Class 11 Maths Probability and Statistics Test 1 – 2026 and Solution

Have you solved Class 11 Maths Probability and Statistics Test 1 and are looking for a solutions to the questions? Here you’ll find complete, step-by-step solutions for every question. Designed for both students and teachers, these answers make board exam revision easier with clear explanations, formula highlights, and downloadable PDF options.

2 Marks Questions

Q1: Find the mean deviation about the mean for the following data: 6, 7, 10, 12, 13, 4, 8, 12.

Solution:
Step 1: Understanding the Formula
To solve this, we first need to recall the formula for Mean Deviation about the Mean M.D.(\bar{x}) for ungrouped data:
M.D.(\bar{x}) = \frac{\sum |x_i - \bar{x}|}{N}

Where:
x_i represents the individual observations.
\bar{x} represents the Arithmetic Mean.
N represents the total number of observations.

Step 2: Calculate the Arithmetic Mean ( \bar{x} )
First, we calculate the sum of all observations and divide it by the total number of items.

Given data: 6, 7, 10, 12, 13, 4, 8, 12
Total number of observations ( N ) = 8

\bar{x} = \frac{6 + 7 + 10 + 12 + 13 + 4 + 8 + 12}{8}

\bar{x} = \frac{72}{8}

\bar{x} = 9

So, the Mean is 9.

Step 3: Calculate Absolute Deviations ( |x_i - \bar{x}| )
Now, we find the difference between each observation and the mean, taking only the absolute value (positive value).

Step 4: Calculate the Mean Deviation
Now, substitute the values back into the main formula.
M.D.(\bar{x}) = \frac{22}{8}

M.D.(\bar{x}) = 2.75
Final Answer: The Mean Deviation about the mean for the given data is \boxed { 2.75 } .

Q2: The mean of 200 items is 48 and their standard deviation is 3. Find the sum of items and sum of squares of all items.

Solution:
Step 1: Identify Given Information
We are given the following values:

• Number of items n = 200
• Mean \bar{x} = 48
• Standard Deviation \sigma = 3

Step 2: Find the Sum of Items ( \sum x_i )
We know that the formula for the mean is:
\bar{x} = \frac{\sum x_i}{n}

Substituting the values:
48 = \frac{\sum x_i}{200}

\sum x_i = 48 \times 200

\sum x_i = 9600

So, the sum of all items is 9600.

Step 3: Find the Sum of Squares of Items ( \sum x_i^2 )
We use the formula for Standard Deviation:

\sigma = \sqrt{\frac{\sum x_i^2}{n} - (\bar{x})^2 }

Squaring both sides to remove the square root (working with Variance):
\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2

Substitute the known values:
(3)^2 = \frac{\sum x_i^2}{200} - (48)^2

9 = \frac{\sum x_i^2}{200} - 2304

\frac{\sum x_i^2}{200} = 9 + 2304

\frac{\sum x_i^2}{200} = 2313

\sum x_i^2 = 2313 \times 200

\sum x_i^2 = 462600

Final Answer: Sum of items = \boxed{9600} Sum of squares of items = \boxed{462600}

Q3: A class consists of 15 girls and 10 boys. The class teacher wants to choose two students from amongst them as class monitors. Find the probability that one boy and one girl are chosen.

Solution:
Step 1: Calculate Total Outcomes
Total number of students = 15 Girls + 10 Boys = 25 Students.
We need to choose 2 students.
Total number of ways to choose 2 students from 25 is given by combinations ( {}^{n}C_r ):

n(S) = {}^{25}C_2 = \frac{25 \times 24}{2 \times 1} = 300

Step 2: Calculate Favorable Outcomes We need to choose exactly one boy and one girl.

Number of ways to choose 1 girl from 15 = {}^{15}C_1 = 15

Number of ways to choose 1 boy from 10 = {}^{10}C_1 = 10

By the Fundamental Principle of Counting, total favorable outcomes:
n(E) = 15 \times 10 =150

Step 3: Calculate Probability
P(E) = \frac{n(E)}{n(S)}

P(E) = \frac{150}{300}

P(E) = \frac{1}{2}

Final Answer: The probability is \boxed{0.5} or \frac{1}{2} .

Q4: Two students Anil and Ashima appeared in an examination. The probability that Anil will qualify is 0.05 and Ashima will qualify is 0.10. The probability that both will qualify is 0.02. Find the probability that at least one of them will not qualify.

Solution:
Step 1: Define Events
Let A be the event that Anil qualifies. Let B be the event that Ashima qualifies.
Given values:
P(A) = 0.05
P(B) = 0.10
P(A \cap B) = 0.02 (Both qualify)

Step 2: Understand the Required Probability
We need to find the probability that “at least one of them will not qualify”.
In set theory notation, this is P(A' \cup B') (Not A or Not B).

Step 3: Apply De Morgan’s Law
According to De Morgan’s Law, P(A' \cup B') = P(A \cap B)' .
This means “At least one not” is the complement of “Both occur”.

