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, is
Answer & explanation
Correct answer: option 3
Let A be the event of drawing a diamond card in the first draw and B be the event of drawing a diamond card in the second draw. Then,
$P(A)=\frac{^{13}C_1}{^{52}C_1}=\frac{13}{52}=\frac{1}{4}$
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 second draw when a diamond card has already been drawn in first draw
$P(B/A)=\frac{^{12}C_1}{^{51}C_1}=\frac{12}{52}=\frac{4}{17}$
Hence,
Required probability = $P( A ∩ B)$
⇒ Required probability = $P(A) P(B/A)$
⇒ Required probability = $\frac{1}{4}×\frac{4}{17}=\frac{1}{17}$