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 : (i) Find the probability that the drawn card is a face card. (ii) Find the probability that the drawn card is either a king or a queen. (iii)(a) Do you think that the probability of getting a queen was higher if none of the cards were dropped? Justify your answer. OR (iii)(b) Find the probability that the drawn card is a jack. Compare it with the probability when none of the cards were dropped. In which case is the probability of getting a jack higher?
Show Hint
When comparing fractions like \(\frac{8}{101}\) and \(\frac{8}{104}\), remember that if the numerators are identical, the fraction with the smaller denominator is always larger.
Step 1: Understanding the Question:
This case-study problem involves a combined deck consisting of two identical standard packs of playing cards.
- One pack has \(52\) cards, so two packs have \(52 \times 2 = 104\) cards.
- Three specific cards are dropped from the total: a Queen of Hearts, a Ten of Spades, and an Ace of Clubs.
- The remaining number of cards is \(104 - 3 = 101\).
We need to answer probability questions based on this remaining deck.
Step 2: Key Formula or Approach:
Use the standard probability definition:
\[ P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of remaining outcomes}} \]
Step 3: Detailed Explanation:
1. Total remaining cards:
- Two full decks have \(104\) cards. Since \(3\) cards dropped, total possible outcomes \(N = 101\).
2. Part (i): Probability of a face card:
- Face cards are Kings, Queens, and Jacks. There are \(3 \times 4 = 12\) face cards in one deck, and thus \(24\) in two decks.
- One face card (Queen of Hearts) was dropped.
- Favorable outcomes = \(24 - 1 = 23\).
- Probability = \(\frac{23}{101}\).
3. Part (ii): Probability of either a King or a Queen:
- Total Kings and Queens in two decks = \(8\text{ Kings} + 8\text{ Queens} = 16\).
- One Queen (Queen of Hearts) was dropped.
- Favorable outcomes = \(16 - 1 = 15\).
- Probability = \(\frac{15}{101}\).
4. Part (iii)(a): Probability of Queen with and without dropped cards:
- Case 1 (No cards dropped):
- Total cards = \(104\), Queens = \(8\).
- \(P(\text{Queen}) = \frac{8}{104} = \frac{1}{13} \approx 0.0769\).
- Case 2 (3 cards dropped):
- Total cards = \(101\), Queens = \(7\) (since one Queen dropped).
- \(P(\text{Queen}) = \frac{7}{101} \approx 0.0693\).
- Comparison: Since \(0.0769 \gt 0.0693\), the probability of getting a queen was indeed higher if no cards were dropped.
5. Part (iii)(b) (Alternative): Probability of Jack with and without dropped cards:
- Case 1 (No cards dropped):
- Total cards = \(104\), Jacks = \(8\).
- \(P(\text{Jack}) = \frac{8}{104} \approx 0.0769\).
- Case 2 (3 cards dropped):
- Total cards = \(101\), Jacks = \(8\) (since no Jacks were dropped).
- \(P(\text{Jack}) = \frac{8}{101} \approx 0.0792\).
- Comparison: Since \(\frac{8}{101} \gt \frac{8}{104}\), the probability of getting a Jack is higher in the case where the three cards were dropped.
Step 4: Final Answer:
(i) The probability of drawing a face card is \(\frac{23}{101}\).
(ii) The probability of drawing either a King or a Queen is \(\frac{15}{101}\).
(iii)(a) Yes, the probability of drawing a queen is higher if no cards are dropped (\(\frac{8}{104} \gt \frac{7}{101}\)).
(iii)(b) The probability of drawing a Jack is \(\frac{8}{101}\). This is higher than when no cards are dropped (\(\frac{8}{101} \gt \frac{8}{104}\)).
