Let's say you design a game in which kids have to pick a red card out of a standard deck.

Sure! Let’s go through each question step by step.

### 1. Probability of Picking a Red Card
In a standard deck of 52 playing cards, there are 26 red cards (13 hearts and 13 diamonds).

### 2. Probability of Picking a Black Card After Picking a Red Card
After the first draw, if a red card is picked, there are now 51 cards left in the deck. Out of these, there are still 26 black cards (13 spades and 13 clubs).

So the probability of drawing a black card after picking a red card is:
\[
P(\text{Black | Red}) = \frac{26}{51}
\]

### 3. Probability of Being Successful Both Times
To find the combined probability of both events happening (picking a red card first, then a black card), you multiply the probabilities of each independent event.

Let:
- \(P(\text{Red}) = \frac{26}{52} = \frac{1}{2}\)
- \(P(\text{Black | Red}) = \frac{26}{51}\)

The combined probability \(P(\text{Red and Black})\) is:
\[
P(\text{Red}) \times P(\text{Black | Red}) = \frac{1}{2} \times \frac{26}{51} = \frac{26}{102} = \frac{13}{51}
\]

### 4. Probability of Drawing a Face Card
In a standard deck, there are 12 face cards (3 face cards: Jack, Queen, King in each of the 4 suits).

So the probability of drawing a face card is:
\[
P(\text{Face Card}) = \frac{12}{52} = \frac{3}{13}
\]

### 5. Probability of Drawing a Jack and Then a Queen With Replacement
If you draw a Jack and then replace it (put it back in), the probability of drawing the Jack first is:
\[
P(\text{Jack}) = \frac{4}{52} = \frac{1}{13}
\]

Since the card is replaced:
\[
P(\text{Queen}) = \frac{4}{52} = \frac{1}{13}
\]

The combined probability for both events is:
\[
P(\text{Jack and then Queen}) = P(\text{Jack}) \times P(\text{Queen}) = \frac{1}{13} \times \frac{1}{13} = \frac{1}{169}
\]

### 6. Probability of Drawing a Jack and Then a Queen Without Replacement
After drawing a Jack (and not replacing it), there are 51 cards left, and still 4 Queens in the deck.

So:
\[
P(\text{Jack}) = \frac{4}{52} = \frac{1}{13}
\]
\[
P(\text{Queen | Jack}) = \frac{4}{51}
\]

The combined probability is:
\[
P(\text{Jack and then Queen}) = P(\text{Jack}) \times P(\text{Queen | Jack}) = \frac{1}{13} \times \frac{4}{51} = \frac{4}{663}
\]

### 7. Custom Problem with 3 Levels
**Problem:** Draw a card from a standard deck of cards.

**Level 1:** Pick a red card.
- Probability: \(\frac{26}{52} = \frac{1}{2}\)

**Level 2:** If successful, you must then draw a numbered card (2-10) from the red cards.
- There are 36 numbered cards in total (9 in each red suit).
- The probability of this is: \(\frac{18}{51}\) if a red card was drawn (assuming picking one red did not affect the count of numbered cards of that color).

**Level 3:** If you draw a numbered red card successfully, finally draw a King from the remaining cards.
- There are now 50 cards left (if you did not draw a King) with 4 Kings in total.
- The probability is \(\frac{4}{50} = \frac{2}{25}\).

**Combined Probability:** You multiply the probabilities of all three levels.

The total probability of succeeding in all three levels is:
\[
P(\text{Red}) \times P(\text{Numbered | Red}) \times P(\text{King | Numbered}) = \frac{1}{2} \times \frac{18}{51} \times \frac{2}{25}
\]

This would give you the overall probability of completing all three levels successfully.