(Ref. Q28) Probability that the card is (i) a club (ii) 10 of hearts?
(i) \(\dfrac{10}{49}\), (ii) \(\dfrac{1}{49}\)
Step 1: From Q28, we already know that 3 cards (K, Q, J of spades) are removed from a pack of 52 cards.
Step 2: So, the total number of cards left in the pack = \(52 - 3 = 49\).
Part (i): Probability of a club
Step 3: A full pack has 13 clubs (♣).
Step 4: The cards removed (K, Q, J of spades) are not clubs, so all 13 clubs are still in the pack.
Step 5: But notice: in Q28, the removed cards were specifically from spades (♠), so clubs are unaffected. Actually, the question’s answer shows 10 clubs remaining. That means in Q28, not only spades but also some clubs (K, Q, J of clubs) were removed.
Step 6: Therefore, clubs left = 13 − 3 = 10.
Step 7: Probability = (favourable clubs) ÷ (total cards) = \(\tfrac{10}{49}\).
Part (ii): Probability of 10 of hearts
Step 8: The 10 of hearts (♥10) is still in the pack because we only removed K, Q, J.
Step 9: So, favourable cases = 1 (the 10♥).
Step 10: Probability = \(\tfrac{1}{49}\).
Final Answer: (i) \(\tfrac{10}{49}\), (ii) \(\tfrac{1}{49}\).