A) \[\frac{48}{91}\]
B) \[\frac{33}{91}\]
C) \[\frac{88}{91}\]
D) \[\frac{24}{91}\]
Correct Answer: A
Solution :
\[P(E)=P(RRBW\,\,or\,\,BBRW\,\,or\,\,WWRB)\] \[n(E){{=}^{6}}{{C}_{2}}{{\cdot }^{5}}{{C}_{1}}{{\cdot }^{4}}{{C}_{1}}{{+}^{5}}{{C}_{2}}{{\cdot }^{6}}{{C}_{1}}{{\cdot }^{4}}{{C}_{1}}{{+}^{4}}{{C}_{2}}{{\cdot }^{6}}{{C}_{1}}{{\cdot }^{5}}{{C}_{1}}\]\[n(S){{=}^{15}}{{C}_{4}}\] \[\therefore \]\[P(E)=\frac{720\cdot 4!}{15\cdot 14\cdot 13\cdot 12}=\frac{48}{91}\]
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