question_answer

A box contains \[6\] red, \[5\] blue and \[4\] white marbles. Four marbles are chosen at random without replacement. The probability that there is atleast one marble of each colour among the four chosen, is -

A) \[\frac{48}{91}\]                                

B) \[\frac{33}{91}\]

C) \[\frac{88}{91}\]                                

D) \[\frac{24}{91}\]

Correct Answer: A

Solution :

\[P(E)=P(RRBW\,\,or\,\,BBRW\,\,or\,\,WWRB)\] \[n(E){{=}^{6}}{{C}_{2}}{{\cdot }^{5}}{{C}_{1}}{{\cdot }^{4}}{{C}_{1}}{{+}^{5}}{{C}_{2}}{{\cdot }^{6}}{{C}_{1}}{{\cdot }^{4}}{{C}_{1}}{{+}^{4}}{{C}_{2}}{{\cdot }^{6}}{{C}_{1}}{{\cdot }^{5}}{{C}_{1}}\]\[n(S){{=}^{15}}{{C}_{4}}\] \[\therefore \]\[P(E)=\frac{720\cdot 4!}{15\cdot 14\cdot 13\cdot 12}=\frac{48}{91}\]