Find the probability of drawing a diamond card in each of the two consecutive draws from a well shuffled pack of cards, if the card drawn is not replaced after the first draw

A) 1/16    B) 1/17     C) 1/18     D) 1/19

Let A be the event of drawing a diamond card in the first draw and B be the even of drawing a diamond card in the second draw. Then,

P(A)=¹³c₁/¹³c₁ =13/52=1/4

After drawing a diamond card in first draw 51 cards are left out of which 12 cards are diamond cards.

∴P(B/A)=Probability of  drawing a diamond card in the second draw  when a diamond card has already been drawn in first draw

?P(B/A)=¹²c₁/⁵¹c₁=4/17

Now, Required probability =P(A∩B)=P(A)P(B/A)=1/4 x 4/17=1/17