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:
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Remember the formula for conditional probability: \(P(A|B) = \frac{P(A \cap B)}{P(B)}\).
Let B1 be the event that the first ball is black. Let B2 be the event that the second ball is black. We want to find \(P(B1 | B2)\), which is the probability that the first ball is black given that the second ball is black.
Step 2: Calculate the probabilities.
Total number of balls = 4 white + 6 black = 10 balls. \(P(B1 \cap B2)\) is the probability that both balls are black. \(P(B1 \cap B2) = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3}\) \(P(B2)\) is the probability that the second ball is black. This can happen in two ways: 1. First ball is white and second ball is black: \(P(W1 \cap B2) = \frac{4}{10} \times \frac{6}{9} = \frac{24}{90} = \frac{4}{15}\) 2. First ball is black and second ball is black: \(P(B1 \cap B2) = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3}\) So, \(P(B2) = P(W1 \cap B2) + P(B1 \cap B2) = \frac{24}{90} + \frac{30}{90} = \frac{54}{90} = \frac{3}{5}\)
Step 3: Apply conditional probability formula.
\(P(B1 | B2) = \frac{P(B1 \cap B2)}{P(B2)} = \frac{\frac{30}{90}}{\frac{54}{90}} = \frac{30}{54} = \frac{5}{9}\) So, \(\frac{m}{n} = \frac{5}{9}\). Since gcd(5, 9) = 1, m = 5 and n = 9. Therefore, m + n = 5 + 9 = 14.
