A) \[\frac{1}{42}\]
B) \[\frac{41}{42}\]
C) \[\frac{2}{63}\]
D) \[\frac{1}{7}\]
Correct Answer: B
Solution :
Total number of ways \[={}^{4}{{C}_{1}}\times {}^{6}{{C}_{4}}+{}^{4}{{C}_{2}}\times {}^{6}{{C}_{3}}+{}^{4}{{C}_{3}}\times {}^{6}{{C}_{2}}+{}^{4}{{C}_{4}}\times {}^{6}{{C}_{1}}+{}^{6}{{C}_{5}}\] \[=60+120+60+6+6=252\] No. of ways in which at least one woman exist are \[={}^{4}{{C}_{1}}\times {}^{6}{{C}_{4}}+{}^{4}{{C}_{2}}\times {}^{6}{{C}_{3}}+{}^{4}{{C}_{3}}\times {}^{6}{{C}_{2}}+{}^{4}{{C}_{4}}\times {}^{6}{{C}_{1}}=246\] Hence required probability \[=\frac{246}{252}=\frac{41}{42}\].
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