$$P(W) = \sum_{n=1}^{6} P(E_n) \cdot P(W|E_n)$$
$$P(W) = \sum_{n=1}^{6} \frac{1}{6} \cdot 2 \left( \frac{1}{n+3} - \frac{1}{n+4} \right) = \frac{1}{3} \sum_{n=1}^{6} \left( \frac{1}{n+3} - \frac{1}{n+4} \right)$$

6. Expand the summation to evaluate it as a telescoping series:
$$\text{Sum} = \left(\frac{1}{4} - \frac{1}{5}\right) + \left(\frac{1}{5} - \frac{1}{6}\right) + \dots + \left(\frac{1}{9} - \frac{1}{10}\right)$$
All the middle terms cancel out perfectly, leaving only the first and last components:
$$\text{Sum} = \frac{1}{4} - \frac{1}{10} = \frac{5 - 2}{20} = \frac{3}{20}$$

7. Compute the final value of the total probability $$P$$:
$$P = \frac{1}{3} \cdot \frac{3}{20} = \frac{1}{20}$$

8. Calculate the requested reciprocal value:
$$\frac{1}{P} = 20$$