Law of Total Probability

We need to find $$ P(\text{first black} \mid \text{second black}) $$

We know, for conditional probability, $$ P(A|B) = \frac{P(A \cap B)}{P(B)} $$

Let, $$ A$$ be the event that the first ball is black, and $$ B$$ be that the second ball is black

So, probability both balls are black (without replacement) is $$ P(A \cap B) = \dfrac{6}{10} \cdot \dfrac{5}{9} = \dfrac{1}{3} $$

Probability second ball is black is 

$$ P(B) = P(\text{second black | first black}) \cdot P(\text{first black}) + P(\text{second black | first white}) \cdot P(\text{first white}) $$

$$ = \dfrac{5}{9} \cdot \dfrac{6}{10} + \dfrac{6}{9} \cdot \dfrac{4}{10} = \dfrac{30}{90} + \dfrac{24}{90} = \dfrac{3}{5} $$

Now,  $$ P(A|B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{1/3}{3/5} = \dfrac{5}{9} $$

Thus, $$ m+n = 5+9 =14$$ .

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

• $$R_1$$: The first ball drawn is Red.
• $$B_1$$: The first ball drawn is Black.
• $$B_2$$: The second ball drawn is Black.

We need to find the conditional probability $$P(R_1 | B_2)$$. By Bayes' Theorem,

$$P(R_1 | B_2) = \frac{P(R_1) \cdot P(B_2 | R_1)}{P(B_2)}$$

Initially, the urn has 3 Red and 7 Black balls (Total = 10 balls).

• $$P(R_1) = \frac{3}{10}$$
• $$P(B_1) = \frac{7}{10}$$

• Case 1: First ball was Red ($$R_1$$)
  • One red ball is removed (leaving 2 Red, 7 Black).
  • Two additional red balls are added (leaving 2 + 2 = 4 Red, 7 Black).
  • Total balls now = 11.
  • $$P(B_2 | R_1) = \frac{7}{11}$$
• Case 2: First ball was Black ($$B_1$$)
  • One black ball is removed (leaving 3 Red, 6 Black).
  • Two additional black balls are added (leaving 3 Red, 6 + 2 = 8 Black).
  • Total balls now = 11.
  • $$P(B_2 | B_1) = \frac{8}{11}$$

Using the Law of Total Probability:

$$P(B_2) = P(R_1) \cdot P(B_2 | R_1) + P(B_1) \cdot P(B_2 | B_1)$$

$$P(B_2) = \left(\frac{3}{10} \times \frac{7}{11}\right) + \left(\frac{7}{10} \times \frac{8}{11}\right)$$

$$P(B_2) = \frac{21}{110} + \frac{56}{110} = \frac{77}{110}$$

$$P(R_1 | B_2) = \frac{\frac{21}{110}}{\frac{77}{110}}$$

$$P(R_1 | B_2) = \frac{21}{77} = \frac{3}{11}$$