Question

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?

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.

Answer 1

Step 1: Set the total sample space.
Two identical packs together have $52 \times 2 = 104$ cards. After the three cards (queen of hearts, ten of spades, ace of clubs) are dropped, the sample space has:
\[ N = 104 - 3 = 101 \text{ cards} \]

Step 2: Part (i), probability of a face card, found through a complement check.
There are $12$ face cards (4 jacks, 4 queens, 4 kings) in one pack, so $24$ face cards across both packs. Non-face cards number $104-24=80$ across both packs. Only one of the three dropped cards, the queen of hearts, is a face card, so the remaining face cards are $24-1=23$ and the remaining non-face cards stay at $80-2=78$ (since the ten of spades and ace of clubs are not face cards). Check: $78+23=101$, which matches $N$.
So instead of counting favourable face cards directly, we can also get the same number from $101-78=23$. Either way:
\[ P(\text{face card}) = \frac{23}{101} \]

Step 3: Part (ii), probability of a king or queen.
Across both packs there are $8$ kings and $8$ queens, so $16$ cards in total. One queen was dropped, leaving $15$.
\[ P(\text{king or queen}) = \frac{15}{101} \]

Step 4: Part (iii)(a), comparing the queen's probability with and without the drop, using percentages instead of decimals.
Without any drop: $P = \dfrac{8}{104} = \dfrac{1}{13}$, which as a percentage is about $7.69\%$.
With the drop: $P = \dfrac{7}{101}$, which as a percentage is about $6.93\%$.
Since $7.69\% > 6.93\%$, the chance of drawing a queen was higher before any card was dropped, because losing a queen from the deck removes one of the very outcomes we are counting, while the total shrinks by only three, not enough to make up for that loss.

Step 5: Part (iii)(b) (alternative), the jack case.
None of the three dropped cards are jacks, so all $8$ jacks remain, while the total shrinks to $101$.
Without any drop: $P = \dfrac{8}{104} \approx 7.69\%$.
With the drop: $P = \dfrac{8}{101} \approx 7.92\%$.
Since the jack count did not fall but the total pool did, the probability of drawing a jack is higher after the drop.

Final Answer:
(i) $P(\text{face card}) = \dfrac{23}{101}$. (ii) $P(\text{king or queen}) = \dfrac{15}{101}$. (iii)(a) The queen's probability was higher with no cards dropped. (iii)(b) The jack's probability is higher after the drop, at $\dfrac{8}{101}$.
\[ \boxed{\tfrac{23}{101},\ \tfrac{15}{101}} \]