You are to pick 4 cards randomly from a condensed deck of cards that contains four suits  and the following denomations: Ace, 2, 3, 4, 5, 6, 7, 8, 9, and 10. There are no face-cards in this deck.  What is the probability that you will get Part (a) 2 aces and 2 hearts

### Step-by-step solution:

1. **Understanding the Deck:**
   The deck consists of 4 suits (hearts, diamonds, clubs, spades) and 10 denominations (Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10). There are a total of:
   
   \[
   4 \text{ suits} \times 10 \text{ cards per suit} = 40 \text{ cards}.
   \]

2. **The Total Number of Ways to Pick 4 Cards:**
   The total number of ways to pick 4 cards from the deck of 40 cards can be calculated using combinations:
   
   \[
   \binom{40}{4} = \frac{40 \times 39 \times 38 \times 37}{4 \times 3 \times 2 \times 1} = 91,390.
   \]

3. **Conditions for the Desired Event:**
   We need to calculate the probability of drawing **2 aces and 2 hearts**.
   
   - **Number of aces in the deck**: There are 4 aces, one for each suit.
   - **Number of hearts in the deck**: There are 10 hearts (since there are no face cards, all denominations are present).
   - We want **2 aces** and **2 hearts**. Since these conditions can overlap, let's break it down.

4. **Selecting 2 Aces:**
   The number of ways to choose 2 aces from the 4 available aces is:
   
   \[
   \binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6.
   \]

5. **Selecting 2 Hearts:**
   The number of ways to choose 2 hearts from the 10 available hearts is:
   
   \[
   \binom{10}{2} = \frac{10 \times 9}{2 \times 1} = 45.
   \]

6. **Correcting for Overlap:**
   However, some of the chosen cards may satisfy both conditions (being an ace and a heart). Specifically, **the ace of hearts** is both an ace and a heart. So we need to adjust for overlap:

   - If the ace of hearts is chosen as one of the aces, we still need 1 more ace (from the 3 remaining non-heart aces) and 1 more heart (from the 9 remaining non-ace hearts).

   The number of ways to choose 1 ace from the remaining 3 non-heart aces and 1 heart from the remaining 9 non-ace hearts is:
   
   \[
   \binom{3}{1} \times \binom{9}{1} = 3 \times 9 = 27.
   \]

7. **Total Number of Favorable Outcomes:**
   The total number of ways to meet the conditions (2 aces and 2 hearts) is the sum of:
   
   - Case 1: 2 aces (but none of them are hearts) and 2 non-ace hearts:
     \[
     \binom{3}{2} \times \binom{9}{2} = 3 \times 36 = 108.
     \]
   
   - Case 2: 1 ace is the ace of hearts, 1 more ace from the remaining non-heart aces, and 1 more heart from the remaining non-ace hearts (calculated above):
     \[
     27.
     \]

   Thus, the total number of favorable outcomes is:
   
   \[
   108 + 27 = 135.
   \]

8. **Final Probability:**
   The probability of drawing 2 aces and 2 hearts is the ratio of favorable outcomes to total outcomes:
   
   \[
   \text{Probability} = \frac{135}{91,390}.
   \]

   Simplifying, this fraction is approximately:

   \[
   \text{Probability} \approx 0.00148 \text{ (or about 0.148%)}.
   \]

### Conclusion:
The probability of drawing 2 aces and 2 hearts from this deck is approximately 0.00148 or 0.148%.

---

Would you like more details or have any questions?

Here are 5 related questions to explore further:
1. What is the probability of drawing exactly 3 hearts from this deck?
2. How does the probability change if you wanted 3 aces and 1 heart?
3. What is the probability of drawing 4 aces from the deck?
4. How many different ways can you pick 2 cards from the hearts suit?
5. What is the probability of drawing 2 cards of the same suit from this deck?

**Tip:** When calculating probabilities with overlapping conditions, be mindful of adjusting for any possible overlaps.

You are to pick 4 cards randomly from a condensed deck of cards that contains four suits and the following denominations: Ace, 2, 3, 4, 5, 6, 7, 8, 9, and 10. There are no face-cards in this deck. What is the probability that you will get Part (a) 2 aces and 2 hearts?

Calculate the probability of drawing 2 aces and 2 hearts from a 40-card deck with no face cards. Explore combinatorics and probability theory to solve this detailed step-by-step problem, perfect for students in Grades 10-12.

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