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 at least one marble of each colour among the four chosen, is

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) \[n(E){{=}^{6}}{{C}_{2}}{{\cdot }^{5}}{{C}_{1}}{{\cdot }^{4}}{{C}_{1}}{{+}^{5}}{{C}_{2}}{{\cdot }^{6}}{{C}_{1}}{{\cdot }^{4}}{{C}_{1}}\] \[{{+}^{4}}{{C}_{2}}{{\cdot }^{6}}{{C}_{1}}{{\cdot }^{5}}{{C}_{1}}n(S){{=}^{15}}{{C}_{4}}\] \[\therefore \]\[P(E)=\frac{720.4!}{15\cdot 14\cdot 13\cdot 12}=\frac{48}{91}\]