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)=^{13}c_{1}/^{13}c_{1} =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)=^{12}c_{1}/^{51}c_{1}=4/17
Now, Required probability =P(A∩B)=P(A)P(B/A)=1/4 x 4/17=1/17