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 (m/n), where gcd(m, n) = 1, then m + n is equal to : (1) 14 (2) 4 (3) 11 (4) 13

Key Concepts You Must Know

1. Total balls: 4 white + 6 black = 10 balls.
2. Without replacement: After drawing the first ball, we don’t put it back, so only 9 balls remain for the second draw.
3. Conditional Probability:
   $P(A\mid B) = \frac{P(A \cap B)}{P(B)}$
   Here,
   • $A$ is the event “first ball is black.”
   • $B$ is the event “second ball is black.”
4. Order symmetry: In any random two-draw situation without replacement, the probability for the 1st draw being black equals the probability for the 2nd draw being black because each position is equally likely to be any one of the original balls.

Logical Chain a Student Should Follow

1. Compute $P(B)$ – the chance the second ball is black.
   Instead of a long calculation, just recall the symmetry idea: it is the same as the chance that any single randomly chosen ball from 10 is black, which is $6/10$.
2. Compute $P(A \cap B)$ – the chance both first and second balls are black.
   • Probability first ball is black: $6/10$.
   • Given the first was black, 5 black out of 9 balls remain: $5/9$.
   • Multiply: $6/10 \times 5/9$.
3. Apply conditional formula: divide $P(A \cap B)$ by $P(B)$.
4. Simplify the fraction and get the answer in lowest terms.

What is the question?

We have 10 balls in a bag: 4 white and 6 black. We pick 2 balls one after another without putting the first one back.

The question asks: “If we already know that the second ball we picked is black, what is the chance that the first ball we picked was also black?”

Think of it like this

Imagine you and your friend take two candies one after another from a box. You peek at the second candy and see it is black. Now you wonder: “Was the candy my friend took first also black?”
We want to find that chance.

To do that we use a special "if-this-then-what" rule in probability called conditional probability. It is like asking, “Out of all the cases where the second candy is black, in how many of those cases was the first candy black?”

Step-by-Step Calculation

1. Define events
   • $A$: first ball is black.
   • $B$: second ball is black.

2. Probability that both balls are black

   $$P(A \cap B) = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3}$$

3. Probability that the second ball is black (symmetry argument):

   $$P(B) = \frac{6}{10} = \frac{3}{5}$$

4. Conditional probability

   $$P(A\mid B) = \frac{P(A \cap B)}{P(B)} = \frac{\tfrac{1}{3}}{\tfrac{3}{5}} = \frac{1}{3} \times \frac{5}{3} = \frac{5}{9}$$

5. Simplest form: $\frac{5}{9}$ (gcd = 1).
   So $m = 5$, $n = 9$, giving $m + n = 5 + 9 = 14$.

Final Answer: 14 (Option 1)