In a school 25% students are failed in Mathematics, 15% in Physics and 5% students fail in both Mathematics and Physics. A student is selected at random.
(i) If he is fail in Physics, find the probability that he is fail in Mathematics.
(ii) If he is fail in Mathematics, find the probability that he in fail in Physics.
(iii) Find the probability that he is fail either in Mathematics or in Physics.

Explanation

To solve the problem step by step, we need to define the events and use the information provided to calculate the required probabilities.

### Given Data:
- Probability of failing in Mathematics (P(M)) = 25% = 25/100 = 1/4
- Probability of failing in Physics (P(PH)) = 15% = 15/100 = 3/20
- Probability of failing in both Mathematics and Physics (P(M ∩ PH)) = 5% = 5/100 = 1/20

### (i) If he fails in Physics, find the probability that he fails in Mathematics.

We need to find the conditional probability P(M | PH), which is given by the formula:
\[ P(M | PH) = \frac{P(M ∩ PH)}{P(PH)} \]

Substituting the known values:
\[ P(M | PH) = \frac{P(M ∩ PH)}{P(PH)} = \frac{1/20}{3/20} \]

Calculating:
\[ P(M | PH) = \frac{1}{3} \]

### (ii) If he fails in Mathematics, find the probability that he fails in Physics.

Now we need to find the conditional probability P(PH | M):
\[ P(PH | M) = \frac{P(M ∩ PH)}{P(M)} \]

Substituting the known values:
\[ P(PH | M) = \frac{1/20}{1/4} \]

Calculating:
\[ P(PH | M) = \frac{1/20}{5/20} = \frac{1}{5} \]

### (iii) Find the probability that he fails either in Mathematics or in Physics.

We need to find the probability P(M ∪ PH), which is given by the formula:
\[ P(M ∪ PH) = P(M) + P(PH) - P(M ∩ PH) \]

Substituting the known values:
\[ P(M ∪ PH) = \frac{1}{4} + \frac{3}{20} - \frac{1}{20} \]

Finding a common denominator (which is 20):
\[ P(M ∪ PH) = \frac{5}{20} + \frac{3}{20} - \frac{1}{20} = \frac{5 + 3 - 1}{20} = \frac{7}{20} \]

### Final Answers:
(i) P(M | PH) = 1/3
(ii) P(PH | M) = 1/5
(iii) P(M ∪ PH) = 7/20

---