Total number of card = 5
(a) number of queen = 1.
therefore, P(getting a queen) = \(\frac{number\, of\,favorable\,outcomes}{number\,of \,all\,possible\,outcomes}\) = \(\frac{1}{5}\)
Thus, the probability that the drawn card is the queen is \(\frac{1}{5}\)
(b) When the queen has put aside, number remaining cards = 4
(i) The number of aces = 1.
Therefore, P(getting an ace) = \(\frac{number\, of\,favorable\,outcomes}{number\,of \,all\,possible\,outcomes}\) = \(\frac{1}{4}\)
Thus, the probability that the drawn card is an ace is \(\frac{1}{4}\).
(ii) Number of queen = 0
Therefore, P(getting a queen now) = \(\frac{number\, of\,favorable\,outcomes}{number\,of \,all\,possible\,outcomes}\) = \(\frac{0}{4}\) = 0
Thus, the probability that the drawn card is a queen is 0.