A) \[\frac{4}{13}\]
B) \[\frac{1}{2}\]
C) \[\frac{1}{4}\]
D) \[\frac{3}{4}\]
Correct Answer: A
Solution :
Number of ways to get a king \[{{=}^{4}}{{C}_{1}}\] Number of ways to get a spade \[{{=}^{13}}{{C}_{1}}\] Number of ways to get both \[{{=}^{1}}{{C}_{1}}\] \[\therefore \]Required probability\[=\frac{^{4}{{C}_{1}}{{+}^{13}}{{C}_{1}}{{-}^{1}}{{C}_{1}}}{^{52}{{C}_{1}}}\] \[=\frac{16}{52}=\frac{4}{13}\]
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