A) \[\frac{1}{5}\]
B) \[\frac{3}{16}\]
C) \[\frac{9}{20}\]
D) \[\frac{1}{9}\]
Correct Answer: C
Solution :
Total cards in the pack =16 Number of ways of selecting two cards randomly\[{{=}^{16}}{{C}_{2}}\] \[\therefore \]Probability of getting atleast one ace \[=\frac{^{4}{{C}_{1}}{{\times }^{12}}{{C}_{1}}}{^{16}{{C}_{2}}}+\frac{^{4}{{C}_{2}}}{^{16}{{C}_{2}}}\] \[=\frac{4\times 12\times 2}{16\times 15}+\frac{4\times 3}{16\times 15}\] \[=\frac{96+12}{16\times 15}=\frac{108}{240}\] \[=\frac{9}{20}\]
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