Consider a standard 52 card deck of playing cards. In total there are four cards that are Aces, four cards that are Kings, four cards that are Queens and four cards that are Jacks. The remaining 36 cards are four each of the numbers 2, 3, . . . , 10. That is there are four cards that are twos, four cards that are threes etc. For this question, suppose that we reduce the number of cards in the deck by

• removing two of the Queens

• removing three other cards that are not Queens

The cards that are removed are discarded and are not used for the remainder of this question. As such we now have a deck that consists of just 47 cards.

(a) Suppose that a card is randomly drawn from this reduced sized deck. Let Q1 denote the event that this card is a Queen. What is P = (Q c/1)?

(b) The card that was drawn from the deck of cards in (a) is now discarded and we continue with a deck of just 46 cards. Suppose that a second card is now randomly drawn from this 46-card deck and let Q2 denote the event that this card is a Queen. Given that the first card drawn was not a Queen, what is the probability that this second card is also not a Queen? That is, using our notation, what is P = (Q c/2 | Q c/1)?

(c) What is P = (Q c/2 and Q c/1)?= P(neither card is a Queen)?

(d) What is P(at least one of the cards is a Queen)?