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: Count the cards remaining after the three cards are dropped.
Two identical decks together have $2 \times 52 = 104$ cards. Three cards are dropped: the queen of hearts, the ten of spades and the ace of clubs. So the number of cards left to draw from is:
\[ n(S) = 104 - 3 = 101 \]
Step 2: Find the probability of a face card by counting each rank separately.
Face cards are jacks, queens and kings. In two decks there are $2 \times 4 = 8$ jacks, $8$ queens and $8$ kings, so $24$ face cards in total before anything is dropped.
Look at what actually happened to each rank:
Jacks: none were dropped, so all 8 remain.
Kings: none were dropped, so all 8 remain.
Queens: one queen (the queen of hearts) was dropped, so $8 - 1 = 7$ remain.
Add these back up rank by rank instead of subtracting from the total in one step:
\[ 8 + 8 + 7 = 23 \]
So the probability of drawing a face card is:
\[ P(\text{face card}) = \frac{23}{101} \]
Step 3: Find the probability of a king or a queen the same rank-by-rank way.
Kings remaining: 8 (untouched). Queens remaining: 7 (one dropped). Add these:
\[ 8 + 7 = 15 \]
\[ P(\text{king or queen}) = \frac{15}{101} \]
Step 4: Compare the probability of drawing a queen, with and without the drop.
If none of the cards had been dropped, all 8 queens out of 104 cards would be available:
\[ P_{\text{no drop}}(\text{queen}) = \frac{8}{104} = \frac{1}{13} \approx 0.0769 \]
After the drop, only 7 queens remain out of 101 cards:
\[ P_{\text{after drop}}(\text{queen}) = \frac{7}{101} \approx 0.0693 \]
Since $0.0769$ is bigger than $0.0693$, the chance of drawing a queen was indeed higher before any card was dropped. This makes sense, since a queen was one of the cards removed, so the pool of queens shrank while the total pool shrank by less, proportionally, in the queen's favour.
Step 5: Compare the probability of drawing a jack, with and without the drop.
If none of the cards had been dropped:
\[ P_{\text{no drop}}(\text{jack}) = \frac{8}{104} = \frac{1}{13} \approx 0.0769 \]
After the drop, all 8 jacks are still there (no jack was among the three dropped cards), but the total pool shrank to 101:
\[ P_{\text{after drop}}(\text{jack}) = \frac{8}{101} \approx 0.0792 \]
Since $0.0792$ is bigger than $0.0769$, the chance of drawing a jack is higher after the drop. This happens because no jack was removed, yet the total number of cards to choose from became smaller, which makes each remaining jack a slightly bigger share of the deck.
Final Answer:
(i) The probability of a face card is $\frac{23}{101}$.
(ii) The probability of a king or a queen is $\frac{15}{101}$.
(iii)(a) Yes, the probability of drawing a queen was higher before any cards were dropped ($0.0769 > 0.0693$).
(iii)(b) The probability of drawing a jack is higher after the cards were dropped ($0.0792 > 0.0769$).
\[ \boxed{\frac{23}{101}, \ \frac{15}{101}} \]
