**19.** Two balls are selected at random one by one without replacement from a bag containing 4 white and 6 black balls. If the probability that the first selected ball is black, given that the second selected ball is also black, is \( \frac{m}{n} \), where \( \gcd(m, n) = 1 \), then \( m + n \) is equal to: - (1) 14 - (2) 4 - (3) 11 - (4) 13

Key Ideas to Solve

1. Conditional Probability Formula
   If events $A$ and $B$ both can happen, then $P(A\mid B) = \frac{P(A \cap B)}{P(B)}$ Here,

   • $A$ = First ball is black (denote as $B_1$)
   • $B$ = Second ball is black (denote as $B_2$)

2. Without Replacement
   After you take one ball out, only 9 balls remain. So probabilities for the second draw depend on what happened in the first draw.

3. Step–by–Step Reasoning

   1. Find $P(B_1 \cap B_2)$ – the probability that both draws are black.
   2. Find $P(B_2)$ – the overall probability that the second draw is black (regardless of the first draw).
   3. Compute the ratio using the conditional‐probability formula.

Why each step?
Because conditional probability literally tells us: out of all the times the second ball is black, how many of those have the first ball black too?

What’s happening?

Imagine you have a bag with 10 balls: 6 are black, 4 are white. You pick one ball, keep it in your hand, then pick another without putting the first back.

The question asks: “If you already know the second ball came out black, what’s the chance the first ball was also black?”
So we are looking at: Probability(First is Black | Second is Black).

Think of it like peeking at the second card in a magic trick. Since you saw it’s black, how likely is it that the first card is black too? That’s all we need to figure out!

Step-by-Step Solution

1. Define Events
   $B_1$: first ball is black
   $B_2$: second ball is black

2. Find $P(B_1 \cap B_2)$

   After drawing one black, 5 black remain out of 9 balls.

   P(B_1 \cap B_2) &= P(\text{1st black}) \times P(\text{2nd black} \mid \text{1st black})\\[6pt] &= \frac{6}{10} \times \frac{5}{9} \\[6pt] &= \frac{30}{90} \\[6pt] &= \frac{1}{3} \end{aligned}$$

3. Find $P(B_2)$

   Two mutually exclusive ways give a black second ball.

   Case 1 (Black then Black): already computed.

   Case 2 (White then Black):

   $\frac{4}{10} \times \frac{6}{9} = \frac{24}{90} = \frac{4}{15}$

   Add both cases:

   $P(B_2) = \frac{1}{3} + \frac{4}{15} = \frac{5}{15} + \frac{4}{15} = \frac{9}{15} = \frac{3}{5}$

4. Compute the Conditional Probability

   P(B_1 \mid B_2) &= \frac{P(B_1 \cap B_2)}{P(B_2)}\\[6pt] &= \frac{\frac{1}{3}}{\frac{3}{5}} = \frac{1}{3} \times \frac{5}{3} = \frac{5}{9} \end{aligned}$$ So $\dfrac{m}{n} = \dfrac{5}{9}$ with $m + n = 5 + 9 = 14$.

Answer: 14 (Option 1).