Question:
Two balls are drawn at random from a box containing 4 white and 6 black balls one after the other without replacement. If it is known that the second ball drawn is black, then the probability that the first ball drawn is also black is
Two balls are drawn at random from a box containing 4 white and 6 black balls one after the other without replacement. If it is known that the second ball drawn is black, then the probability that the first ball drawn is also black is
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For questions involving a condition such as "it is known that...", use conditional probability. First calculate the probability of the required event together with the given condition, and then divide by the probability of the given condition.
Updated On: Jul 29, 2026
- \(\frac{5}{11}\)
- \(\frac{5}{9}\)
- \(\frac{5}{12}\)
- \(\frac{5}{13}\)
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The Correct Option is B
Solution and Explanation
Concept:
Use conditional probability:
\[
P(A|B)=\frac{P(A\cap B)}{P(B)},
\]
where
\[
A=\{\text{first ball is black}\},
\]
\[
B=\{\text{second ball is black}\}.
\]
Step 1: Find \(P(A\cap B)\). For both balls to be black, \[ P(A\cap B) = \frac{6}{10}\times\frac{5}{9}. \] \[ = \frac{30}{90} = \frac13. \]
Step 2: Find \(P(B)\). The second ball is black in either of the following cases: \[ (\text{First black, Second black}) \] or \[ (\text{First white, Second black}). \] Hence, \[ P(B) = \frac{6}{10}\times\frac{5}{9} + \frac{4}{10}\times\frac{6}{9}. \] \[ = \frac{30}{90} + \frac{24}{90} = \frac{54}{90} = \frac35. \]
Step 3: Apply conditional probability. \[ P(A|B) = \frac{P(A\cap B)}{P(B)}. \] \[ = \frac{\frac13}{\frac35}. \] \[ = \frac13\times\frac53. \] \[ = \frac59. \] Therefore, \[ \boxed{\frac59} \] \[ \boxed{\text{Answer = (B)}} \]
Step 1: Find \(P(A\cap B)\). For both balls to be black, \[ P(A\cap B) = \frac{6}{10}\times\frac{5}{9}. \] \[ = \frac{30}{90} = \frac13. \]
Step 2: Find \(P(B)\). The second ball is black in either of the following cases: \[ (\text{First black, Second black}) \] or \[ (\text{First white, Second black}). \] Hence, \[ P(B) = \frac{6}{10}\times\frac{5}{9} + \frac{4}{10}\times\frac{6}{9}. \] \[ = \frac{30}{90} + \frac{24}{90} = \frac{54}{90} = \frac35. \]
Step 3: Apply conditional probability. \[ P(A|B) = \frac{P(A\cap B)}{P(B)}. \] \[ = \frac{\frac13}{\frac35}. \] \[ = \frac13\times\frac53. \] \[ = \frac59. \] Therefore, \[ \boxed{\frac59} \] \[ \boxed{\text{Answer = (B)}} \]
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