A group of friends wanted to play cards with two identical packs together. While shuffling the cards, three cards are dropped. Rest of the cards are shuffled and one card is drawn at random. Assuming that the dropped cards were a queen of hearts, a ten of spades and an ace of clubs, answer the following questions:
Assuming that the dropped cards were a queen of hearts, a ten of spades and an ace of clubs, answer the following questions:
Generated by claude-sonnet-4-6 · 2026-06-15 10:31 · grounding stimulus
Model Answer
Given: Two identical packs = 104 cards total. Cards dropped: queen of hearts, ten of spades, ace of clubs → Remaining cards = 104 − 3 = 101 cards
(i) Probability of drawing a face card:
Two packs have 2 × 12 = 24 face cards. One queen is dropped, so face cards left = 23.
$$P(\text{face card}) = \frac{23}{101}$$
(ii) Probability of drawing a king or a queen:
Kings in two packs = 8; Queens in two packs = 8, but one queen dropped → queens left = 7.
Total kings or queens = 8 + 7 = 15.
$$P(\text{king or queen}) = \frac{15}{101}$$
(iii) If no cards were dropped, total cards = 104 and queens = 8.
$$P(\text{queen}) = \frac{8}{104} = \frac{1}{13} \approx 0.077$$
With dropped cards, queens = 7 and total = 101.
$$P(\text{queen}) = \frac{7}{101} \approx 0.069$$
Since $\frac{1}{13} > \frac{7}{101}$, yes, the probability of getting a queen was higher when no cards were dropped.
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Explanation
- Always start by calculating the new total (104 − 3 = 101) and updating counts for each suit/type after the drop.
- Face cards = Jack, Queen, King (12 per pack × 2 = 24 total).
- For part (iii), examiners expect both probabilities to be calculated and compared numerically — just saying "yes" without justification will lose a mark.
- Keep fractions unreduced unless simplification is straightforward, to avoid arithmetic errors.