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.

Options

A. `(4)/(17)`
B. `(13)/(17)`
C. `(1)/(17)`
D. none of these

Answer

C

Explanation

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)=(.^(13)C_(1))/(.^(52)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.
`therefore 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)=(.^(12)C_(1))/(.^(51)C_(1))=(12)/(51)=(4)/(17)`
Hence,
Required probability `=P(A cap B)`
`implies` Required probability `=P(A)P(B//A)`
`implies` Required probability `=(1)/(4)xx(4)/(17)=(1)/(17)`