In a class of 75 students, 15 are above average, 45 are average and the rest below average achievers. The probability that an above average achieving student fails is 0.005, that an average achieving student fails is 0.05 and the probability of a below average achieving student failing is 0.15. If a student is known to have passed, what is the probability that he is a below average achiever?

Answer

To solve the problem step by step, we will use the concepts of probability and conditional probability.

### Step 1: Define the events
Let:
- \( A_1 \): Event that a student is above average.
- \( A_2 \): Event that a student is average.
- \( A_3 \): Event that a student is below average.
- \( E \): Event that a student passes.

### Step 2: Calculate the probabilities of each event
- Total number of students = 75
- Number of above average students = 15
- Number of average students = 45
- Number of below average students = \( 75 - (15 + 45) = 15 \)

Now, we can calculate the probabilities:
- \( P(A_1) = \frac{15}{75} = \frac{1}{5} \)
- \( P(A_2) = \frac{45}{75} = \frac{3}{5} \)
- \( P(A_3) = \frac{15}{75} = \frac{1}{5} \)

### Step 3: Calculate the probabilities of passing given each event
- Probability of passing for above average students:
\[ P(E | A_1) = 1 - P(\text{Fail} | A_1) = 1 - 0.005 = 0.995 \]
- Probability of passing for average students:
\[ P(E | A_2) = 1 - P(\text{Fail} | A_2) = 1 - 0.05 = 0.95 \]
- Probability of passing for below average students:
\[ P(E | A_3) = 1 - P(\text{Fail} | A_3) = 1 - 0.15 = 0.85 \]

### Step 4: Calculate the total probability of passing
Using the law of total probability:
\[ P(E) = P(A_1) \cdot P(E | A_1) + P(A_2) \cdot P(E | A_2) + P(A_3) \cdot P(E | A_3) \]
Substituting the values:
\[ P(E) = \left(\frac{1}{5} \cdot 0.995\right) + \left(\frac{3}{5} \cdot 0.95\right) + \left(\frac{1}{5} \cdot 0.85\right) \]
Calculating each term:
- \( \frac{1}{5} \cdot 0.995 = 0.199 \)
- \( \frac{3}{5} \cdot 0.95 = 0.57 \)
- \( \frac{1}{5} \cdot 0.85 = 0.17 \)

Now summing them up:
\[ P(E) = 0.199 + 0.57 + 0.17 = 0.939 \]

### Step 5: Calculate the conditional probability that a student is below average given that they passed
We want to find \( P(A_3 | E) \):
\[ P(A_3 | E) = \frac{P(A_3) \cdot P(E | A_3)}{P(E)} \]
Substituting the values:
\[ P(A_3 | E) = \frac{\left(\frac{1}{5} \cdot 0.85\right)}{0.939} \]
Calculating the numerator:
\[ \frac{1}{5} \cdot 0.85 = 0.17 \]

Now substituting back:
\[ P(A_3 | E) = \frac{0.17}{0.939} \approx 0.181 \]

### Final Answer
The probability that a student is a below average achiever given that they passed is approximately \( 0.181 \).