The problem provides a table of marks scored by 40 students in an objective test. The task is to find the probability that a student selected at random from the class scored: a) marks greater than 28 b) marks less than 28 c) 33 marks d) marks between 3 and 33 (inclusive)

1. Problem Description

The problem provides a table of marks scored by 40 students in an objective test. The task is to find the probability that a student selected at random from the class scored:

a) marks greater than 28

b) marks less than 28

c) 33 marks

d) marks between 3 and 33 (inclusive)

2. Solution Steps

First, let's find the total number of students, which is given as

4

0.

a) Marks greater than 28:

The marks greater than 28 are 33 and

3

8. The number of students who scored 33 is

5. The number of students who scored 38 is

1. Total number of students who scored greater than 28 is $5 + 1 = 6$.

The probability is the number of students who scored greater than 28 divided by the total number of students.

$P(\text{marks} > 28) = \frac{6}{40} = \frac{3}{20}$

b) Marks less than 28:

The marks less than 28 are 3, 8, 13, 18, and

2

3. The number of students who scored 3 is

2. The number of students who scored 8 is

5. The number of students who scored 13 is

9. The number of students who scored 18 is

6. The number of students who scored 23 is

3. Total number of students who scored less than 28 is $2 + 5 + 9 + 6 + 3 = 25$.

$P(\text{marks} < 28) = \frac{25}{40} = \frac{5}{8}$

c) 33 marks:

The number of students who scored 33 is

5. $P(\text{marks} = 33) = \frac{5}{40} = \frac{1}{8}$

d) Marks between 3 and 33 (inclusive):

The marks between 3 and 33 are 3, 8, 13, 18, 23, 28 and

3

3. The number of students who scored 3 is

2. The number of students who scored 8 is

5. The number of students who scored 13 is

9. The number of students who scored 18 is

6. The number of students who scored 23 is

3. The number of students who scored 28 is

4. The number of students who scored 33 is

5. Total number of students who scored between 3 and 33 is $2 + 5 + 9 + 6 + 3 + 4 + 5 = 34$.

$P(3 \leq \text{marks} \leq 33) = \frac{34}{40} = \frac{17}{20}$

3. Final Answer

a) $P(\text{marks} > 28) = \frac{3}{20}$

b) $P(\text{marks} < 28) = \frac{5}{8}$

c) $P(\text{marks} = 33) = \frac{1}{8}$

d) $P(3 \leq \text{marks} \leq 33) = \frac{17}{20}$