Question:
A box \(P\) contains 3 white and 7 red balls. A bag \(Q\) contains 4 green and 5 blue balls. Two balls are randomly drawn from box \(P\). If both are of the same color, one ball is drawn from bag \(Q\), and if the two balls are of different colors, 2 balls are drawn from bag \(Q\). If it is known that there is exactly one green ball among the balls drawn from bag \(Q\), then the probability that the two balls drawn from box \(P\) are of different colors is
A box \(P\) contains 3 white and 7 red balls. A bag \(Q\) contains 4 green and 5 blue balls. Two balls are randomly drawn from box \(P\). If both are of the same color, one ball is drawn from bag \(Q\), and if the two balls are of different colors, 2 balls are drawn from bag \(Q\). If it is known that there is exactly one green ball among the balls drawn from bag \(Q\), then the probability that the two balls drawn from box \(P\) are of different colors is
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When a condition is given after an experiment, Bayes' theorem is usually the right approach. First calculate the conditional probabilities for the given event under each possible case, then update the probability accordingly.
Updated On: Jul 29, 2026
- \(\frac{35}{67}\)
- \(\frac{21}{62}\)
- \(\frac{20}{43}\)
- \(\frac{32}{67}\)
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The Correct Option is A
Solution and Explanation
Concept:
Use Bayes' theorem:
\[
P(A|B)=\frac{P(A)P(B|A)}
{P(A)P(B|A)+P(A')P(B|A')}.
\]
Step 1: Define the events. Let \[ A=\{\text{two balls drawn from }P\text{ are of different colors}\}, \] \[ A'=\{\text{two balls drawn from }P\text{ are of the same color}\}. \] Let \[ B=\{\text{exactly one green ball is drawn from }Q\}. \] We need \[ P(A|B). \]
Step 2: Find \(P(A)\) and \(P(A')\). Total ways of drawing 2 balls from box \(P\): \[ {10\choose2}=45. \] Different colors: \[ {3\choose1}{7\choose1}=21. \] Hence, \[ P(A)=\frac{21}{45}=\frac{7}{15}. \] Therefore, \[ P(A')=1-\frac{7}{15} =\frac{8}{15}. \]
Step 3: Find \(P(B|A)\). If \(A\) occurs, 2 balls are drawn from \(Q\). Exactly one green ball means \[ {4\choose1}{5\choose1} \] favourable ways. Total ways: \[ {9\choose2}. \] Thus, \[ P(B|A) = \frac{{4\choose1}{5\choose1}} {{9\choose2}} = \frac{20}{36} = \frac{5}{9}. \]
Step 4: Find \(P(B|A')\). If \(A'\) occurs, only one ball is drawn from \(Q\). To have exactly one green ball, that ball must be green. Hence, \[ P(B|A') = \frac{4}{9}. \]
Step 5: Apply Bayes' theorem. \[ P(A|B) = \frac{P(A)P(B|A)} {P(A)P(B|A)+P(A')P(B|A')}. \] \[ = \frac{\frac{7}{15}\cdot\frac{5}{9}} {\frac{7}{15}\cdot\frac{5}{9} +\frac{8}{15}\cdot\frac{4}{9}}. \] \[ = \frac{35}{35+32}. \] \[ = \frac{35}{67}. \] Therefore, \[ \boxed{\frac{35}{67}} \] \[ \boxed{\text{Answer = (A)}} \]
Step 1: Define the events. Let \[ A=\{\text{two balls drawn from }P\text{ are of different colors}\}, \] \[ A'=\{\text{two balls drawn from }P\text{ are of the same color}\}. \] Let \[ B=\{\text{exactly one green ball is drawn from }Q\}. \] We need \[ P(A|B). \]
Step 2: Find \(P(A)\) and \(P(A')\). Total ways of drawing 2 balls from box \(P\): \[ {10\choose2}=45. \] Different colors: \[ {3\choose1}{7\choose1}=21. \] Hence, \[ P(A)=\frac{21}{45}=\frac{7}{15}. \] Therefore, \[ P(A')=1-\frac{7}{15} =\frac{8}{15}. \]
Step 3: Find \(P(B|A)\). If \(A\) occurs, 2 balls are drawn from \(Q\). Exactly one green ball means \[ {4\choose1}{5\choose1} \] favourable ways. Total ways: \[ {9\choose2}. \] Thus, \[ P(B|A) = \frac{{4\choose1}{5\choose1}} {{9\choose2}} = \frac{20}{36} = \frac{5}{9}. \]
Step 4: Find \(P(B|A')\). If \(A'\) occurs, only one ball is drawn from \(Q\). To have exactly one green ball, that ball must be green. Hence, \[ P(B|A') = \frac{4}{9}. \]
Step 5: Apply Bayes' theorem. \[ P(A|B) = \frac{P(A)P(B|A)} {P(A)P(B|A)+P(A')P(B|A')}. \] \[ = \frac{\frac{7}{15}\cdot\frac{5}{9}} {\frac{7}{15}\cdot\frac{5}{9} +\frac{8}{15}\cdot\frac{4}{9}}. \] \[ = \frac{35}{35+32}. \] \[ = \frac{35}{67}. \] Therefore, \[ \boxed{\frac{35}{67}} \] \[ \boxed{\text{Answer = (A)}} \]
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