Question

Sasha is working on a magic trick with a standard deck of cards. She chooses an ace, puts it back, then chooses a red card. What is P(ace|red) ? Round to the nearest tenth.

Answer 1

To find \( P(\text{ace} | \text{red}) \), we can use Bayes' Theorem, which can be formulated as follows:

\[
P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)}
\]

Where:
- \( A \) is the event of drawing an ace.
- \( B \) is the event of drawing a red card.

### Step 1: Calculate \( P(A) \)

There are 4 aces in a standard deck of 52 cards. Therefore:

\[
P(A) = \frac{4}{52} = \frac{1}{13}
\]

### Step 2: Calculate \( P(B) \)

There are 26 red cards in a standard deck (13 hearts and 13 diamonds). Therefore:

\[
P(B) = \frac{26}{52} = \frac{1}{2}
\]

### Step 3: Calculate \( P(B|A) \)

If we know that an ace has been drawn, the only red aces are the Ace of Hearts and the Ace of Diamonds, which gives us 2 favorable outcomes out of 4 possible aces. Therefore:

\[
P(B|A) = \frac{2}{4} = \frac{1}{2}
\]

### Step 4: Use Bayes' Theorem

Now we can substitute these values into Bayes’ theorem:

\[
P(\text{ace} | \text{red}) = \frac{P(\text{red} | \text{ace}) \cdot P(\text{ace})}{P(\text{red})}
\]
\[
P(\text{ace} | \text{red}) = \frac{P(B|A) \cdot P(A)}{P(B)} = \frac{\left(\frac{1}{2}\right) \cdot \left(\frac{1}{13}\right)}{\frac{1}{2}}
\]

Since \(P(B)\) cancels out:

\[
P(\text{ace} | \text{red}) = \frac{1}{13}
\]

### Step 5: Convert to decimal and round

Converting \( \frac{1}{13} \) to a decimal gives approximately 0.0769. Rounding to the nearest tenth gives:

\[
\approx 0.1
\]

Thus, \( P(\text{ace} | \text{red}) \) is approximately \( \boxed{0.1} \).