A) \[\frac{13}{32}\]
B) \[\frac{1}{4}\]
C) \[\frac{1}{32}\]
D) \[\frac{3}{16}\]
Correct Answer: A
Solution :
[a] P (2 white and 1 black) \[=P({{W}_{1}}{{W}_{2}}{{B}_{3}}\,\,or\,\,{{W}_{1}}{{B}_{2}}{{W}_{3}}\,\,or\,\,{{B}_{1}}{{W}_{2}}{{W}_{3}})\] \[=P({{W}_{1}}{{W}_{2}}{{B}_{3}})+P({{W}_{1}}{{B}_{2}}{{W}_{3}})+P({{B}_{1}}{{W}_{2}}{{W}_{3}})\] \[=P({{W}_{1}})P({{W}_{2}})P({{B}_{3}})+P({{W}_{1}})P({{B}_{2}})P({{W}_{3}})+\] \[P({{W}_{1}})({{W}_{2}})({{W}_{3}})\] \[=\frac{3}{4}.\frac{2}{4}.\frac{3}{4}+\frac{3}{4}.\frac{2}{4}.\frac{1}{4}+\frac{1}{4}.\frac{2}{4}.\frac{1}{4}=\frac{1}{32}(9+3+1)=\frac{13}{32}\]You need to login to perform this action.
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