A doctor is called to see a sick child. The doctor knows (prior to the visit) that 90% of the sick children in that neighbourhood are sick with the flu, denoted by `F ,` while 10% are sick with the measles, denoted by `Mdot` A well-known symptom of measles is a rash, denoted by R. The probability having a rash for a child sick with the measles is 0.95. however, occasionally children with the flu also develop a rash, with conditional probability 0.08. upon examination the child, the doctor finds a rash. The what is the probability that the child has the measles? `91//165` b. `90//163` c. `82//161` d. `95//167`

Options

A. `91//165`
B. `90//163`
C. `82//161`
D. `95//167`

Explanation

To solve the problem, we will use Bayes' theorem, which allows us to find the probability of an event based on prior knowledge of conditions related to the event.

### Step-by-Step Solution:

1. **Define Events:**
- Let \( M \) be the event that the child has measles.
- Let \( F \) be the event that the child has the flu.
- Let \( R \) be the event that the child has a rash.

2. **Given Probabilities:**
- \( P(M) = 0.10 \) (Probability that the child has measles)
- \( P(F) = 0.90 \) (Probability that the child has the flu)
- \( P(R|M) = 0.95 \) (Probability of having a rash given measles)
- \( P(R|F) = 0.08 \) (Probability of having a rash given flu)

3. **Use Bayes' Theorem:**
We want to find \( P(M|R) \), the probability that the child has measles given that the child has a rash. According to Bayes' theorem:

\[ P(M|R) = \frac{P(R|M) \cdot P(M)}{P(R)} \]

4. **Calculate \( P(R) \):**
To find \( P(R) \), we use the law of total probability:

\[ P(R) = P(R|M) \cdot P(M) + P(R|F) \cdot P(F) \]

Substituting the known values:

\[ P(R) = (0.95 \cdot 0.10) + (0.08 \cdot 0.90) \]
\[ P(R) = 0.095 + 0.072 = 0.167 \]

5. **Substitute into Bayes' Theorem:**
Now substitute \( P(R) \) back into Bayes' theorem:

\[ P(M|R) = \frac{0.95 \cdot 0.10}{0.167} \]
\[ P(M|R) = \frac{0.095}{0.167} \]

6. **Calculate the Final Probability:**
Now, we can calculate \( \frac{0.095}{0.167} \):

\[ P(M|R) = \frac{95}{167} \]

### Conclusion:
The probability that the child has measles given that the child has a rash is \( \frac{95}{167} \).