A) \[\frac{9}{16}\]
B) \[\frac{1}{16}\]
C) \[\frac{9}{64}\]
D) None of these
Correct Answer: C
Solution :
The required probability is given by \[\frac{{}^{39}{{C}_{1}}}{{}^{52}{{C}_{1}}}\times \frac{{}^{39}{{C}_{1}}}{{}^{52}{{C}_{1}}}\times \frac{{}^{13}{{C}_{1}}}{{}^{52}{{C}_{1}}}=\frac{3}{4}\times \frac{3}{4}\times \frac{1}{4}=\frac{9}{64}.\]
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