P(\text{at least one not}) = 1 - P(\text{Both qualify})
P(A' \cup B') = 1 - P(A \cap B)

Step 4: Calculation
P(A' \cup B') = 0.02

P(A' \cup B') = 0.98

Final Answer: The probability that at least one of them will not qualify is \boxed{0.98} .

3 Marks Questions

Q5: Find the standard deviation of the first 10 natural numbers.

Solution:
Step 1: Identify the formula
The standard deviation ( \sigma ) for the first n natural numbers can be calculated using the the standard formula:
\sigma = \sqrt{\frac{\sum x_i^2}{n} - (\frac{\sum x_i}{n})^2} … (1)

Using the formulas
\sum x_i = \frac{n(n+1)}{2}

\sum x_i^2 = \frac{n(n+1)(2n+1)}{6}

Substituting in (1)
\sigma = \sqrt{\frac{\frac{n(n+1)(2n+1)}{6}}{n} - (\frac{\frac{n(n+1)}{2}}{n})^2}

\sigma = \sqrt{\frac{n(n+1)(2n+1)}{6n} - (\frac{n(n+1)}{2n})^2}

\sigma = \sqrt{\frac{n(n+1)(2n+1)}{6n} - \frac{n^2(n+1)^2}{4n^2}}

\sigma = \sqrt{\frac{n(n+1)}{2n} \left[ \frac {(2n+1)}{3} - \frac {n(n+1)}{2n} \right] }

\sigma = \sqrt{\frac{n(n+1)}{2n} \left[ \frac {2n(2n+1)}{6n} - \frac {3n(n+1)}{6n} \right] }

\sigma = \sqrt{\frac{n(n+1)}{2n} \left[ \frac {(4n^2+2n) - (3n^2 + 3n)}{6n} \right] }

\sigma = \sqrt{\frac{n(n+1)}{2n} \left[ \frac {n^2 - n}{6n} \right] }

\sigma = \sqrt{\frac{n(n+1)}{2n} \left[ \frac {n(n - 1)}{6n} \right] }

\sigma = \sqrt{\frac{n(n+1)}{2n} \left[ \frac {(n - 1)}{6} \right] }

\sigma = \sqrt{\frac{(n^2-1)}{12}}

Step 2: Apply the values
Here, n = 10 .

Using the direct formula:
\sigma = \sqrt{\frac{(10)^2 - 1}{12}}

\sigma = \sqrt{\frac{100 - 1}{12}}

\sigma = \sqrt{\frac{99}{12}}

\sigma = \sqrt{8.25}

\sigma \approx 2.87

Final Answer: The standard deviation of the first 10 natural numbers is 2.87.

Q6: If P(A)=0.35 , P(A\cap B)=0.25 , P(A\cup B)=0.6 , then find P(B) and P(\text{not } B) .

Solution:
Step 1: Find P(B)
We use the addition theorem of probability:
P(A \cup B) = P(A) + P(B) - P(A \cap B)

Substitute the given values:
0.6 = 0.35 + P(B) - 0.25

0.6 = 0.10 + P(B)

P(B) = 0.6 - 0.10

P(B) = 0.5

Step 2: Find P(\text{not } B)
The probability of “not B” is the complement of event B:
P(\text{not } B) = 1 - P(B)

P(\text{not } B) = 1 - 0.5

P(\text{not } B) = 0.5

Final Answer: P(B) = 0.5 and P(\text{not } B) = 0.5 .

Q7: A bag contains 8 red and 5 white balls. Three balls are drawn at random. Find the probability that:
(a) All the three balls are white.
(b) All the three balls are red.
(c) One ball is red and two balls are white.

Solution:
Step 1: Calculate Total Outcomes
Total balls = 8 Red + 5 White = 13 balls.
Number of balls drawn = 3.
Total possible outcomes ( n(S) ):
n(S) = {}^{13}C_3 = \frac{13 \times 12 \times 11}{3 \times 2 \times 1} = 286

Step 2: Part (a) – All 3 balls are white
Favorable outcomes (choosing 3 from 5 white):
n(E_1) = {}^{5}C_3 = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10

Probability:
P(\text{3 White}) = \frac{10}{286} = \frac{5}{143}

Step 3: Part (b) – All 3 balls are red
Favorable outcomes (choosing 3 from 8 red):
n(E_2) = {}^{8}C_3 = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56

Probability:
P(\text{3 Red}) = \frac{56}{286} = \frac{28}{143}

Step 4: Part (c) – One ball is red and two balls are white
Favorable outcomes (1 from 8 Red AND 2 from 5 White):
n(E_3) = {}^{8}C_1 \times {}^{5}C_2

n(E_3) = 8 \times 10 = 80

Probability: P(\text{1 Red, 2 White}) = \frac{80}{286} = \frac{40}{143}

5 Marks Questions

Q8: Determine mean and standard deviation of the following distribution of coffee jar weights:

