A) \[\frac{7}{{{\,}^{11}}{{C}_{7}}}\]
B) \[\frac{{{\,}^{5}}{{C}_{3}}+{{\,}^{6}}{{C}_{4}}}{{{\,}^{11}}{{C}_{7}}}\]
C) \[\frac{{{\,}^{5}}{{C}_{2}}+{{\,}^{6}}{{C}_{2}}}{{{\,}^{11}}{{C}_{7}}}\]
D) \[\frac{{{\,}^{6}}{{C}_{3}}+{{\,}^{5}}{{C}_{4}}}{{{\,}^{11}}{{C}_{7}}}\]
Correct Answer: C
Solution :
Number of ways to get 3 white and 4 green balls from 5 white and 6 green balls \[={{\,}^{5}}{{C}_{3}}{{\times }^{6}}{{C}_{4}}{{=}^{5}}{{C}_{2}}{{\times }^{6}}{{C}_{2}}\] and total number of ways\[={{\,}^{11}}{{C}_{7}}\] \[\therefore \] Required probability \[=\frac{{{\,}^{5}}{{C}_{2}}\times {{\,}^{6}}{{C}_{2}}}{{{\,}^{11}}{{C}_{7}}}\]
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