You are to pick 4 cards randomly from a condensed deck of cards that contain 4 suits and the following denominations, ace,2,3,4,5,6,7,8,9,10. There are no face-cards in the deck.
What is probability you get two aces and two hearts

The problem involves calculating the probability of drawing 4 cards randomly from a modified deck with 4 suits and the following denominations: Ace, 2, 3, 4, 5, 6, 7, 8, 9, and 10, and there are no face cards. The total number of cards in the deck is:

\[
10 \text{ denominations} \times 4 \text{ suits} = 40 \text{ cards}
\]

We are asked to find the probability of drawing **two aces** and **two hearts** from this deck. Let's break down the solution:

### Step 1: Total number of ways to draw 4 cards
The total number of ways to draw 4 cards from 40 is given by the combination formula:

\[
\binom{40}{4} = \frac{40!}{4!(40 - 4)!} = \frac{40 \times 39 \times 38 \times 37}{4 \times 3 \times 2 \times 1} = 91,390
\]

### Step 2: Number of ways to get two aces
There are 4 aces in the deck (one in each suit). The number of ways to choose 2 aces from these 4 is:

\[
\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6
\]

### Step 3: Number of ways to get two hearts
There are 10 hearts in the deck (one for each denomination: Ace, 2, 3, ..., 10). Since we need two hearts and already have possibly one or two aces, we need to break this into cases:

#### Case 1: If both of the aces selected in Step 2 are **not hearts**
In this case, we need to select both of the two hearts from the remaining 9 non-ace hearts. The number of ways to do this is:

\[
\binom{9}{2} = \frac{9 \times 8}{2 \times 1} = 36
\]

#### Case 2: If one of the aces selected is a **heart** and one is not
If one ace is a heart, we need to select only 1 more heart from the remaining 9 hearts. The number of ways to do this is:

\[
\binom{9}{1} = 9
\]

#### Case 3: If both of the aces selected are **hearts**
In this case, we already have 2 hearts (both are aces), so there are no additional hearts to select, and the number of ways is simply:

\[
1 \text{ way}
\]

### Step 4: Total favorable outcomes
Now we add up all the favorable outcomes from the different cases:

\[
\text{Total favorable outcomes} = 6 \times (36 + 9 + 1) = 6 \times 46 = 276
\]

### Step 5: Probability
The probability of getting two aces and two hearts is the ratio of favorable outcomes to the total outcomes:

\[
P(\text{two aces and two hearts}) = \frac{276}{91,390} \approx 0.00302
\]

So, the probability is approximately **0.00302** or **0.302%**.

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Would you like further details or have any questions? 

Here are some related questions:
1. What is the probability of drawing three aces and one heart?
2. What is the probability of drawing no aces and all hearts?
3. How many ways are there to draw exactly two hearts from this deck?
4. How does the probability change if the deck had face cards included?
5. What is the probability of drawing at least one ace in 4 cards?

**Tip:** When calculating probabilities with combinations, always start by identifying total outcomes and break the problem into manageable cases!

You are to pick 4 cards randomly from a condensed deck of cards that contain 4 suits and the following denominations, ace,2,3,4,5,6,7,8,9,10. There are no face-cards in the deck. What is the probability you get two aces and two hearts?

Calculate the probability of drawing two aces and two hearts from a modified deck of 40 cards with 4 suits and no face cards. This problem involves combinatorics and probability theory, offering a step-by-step breakdown of the solution. Suitable for Grades 10-12 students.

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