Question

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:

• A. 14
• B. 4
• C. 11
• D. 13

Hint:

Remember the formula for conditional probability: \(P(A|B) = \frac{P(A \cap B)}{P(B)}\).

Answer 1

The Correct Option is A

To determine the probability \(P(A|B)\), where event \(A\) signifies the first ball drawn is black and event \(B\) signifies the second ball drawn is black, we employ the conditional probability formula:

\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]

Step 1: Calculate \(P(A \cap B)\).

The probability that both the first and second balls drawn are black is computed as follows:

• Probability of drawing a black ball first: \(\frac{6}{10}\).
• Given the first ball was black, the probability of drawing a second black ball from the remaining 9 balls (5 of which are black): \(\frac{5}{9}\).

Thus:

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

Step 2: Calculate \(P(B)\).

\(P(B)\) represents the probability that the second ball drawn is black, irrespective of the first ball's color. This can occur in two mutually exclusive ways:

• Scenario 1: First ball is black (probability \(\frac{6}{10}\)) AND second ball is black (probability \(\frac{5}{9}\)).
• Scenario 2: First ball is white (probability \(\frac{4}{10}\)) AND second ball is black (probability \(\frac{6}{9}\)).

Summing the probabilities of these scenarios:

\[ P(B) = \left(\frac{6}{10} \times \frac{5}{9}\right) + \left(\frac{4}{10} \times \frac{6}{9}\right) = \frac{30}{90} + \frac{24}{90} = \frac{54}{90} = \frac{3}{5} \]

Step 3: Compute \(P(A|B)\).

Applying the conditional probability formula with the computed values:

\[ P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{3}}{\frac{3}{5}} = \frac{1}{3} \times \frac{5}{3} = \frac{5}{9} \]

The fraction \(\frac{5}{9}\) is in its simplest form, as the greatest common divisor of 5 and 9 is 1. Therefore, \(m = 5\) and \(n = 9\), and \(m + n = 14\).