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The question in the image asks:

> "Ali is holding four cards from a deck in his hand. He sees that there are two aces and two threes. He then shuffles the cards and randomly draws one card, followed by another. Determine the probability that at least one of the drawn cards is an ace. Also, justify your calculations by drawing a tree diagram that describes the situation."

Let's break down the problem and calculate the probability that **at least one card is an ace**.

### Step-by-Step Approach:

1. **Total Cards:** There are 2 aces (A) and 2 threes (3) in the hand. The total number of cards is 4.

2. **Outcomes for drawing two cards:**
   - The first card could be an ace or a three.
   - After removing the first card, the second card could either be an ace or a three, depending on what was drawn first.

3. **Tree Diagram:**
   We can represent this situation with a tree diagram that shows the possible outcomes after two draws.

   - The first branch represents drawing an ace (A) or a three (3).
   - Each of those outcomes leads to two further branches: drawing an ace (A) or a three (3) again, depending on what was already drawn.

   **First draw:**
   - Probability of drawing an ace first: \( P(A_1) = \frac{2}{4} = \frac{1}{2} \)
   - Probability of drawing a three first: \( P(3_1) = \frac{2}{4} = \frac{1}{2} \)

   **Second draw (if the first card was an ace):**
   - Probability of drawing an ace again: \( P(A_2|A_1) = \frac{1}{3} \)
   - Probability of drawing a three: \( P(3_2|A_1) = \frac{2}{3} \)

   **Second draw (if the first card was a three):**
   - Probability of drawing an ace: \( P(A_2|3_1) = \frac{2}{3} \)
   - Probability of drawing a three: \( P(3_2|3_1) = \frac{1}{3} \)

### Total Probability of at least one Ace:
There are two ways to have at least one ace:
   1. Drawing an ace first.
   2. Drawing a three first and then an ace.

We can calculate the probability of drawing **no aces** and then subtract that from 1.

**Probability of drawing no aces (only threes):**
- First draw: three (\( P(3_1) = \frac{2}{4} \))
- Second draw: three (\( P(3_2 | 3_1) = \frac{1}{3} \))

Thus, \( P(\text{no aces}) = P(3_1) \times P(3_2 | 3_1) = \frac{2}{4} \times \frac{1}{3} = \frac{1}{6} \).

**Therefore, the probability of at least one ace:**
\[
P(\text{at least one ace}) = 1 - P(\text{no aces}) = 1 - \frac{1}{6} = \frac{5}{6}.
\]

### Answer:
The probability that at least one of the two drawn cards is an ace is \( \frac{5}{6} \).

Would you like me to create the tree diagram for better visualization or provide more details on this? Here are 5 related questions to expand your understanding:

1. What is the probability that **both drawn cards are aces**?
2. How would the outcome change if Ali added more cards to his hand?
3. What would happen to the probabilities if there were more than two types of cards?
4. How would you calculate the probability if you only drew one card?
5. What is the difference between conditional probability and joint probability in this scenario?

**Tip:** When calculating probabilities, it's often easier to calculate the complement (opposite event) first, as we did here for "no aces," and then subtract it from 1.

Ali håller fyra kort från en kortlek i sin hand. Han ser att det är två ess och två treor. Han blandar sedan korten och drar slumpmässigt först ett kort och sedan ett kort till. Bestäm sannolikheten att åtminstone ett av korten är ett ess. Motivera också dina uträkningar genom att rita ett träddiagram som beskriver situationen.

Learn how to calculate the probability of drawing at least one ace from a set of four cards, using a tree diagram to explain the process. This problem covers key probability concepts, including the complement rule and conditional probability.

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