InterviewSolution
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InterviewSolution
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A box `B_(1)` contains 1 white ball, 3 red balls, and 2 black balls. An- other box `B_(2)` contains 2 white balls, 3 red balls. A third box `B_(3)` contains 3 white balls, 4 red balls, and 5 black balls. If 2 balls are drawn (without replecement) from a randomly selected box and one of the balls is white and the other ball is red the probability that these 2 balls are drawn from box `B_(2)` isA. `116//182`B. `126//181`C. `65//181`D. `55//181` |
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Answer» Correct Answer - D Let A: one ball is white and other is red `E_(1):` both balls are from box `B_(1),` `E_(2): ` both balls are from box `B_(2),` `E_(3):` both balls are form box `B_(3)` Hence, P (required) `=P((E_(2))/(A))` `=(P((A)/(E_(2))).P(E_(2)))/(P((A)/(E_(1))).P(E_(1))+P((A)/(E_(2))).P(E_(2))+P((A)/(E_(3))).P(E_(3)))` `=((""^(2)C_(1)xx""^(3)C_(1))/(""^(9)C_(2))xx1/3)/((""^(1)C_(1)xx""^(3)C_(1))/(""^(6)C_(2))xx1/3+(""^(2)C_(1)xx""^(3)C_(1))/(""^(9)C_(2))xx1/3+(""^(3)C_(1)xx""^(4)C_(1))/(""^(12)C_(2))xx1/3)` `(1/6)/(1/5+1/6+2/11)=55/181` |
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