Urn Probability: Drawing Two Black Balls Without Replacement

The problem involves calculating the probability of a sequence of dependent events (drawing balls without replacement).

Initial Setup

• Total balls in the urn: 10 black + 5 white = 15 balls.
• We need to find the probability of drawing two black balls consecutively without putting the first ball back.

Step 1: Probability of the First Ball Being Black

The probability of the first ball drawn being black is the ratio of the number of black balls to the total number of balls.

Let $P(B_1)$ be the probability that the first ball drawn is black.

$P(B_1) = \frac{\text{Number of black balls}}{\text{Total number of balls}} = \frac{10}{15}$

Step 2: Probability of the Second Ball Being Black

After drawing one black ball without replacement, there are now 14 balls left in the urn, and 9 of them are black.

Let $P(B_2|B_1)$ be the probability that the second ball drawn is black, given that the first ball drawn was black.

$P(B_2|B_1) = \frac{\text{Number of remaining black balls}}{\text{Total number of remaining balls}} = \frac{9}{14}$

Step 3: Probability of Both Balls Being Black

The probability of both events happening in sequence is the product of their individual probabilities (using the rule for conditional probability $P(A \cap B) = P(A) \times P(B|A)$).

Let $P(B_1 \cap B_2)$ be the probability that both balls drawn are black.

$P(B_1 \cap B_2) = P(B_1) \times P(B_2|B_1)$

$P(B_1 \cap B_2) = \frac{10}{15} \times \frac{9}{14}$

Simplify the fractions:

$P(B_1 \cap B_2) = \frac{2}{3} \times \frac{9}{14}$

$P(B_1 \cap B_2) = \frac{2 \times 9}{3 \times 14} = \frac{18}{42}$

Reduce the final fraction to its simplest form:

$P(B_1 \cap B_2) = \frac{18 \div 6}{42 \div 6} = \frac{3}{7}$

Conclusion

The probability that both drawn balls are black is $\frac{3}{7}$.