Remove all J, Q, K from a 52-card deck. With Ace valued 1, find probability that a drawn card has value (i) 7 (ii) >7 (iii) <7.
(i) \(\dfrac{1}{10}\), (ii) \(\dfrac{3}{10}\), (iii) \(\dfrac{3}{5}\)
Step 1: Total cards after removing J, Q, K
A standard deck has 52 cards. Each suit (♠, ♥, ♦, ♣) has 13 cards.
We remove J, Q, K from all 4 suits. That means 3 × 4 = 12 cards are removed.
So, total cards left = 52 − 12 = 40.
Step 2: What values are left?
The remaining cards are Ace (1), 2, 3, 4, 5, 6, 7, 8, 9, 10. That makes 10 different values.
Each value has 4 cards (one from each suit). So, each value count = 4.
Step 3: Case (i) Probability that the card value = 7
Number of cards with value 7 = 4.
Total cards = 40.
So probability = 4 ÷ 40 = 1/10.
Step 4: Case (ii) Probability that the card value > 7
Values greater than 7 are: 8, 9, 10.
Each has 4 cards. So total = 3 × 4 = 12 cards.
So probability = 12 ÷ 40 = 3/10.
Step 5: Case (iii) Probability that the card value < 7
Values less than 7 are: Ace (1), 2, 3, 4, 5, 6.
That is 6 values, each with 4 cards. So total = 6 × 4 = 24 cards.
So probability = 24 ÷ 40 = 3/5.
Final Answer: (i) 1/10, (ii) 3/10, (iii) 3/5.