A person draws a card from a pack of 52 playing cards, replaces it and shuffles the pack. He continues doing this until he draws a spade, the chance that he will fail in the first two draws is

$\frac{9}{16}$

Let $A_i$ be the event that the person ails in i^th draw.

Then,

$P(A_i)= 1 -\frac{13}{52}=\frac{3}{4}, i = 1, 2, 3, .....$

∴ Required probability $= P(A_1 ∩ A_2)= P(A_1)P(A_2)$

⇒ Required probability $=\frac{3}{4}× \frac{3}{4}=\frac{9}{16}$