A) \[\frac{48}{91}\]
B) \[\frac{44}{91}\]
C) \[\frac{88}{91}\]
D) \[\frac{24}{91}\]
Correct Answer: A
Solution :
P (E) = P(R R B W or B B R W or W W R B) \[n(E){{=}^{6}}{{C}_{2}}{{\cdot }^{5}}{{C}_{1}}{{\cdot }^{4}}{{C}_{1}}{{+}^{5}}{{C}_{2}}{{\cdot }^{6}}{{C}_{1}}{{\cdot }^{4}}{{C}_{1}}\] \[{{+}^{4}}{{C}_{2}}{{\cdot }^{6}}{{C}_{1}}{{\cdot }^{5}}{{C}_{1}}n(S){{=}^{15}}{{C}_{4}}\] \[\therefore \]\[P(E)=\frac{720.4!}{15\cdot 14\cdot 13\cdot 12}=\frac{48}{91}\]You need to login to perform this action.
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