Let's break down and solve the probability questions from the image:

Given:

• A deck of 52 standard playing cards (no jokers).
• Cards are ranked from 2 to Ace (where Ace is the highest, above a King).

We'll solve the problems step by step:

(a) Probability of drawing a card above a King:

• The only card above a King is the Ace.
• There are 4 Aces (one for each suit: spades, hearts, diamonds, clubs).

So, the probability $P(\text{above a King})$ is: $P(\text{above a King}) = \frac{4}{52} = \frac{1}{13}$

(b) Probability of drawing a card below a 4:

• Cards below a 4 are 2 and 3. Each suit has one 2 and one 3, so there are 4 twos and 4 threes.

The total number of cards below a 4 is $4 + 4 = 8$.

So, the probability $P(\text{below a 4})$ is: $P(\text{below a 4}) = \frac{8}{52} = \frac{2}{13}$

(c) Probability of drawing a card both above a King and below a 4:

• A card cannot be both above a King and below a 4 simultaneously, as they are mutually exclusive events (no card satisfies both conditions).

Thus, the probability is: $P(\text{both above a King and below a 4}) = 0$

(d) Probability of drawing a card either above a King or below a 4:

• We already know there are 4 cards above a King (the Aces) and 8 cards below a 4 (the 2s and 3s).
• Since there is no overlap between the two groups (disjoint events), we can simply add the probabilities.

Thus, the probability $P(\text{above a King or below a 4})$ is: $P(\text{either above a King or below a 4}) = \frac{4}{52} + \frac{8}{52} = \frac{12}{52} = \frac{3}{13}$

Final Answers:

• (a) $\frac{1}{13}$
• (b) $\frac{2}{13}$
• (c) 0
• (d) $\frac{3}{13}$

Would you like further details on any of these steps, or have any questions?

Here are five related questions:

1. What is the probability of drawing a face card (Jack, Queen, or King)?
2. How would the probabilities change if jokers were included in the deck?
3. What is the probability of drawing a card between 4 and King, inclusive?
4. How can we use complement probability to solve some of these problems?
5. If two cards are drawn without replacement, what is the probability both are Aces?

Tip: Always check for mutually exclusive events when calculating probabilities involving "and" or "or" conditions.