In a school, 300 students were surveyed. The survey showed that 186 students liked watching cricket match while the rest disliked it. Out of these students, if one student is chosen at random, what is the probability that the chosen student :
(i) likes watching cricket match?
(ii) dislikes watching cricket match?

Answer

To solve the problem step by step, we will calculate the probability of two events: (i) a student liking cricket matches and (ii) a student disliking cricket matches.

### Step-by-Step Solution:

1. **Identify the total number of students surveyed:**
\[ \text{Total number of students} = 300 \]

2. **Identify the number of students who like watching cricket matches:**
\[ \text{Number of students who like cricket} = 186 \]

3. **Calculate the number of students who dislike watching cricket matches:**
\[ \text{Number of students who dislike cricket} = \text{Total students} - \text{Students who like cricket} \]
\[ = 300 - 186 = 114 \]

4. **Calculate the probability that a randomly chosen student likes watching cricket matches:**
\[ P(\text{likes cricket}) = \frac{\text{Number of students who like cricket}}{\text{Total number of students}} \]
\[ = \frac{186}{300} \]
To simplify this fraction, we can divide both the numerator and the denominator by their greatest common divisor (GCD), which is 6:
\[ = \frac{186 \div 6}{300 \div 6} = \frac{31}{50} \]

5. **Calculate the probability that a randomly chosen student dislikes watching cricket matches:**
\[ P(\text{dislikes cricket}) = \frac{\text{Number of students who dislike cricket}}{\text{Total number of students}} \]
\[ = \frac{114}{300} \]
Again, we simplify this fraction by dividing both the numerator and the denominator by their GCD, which is 6:
\[ = \frac{114 \div 6}{300 \div 6} = \frac{19}{50} \]

### Final Answers:
(i) The probability that the chosen student likes watching cricket match is \( \frac{31}{50} \).

(ii) The probability that the chosen student dislikes watching cricket match is \( \frac{19}{50} \).