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Let's analyze the problem and find the solutions to the given questions.

**Problem Summary:**
- A hospital has 70 patients with a certain disease.
- 66 of these patients are randomly selected to receive a new treatment that increases the probability of cure from 0.4 to 0.45.

We need to calculate the following probabilities:
1. The overall probability of cure among all patients.
2. The probability that a cured patient received the new treatment.
3. The probability that a cured patient received the old treatment.

**Solution:**

### 1. Probability of Cure Among All Patients
We have:
- \( 66 \) patients receive the new treatment (probability of cure \( = 0.45 \)).
- \( 4 \) patients receive the old treatment (probability of cure \( = 0.4 \)).

The expected number of cures from each group is:
- New treatment: \( 66 \times 0.45 = 29.7 \).
- Old treatment: \( 4 \times 0.4 = 1.6 \).

Total expected cures:
\[ 29.7 + 1.6 = 31.3. \]

Therefore, the probability of cure among all patients:
\[
\frac{31.3}{70} \approx 0.447.
\]

### 2. Probability of a Patient Receiving the New Treatment, Given They Are Cured
Using Bayes' theorem:
\[
P(\text{New Treatment} \mid \text{Cured}) = \frac{P(\text{Cured} \mid \text{New Treatment}) \times P(\text{New Treatment})}{P(\text{Cured})}.
\]

Where:
- \( P(\text{Cured} \mid \text{New Treatment}) = 0.45 \).
- \( P(\text{New Treatment}) = \frac{66}{70} \).
- \( P(\text{Cured}) \approx 0.447 \).

Substituting these values:
\[
P(\text{New Treatment} \mid \text{Cured}) = \frac{0.45 \times \frac{66}{70}}{0.447} \approx 0.95.
\]

### 3. Probability of a Patient Receiving the Old Treatment, Given They Are Cured
Similarly, using Bayes' theorem:
\[
P(\text{Old Treatment} \mid \text{Cured}) = \frac{P(\text{Cured} \mid \text{Old Treatment}) \times P(\text{Old Treatment})}{P(\text{Cured})}.
\]

Where:
- \( P(\text{Cured} \mid \text{Old Treatment}) = 0.4 \).
- \( P(\text{Old Treatment}) = \frac{4}{70} \).
- \( P(\text{Cured}) \approx 0.447 \).

Substituting these values:
\[
P(\text{Old Treatment} \mid \text{Cured}) = \frac{0.4 \times \frac{4}{70}}{0.447} \approx 0.05.
\]

### Final Answers:
1. Probability of cure among the patients: **0.447**.
2. Probability of receiving the new treatment, given cured: **0.95**.
3. Probability of receiving the old treatment, given cured: **0.05**.

Would you like more details or have any questions? Here are some related questions:

1. How would the probabilities change if more patients received the old treatment?
2. What if the probability of cure with the old treatment was higher?
3. How does sample size affect the accuracy of these probabilities?
4. What are other methods to verify these probabilities?
5. How can these probabilities guide future treatment decisions?

**Tip:** Always verify the assumptions of your probability models to ensure accurate conclusions.

Em um hospital, há 70 pacientes com uma certa doença. Dentre estes pacientes, 66 são selecionados ao acaso para receberem o novo tratamento que aumenta a probabilidade de cura de 0.4 para 0.45. Qual é a probabilidade de cura entre os pacientes? Qual é a probabilidade de um paciente ter recebido o novo tratamento, dado que foi curado? Qual é a probabilidade de um paciente ter recebido o tratamento antigo, dado que foi curado?

This problem involves calculating the probability of cure in a group of patients treated with two different methods. It uses concepts such as conditional probability, Bayes' Theorem, and expected values to find the solution.

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