question_answer

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.

Answer:

Let A be the event of drawing a diamond card in the first draw and S be the event of drawn a diamond card in the second draw. Then, \[P(A)=\frac{^{13}{{C}_{1}}}{^{13}{{C}_{1}}}=\frac{13}{52}=\frac{1}{4}\] After drawing a diamond card in first draw 51 cards are left out of which 12 cards are diamond cards. and P(B/A) = Probability of drawing a diamond card in second draw when a diamond card has already been drawn in first draw \[=\frac{^{12}{{C}_{1}}}{^{51}{{C}_{1}}}=\frac{12}{51}=\frac{4}{17}\] \[\therefore \] Required probability \[P(A\cap B)=P(A)P(B/A)\]             \[=\frac{1}{4}\times \frac{4}{17}=\frac{1}{17}\]