In 2013, there will be three candidates for the post of professor. Mr.X, Mr.Y and Mr.Z, who's chances of getting the promotion are in the proportions 4:2:3 respectively. The probability that x  if selected would introduce new subjects of today's requirement in the department is 0.35. The probability of Y and Z doing the same are respectively 0.52 and 0.80. What is the probability that there will be new subjects in the department in the year 2014? (Solve using bayes theorem)

We can solve this problem using **Total Probability Theorem**, not Bayes' Theorem, since we are not asked to find conditional probabilities given an event, but rather to find the total probability that new subjects will be introduced by one of the candidates.

### Step 1: Given Data

- The chances of \( X \), \( Y \), and \( Z \) getting selected are in the proportion 4:2:3, which means the probabilities are:
  \[
  P(X) = \frac{4}{9}, \quad P(Y) = \frac{2}{9}, \quad P(Z) = \frac{3}{9}
  \]

- The probability of introducing new subjects if each candidate is selected are:
  \[
  P(N | X) = 0.35, \quad P(N | Y) = 0.52, \quad P(N | Z) = 0.80
  \]
  where \( N \) represents the event of introducing new subjects.

### Step 2: Use of the Law of Total Probability

The total probability that new subjects will be introduced is the weighted sum of the probabilities that each candidate introduces new subjects, weighted by their respective chances of being selected. Mathematically:

\[
P(N) = P(N | X)P(X) + P(N | Y)P(Y) + P(N | Z)P(Z)
\]

### Step 3: Substitute the values

Substituting the given values:

\[
P(N) = (0.35) \times \left(\frac{4}{9}\right) + (0.52) \times \left(\frac{2}{9}\right) + (0.80) \times \left(\frac{3}{9}\right)
\]

### Step 4: Simplify the equation

\[
P(N) = \frac{4 \times 0.35}{9} + \frac{2 \times 0.52}{9} + \frac{3 \times 0.80}{9}
\]

\[
P(N) = \frac{1.4}{9} + \frac{1.04}{9} + \frac{2.4}{9}
\]

\[
P(N) = \frac{1.4 + 1.04 + 2.4}{9} = \frac{4.84}{9}
\]

\[
P(N) \approx 0.538
\]

### Final Answer:
The probability that new subjects will be introduced in the department in 2014 is approximately **0.538**, or **53.8%**.

---

Would you like a more detailed explanation or have any questions about this process?

### 5 Related Questions:
1. How is the law of total probability applied in real-life decision-making scenarios?
2. Can Bayes' Theorem be applied if we were given more conditions in this scenario?
3. What are the differences between Total Probability Theorem and Bayes' Theorem?
4. How would the result change if the proportions of candidates' chances changed?
5. Can this method be used to evaluate outcomes in other types of probabilistic problems?

### Tip:
When using the Law of Total Probability, always ensure the events you are summing over are **mutually exclusive and collectively exhaustive** (i.e., they cover all possible outcomes).

In 2013, there will be three candidates for the post of professor. Mr.X, Mr.Y and Mr.Z, whose chances of getting the promotion are in the proportions 4:2:3 respectively. The probability that X, if selected, would introduce new subjects of today's requirement in the department is 0.35. The probability of Y and Z doing the same are respectively 0.52 and 0.80. What is the probability that there will be new subjects in the department in the year 2014? (Solve using Bayes' theorem)

Learn how to calculate the probability of introducing new subjects in a department for 2014 using the Law of Total Probability. This example evaluates the chances of three candidates being selected and introduces key probability concepts with step-by-step solutions. Suitable for students in Grades 10-12.

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