We are tasked with finding the probability that the man was telling the truth the first time, given that the second card he picked is an Ace, and he reported it to be a Hearts card.

Problem Breakdown

1. The man tells the truth $\frac{3}{4}$ of the time and lies $\frac{1}{4}$ of the time.
2. The first report: He reports the first card he drew as an Ace from a red suit (either hearts or diamonds). Since he could lie or tell the truth, this needs to be considered carefully.
3. The second report: He reports the second card he drew as a Hearts card. The second card is known to be an Ace.

Approach: Applying Bayes' Theorem

We will use Bayes' Theorem to solve this problem. Bayes' Theorem allows us to calculate the probability of an event, given certain conditions.

Let’s define the events:

• $T_1$: The man tells the truth on the first report (he reports the card as an Ace from a red suit).
• $L_1$: The man lies on the first report.
• $A_2$: The second card is an Ace.
• $H_2$: The second card is a Hearts card.

We are interested in finding $P(T_1 | A_2, H_2)$, the probability that the man told the truth on the first report given that the second card is an Ace and is a Hearts card.

Bayes' Theorem Formula

$P(T_1 | A_2, H_2) = \frac{P(A_2, H_2 | T_1) P(T_1)}{P(A_2, H_2)}$

Step 1: Find $P(A_2, H_2 | T_1)$

• Given $T_1$ (he told the truth about the first card), we know the first card was an Ace from a red suit (either Hearts or Diamonds).
• For the second card to be an Ace, it could be any Ace, including the Hearts Ace. If the first card was a red Ace, the probability of the second card being the Ace of Hearts (since one card is removed already) is $\frac{1}{51}$ (since there are 52 cards in a deck and one is already removed).
• Therefore, $P(A_2, H_2 | T_1) = \frac{1}{51}$.

Step 2: Find $P(A_2, H_2 | L_1)$

• If the man lied on the first report, then the card he reported as an Ace from a red suit was not actually an Ace of a red suit. The second card could still be the Ace of Hearts, but in this case, it's not influenced by the first card being a red Ace.
• In this case, the second card being the Ace of Hearts is simply $\frac{1}{51}$, as the first card's identity is irrelevant.
• Therefore, $P(A_2, H_2 | L_1) = \frac{1}{51}$.

Step 3: Find $P(A_2, H_2)$

The total probability of the second card being an Ace and a Hearts card is the weighted sum of the probabilities for $T_1$ and $L_1$:

$P(A_2, H_2) = P(A_2, H_2 | T_1) P(T_1) + P(A_2, H_2 | L_1) P(L_1)$

Substitute the values:

$P(A_2, H_2) = \frac{1}{51} \times \frac{3}{4} + \frac{1}{51} \times \frac{1}{4}$ $P(A_2, H_2) = \frac{1}{51} \times \left( \frac{3}{4} + \frac{1}{4} \right)$ $P(A_2, H_2) = \frac{1}{51} \times 1 = \frac{1}{51}$

Step 4: Calculate $P(T_1 | A_2, H_2)$

Now we can apply Bayes' Theorem:

$P(T_1 | A_2, H_2) = \frac{\frac{1}{51} \times \frac{3}{4}}{\frac{1}{51}} = \frac{3}{4}$

Final Answer

The probability that the man was telling the truth the first time, given that the second card is an Ace and a Hearts card, is $\frac{3}{4}$.

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Would you like further details on any part of this solution? Here are some related questions you might find useful:

1. How does Bayes' Theorem work in probability theory?
2. What is the difference between conditional probability and joint probability?
3. Can you explain how to compute the probability of dependent events?
4. How does the concept of prior and posterior probability apply here?
5. What are some real-world applications of Bayes' Theorem?

Tip: Always be sure to carefully define your events and consider whether they are independent or dependent before applying Bayes' Theorem.