Case Study - 3 : 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 a card of a specific group is dropped, both the favorable outcomes and total outcomes decrease, reducing its probability.
When a card of an unrelated group is dropped, the favorable outcomes remain constant while the total outcomes decrease, increasing its probability.
This simple concept helps you quickly answer comparison questions without doing long division.
Step 1: Understanding the Question:
The topic of this question is Probability.
We are dealing with two identical packs of playing cards combined.
Initially, total cards in two packs is:
\[ \text{Total Cards} = 2 \times 52 = 104 \]
Three specific cards are dropped:
1. Queen of Hearts (red face card, queen)
2. Ten of Spades (black number card)
3. Ace of Clubs (black ace)
The remaining number of cards in the sample space is:
\[ n(S) = 104 - 3 = 101 \]
Step 2: Key Formula or Approach:
The probability of drawing a specific card is:
\[ P(E) = \frac{\text{Number of remaining favorable cards}}{\text{Total remaining cards (101)}} \]
Step 3: Detailed Explanation:
1. Part (i): Find the probability that the drawn card is a face card:
In a single deck, there are 12 face cards (Jacks, Queens, Kings of 4 suits).
In two identical decks, total face cards initially:
\[ 2 \times 12 = 24 \]
One face card (Queen of Hearts) was dropped.
Favorable face cards remaining:
\[ 24 - 1 = 23 \]
\[ P(\text{Face Card}) = \frac{23}{101} \approx 0.228 \]
2. Part (ii): Find the probability that the drawn card is either a king or a queen:
Initially, total kings and queens in two decks:
- Kings: \( 2 \times 4 = 8 \)
- Queens: \( 2 \times 4 = 8 \)
- Total: \( 8 + 8 = 16 \)
One queen (Queen of Hearts) was dropped, and no kings were dropped.
Favorable cards remaining:
\[ 16 - 1 = 15 \]
\[ P(\text{King or Queen}) = \frac{15}{101} \approx 0.1485 \]
3. Part (iii)(a): Probability of getting a queen comparison:
- Case A (No cards dropped):
Total cards = 104, Total queens = 8.
\[ P_A(\text{Queen}) = \frac{8}{104} = \frac{1}{13} \approx 0.07692 \]
- Case B (After cards dropped):
Total cards = 101, Remaining queens = 7.
\[ P_B(\text{Queen}) = \frac{7}{101} \approx 0.06931 \]
Since \( 0.07692 \gt 0.06931 \), yes, the probability of drawing a queen was higher if none of the cards were dropped.
4. Part (iii)(b): Probability of getting a jack comparison:
- Case A (No cards dropped):
Total cards = 104, Total jacks = 8.
\[ P_A(\text{Jack}) = \frac{8}{104} = \frac{1}{13} \approx 0.07692 \]
- Case B (After cards dropped):
No jacks were dropped, so total jacks remaining is still 8. Total remaining cards = 101.
\[ P_B(\text{Jack}) = \frac{8}{101} \approx 0.07921 \]
Comparing the probabilities:
Since \( 0.07921 \gt 0.07692 \), the probability of drawing a jack is higher in the case after the cards were dropped.
Step 4: Final Answer:
(i) Probability of a face card is \(\frac{23}{101}\).
(ii) Probability of a king or a queen is \(\frac{15}{101}\).
(iii)(a) Yes, the probability of drawing a queen was higher before cards were dropped (\(0.0769 \gt 0.0693\)).
(iii)(b) The probability of drawing a jack is higher after the cards were dropped (\(0.0792 \gt 0.0769\)).