Solution:
Step 1: Construct the Calculation Table
We will use the Step Deviation Method to simplify calculations.
Let Assumed Mean ( A ) = 202.5
Class Height ( h ) = 1

Step 2: Calculate the Mean ( \bar{x} )
\bar{x} = A + \frac{\sum f_i u_i}{N} \times h

\bar{x} = 202.5 + \frac{-38}{70} \times 1

\bar{x} = 202.5 - 0.54

\bar{x} = 201.96 g

Step 3: Calculate the Standard Deviation ( \sigma )

\sigma = h \sqrt{\frac{\sum f_i u_i^2}{N} - \left(\frac{\sum f_i u_i}{N}\right)^2}

\sigma = 1 \times \sqrt{\frac{102}{70} - \left(\frac{-38}{70}\right)^2}

\sigma = \sqrt{1.4571 - (-0.5428)^2}

\sigma = \sqrt{1.4571 - 0.2946}

\sigma = \sqrt{1.1625}

\sigma \approx 1.08 g

Final Answer: Mean = \boxed{201.96 \ g} Standard Deviation = \boxed { 1.08 \ g }

Case Study: 4 marks questions

Q9 A teacher wants to analyze the performance of 50 students in a Mathematics test. The marks are organized into class intervals (0-10, 10-20, etc.). Based on the histogram provided, answer the following:
(i) Find the mean deviation about median.
(ii) Find the mean deviation about mean.

Solution:
Step 1: Extract Data from Histogram
Based on the visual representation and the total count of 50 students, the frequency distribution is interpreted as follows:

(i) Mean Deviation about Median
Step 2: Find the Median Class
N = 50 , so \frac{N}{2} = 25 .

Looking at the cumulative frequency (c.f.) column, the value just greater than 25 is 32.
Therefore, the Median Class is 20-30.

l (Lower limit) = 20
f (Frequency of median class) = 16
C (c.f. of preceding class) = 16
h (Class height) = 10

Step 3: Calculate Median ( M )
M = l + \frac{\frac{N}{2} - C}{f} \times h

M = 20 + \frac{25 - 16}{16} \times 10

M = 20 + \frac{9}{16} \times 10

M = 20 + 5.625

M = 25.625

Step 4: Calculate Mean Deviation about Median
We calculate |x_i - M| and multiply by f_i .

M.D.(M) = \frac{\sum f_i |x_i - M|}{N}

M.D.(M) = \frac{468.75}{50}

M.D.(M) = 9.375

(ii) Mean Deviation about Mean
Step 5: Calculate Mean ( \bar{x} )
\bar{x} = \frac{\sum f_i x_i}{N}

Using the mid-points x_i :

\sum f_i x_i = (4 \times 5) + (12 \times 15) + (16 \times 25) + (10 \times 35) + (8 \times 45)

\sum f_i x_i = 20 + 180 + 400 + 350 + 360 = 1310

\bar{x} = \frac{1310}{50} = 26.2

Step 6: Calculate Mean Deviation about Mean

We calculate |x_i - 26.2| and multiply by f_i .

Final Answers: (i) Mean Deviation about Median = \boxed {9.375}
(ii) Mean Deviation about Mean = \boxed {9.536}

Q10: One of the four persons John, Rita, Aslam, or Gurpreet will be promoted. The sample space is S = { \text{John, Rita, Aslam, Gurpreet} } John’s chance is same as Gurpreet’s, Rita’s chance is twice as likely as John’s, Aslam’s chance is four times that of John’s. Find the probability that (a) John, (b) Rita, (c) Aslam, (d) Gurpreet got promoted.

Solution:
Step 1: Set up the Probabilities
Let the probability of John being promoted be x .
P(\text{John}) = x

According to the given conditions:
John’s chance is same as Gurpreet’s: P(\text{Gurpreet}) = x
Rita’s chance is twice John’s: P(\text{Rita}) = 2x
Aslam’s chance is four times John’s: P(\text{Aslam}) = 4x

Step 2: Solve for x
Since these are the only elementary outcomes in the sample space, the sum of their probabilities must be 1.
P(\text{John}) + P(\text{Rita}) + P(\text{Aslam}) + P(\text{Gurpreet}) = 1
x + 2x + 4x + x = 1
8x = 1
x = \frac{1}{8} or 0.125

Step 3: Calculate Individual Probabilities
(a) Probability John got promotion:
P(\text{John}) = x = \frac{1}{8}

(b) Probability Rita got promotion:
P(\text{Rita}) = 2x = 2 \times \frac{1}{8} = \frac{1}{4}

(c) Probability Aslam got promotion:
P(\text{Aslam}) = 4x = 4 \times \frac{1}{8} = \frac{1}{2}

(d) Probability Gurpreet got promotion:
P(\text{Gurpreet}) = x = \frac{1}{8}

Final Answers: (a) \frac{1}{8} (b) \frac{1}{4} (c) \frac{1}{2} (d) \frac{1}{8}