The ace , king queen , jack and ten from both the spades and hearts suits are placed in two separate piles and one card is taken from each pile : draw the sample space diagram and find the probability that at least one card will be an ace ,

Answer

To solve the problem, we need to follow these steps:

### Step 1: Identify the Cards
We have two piles of cards:
- **Pile 1 (Spades)**: Ace (A), King (K), Queen (Q), Jack (J), and Ten (10)
- **Pile 2 (Hearts)**: Ace (A), King (K), Queen (Q), Jack (J), and Ten (10)

### Step 2: Draw the Sample Space
The sample space consists of all possible outcomes when drawing one card from each pile. We can represent this as pairs (x, y) where x is from the Spades pile and y is from the Hearts pile.

The sample space can be listed as follows:
- (K, K), (K, Q), (K, A), (K, J), (K, 10)
- (Q, K), (Q, Q), (Q, A), (Q, J), (Q, 10)
- (A, K), (A, Q), (A, A), (A, J), (A, 10)
- (J, K), (J, Q), (J, A), (J, J), (J, 10)
- (10, K), (10, Q), (10, A), (10, J), (10, 10)

Thus, the complete sample space is:
- KK, KQ, KA, KJ, K10
- QK, QQ, QA, QJ, Q10
- AK, AQ, AA, AJ, A10
- JK, JQ, JA, JJ, J10
- 10K, 10Q, 10A, 10J, 10 10

### Step 3: Count the Total Outcomes
The total number of outcomes in the sample space is:
- There are 5 choices from the Spades pile and 5 choices from the Hearts pile.
- Therefore, the total number of outcomes = 5 * 5 = 25.

### Step 4: Identify Favorable Outcomes
Next, we need to find the outcomes where at least one card is an Ace. We can list these outcomes:
- (A, K), (A, Q), (A, A), (A, J), (A, 10) → 5 outcomes from Spades with Ace
- (K, A), (Q, A), (A, A), (J, A), (10, A) → 5 outcomes from Hearts with Ace
- (A, A) is counted twice, so we need to subtract it once.

Thus, the total number of favorable outcomes is:
- 5 (from Spades) + 5 (from Hearts) - 1 (double counted) = 9.

### Step 5: Calculate the Probability
The probability of getting at least one Ace is given by the formula:
\[ P(\text{at least one Ace}) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} \]

Substituting the values we found:
\[ P(\text{at least one Ace}) = \frac{9}{25} \]

### Final Answer
The probability that at least one card will be an Ace is \( \frac{9}{25} \).

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