Here you can download the Class 10 CBSE Statistics and Probability test in pdf format and check solution for the test.
Let’s break down each problem step-by-step. Designed for both students and teachers, these answers make exam revision easier with clear explanations, formula highlights, and downloadable PDF options.

Multiple Choice Questions: 1 Mark Questions

Q1: If the probability of a player winning a game is 0.79, then the probability of his losing the same game is:
(a) 1.79
(b) 0.31
(c) 0.21 \%
(d) 0.21

Solution:
Let the event of the player winning the game be W and the event of the player losing the game be L.

In probability, winning and losing a game are considered “complementary events” (assuming there are no ties).
The sum of the probabilities of all possible complementary outcomes is always 1.

Therefore, we can write the formula: P(W) + P(L) = 1

We are given that the probability of winning is 0.79
P(W) = 0.79

Now, substitute this value into our formula to find the probability of losing:
0.79 + P(L) = 1
P(L) = 1 - 0.79
P(L) = 0.29

Correct option: (d)

Q2: From the data 1, 4, 7, 9, 16, 21, 25, if all the even numbers are removed, then the probability of getting at random a prime number from the remaining is:

(a) \frac{1}{5}

(b) \frac{2}{5}

(c) \frac{1}{7}

(d) \frac{2}{5}

Solution:
First, let’s look at the initial set of numbers: 1, 4, 7, 9, 16, 21, 25.

The question asks us to remove all the even numbers.
The even numbers in this set are 4 and 16.
After removing them, our new remaining data set is: 1, 7, 9, 21, 25

Total number of possible outcomes (remaining numbers) = 5.

Now, we need to find the probability of selecting a prime number from this remaining set.
Let’s check each number:

• 1 is neither prime nor composite.
• 7 is a prime number.
• 9 is composite (3 \times 3).
• 21 is composite (3 \times 7).
• 25 is composite (5 \times 5).

There is only 1 prime number in our remaining set, which is 7.

Number of favorable outcomes = 1.
\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}

\text{Probability} = \frac{1}{5}

Correct Option: (a)

Q3: Consider the frequency distribution of 45 observations… The upper limit of median class is:
(a) 20
(b) 10
(c) 30
(d) 40

Solution:
To find the median class, we first need to calculate the cumulative frequency (cf) for each class interval.

• Class 0-10: Frequency = 5, Cumulative Frequency = 5
• Class 10-20: Frequency = 9, Cumulative Frequency = 5 + 9 = 14
• Class 20-30: Frequency = 15, Cumulative Frequency = 14 + 15 = 29
• Class 30-40: Frequency = 10, Cumulative Frequency = 29 + 10 = 39
• Class 40-50: Frequency = 6, Cumulative Frequency = 39 + 6 = 45

The total number of observations, N, is 45.
We find the position of the median by calculating \frac{N}{2} = \frac{45}{2} = 22.5

Now, we look for the class interval where the cumulative frequency is just greater than or equal to 22.5

Looking at our list, 29 is the first cumulative frequency greater than 22.5
This corresponds to the class interval 20-30.

Therefore, the median class is 20-30.
The upper limit of this median class is 30.

Correct option: (c)

2 marks Questions

Q4: How many positive three digit integers have the hundredths digit 8 and unit’s digit 5? Find the probability of selecting one such number out of all 3 digit numbers.

Solution:
Let’s first determine how many positive 3-digit numbers fit the given condition.
A 3-digit number has three place values: Hundreds, Tens, and Units.

• The Hundreds digit is fixed as 8 (1 option).
• The Units digit is fixed as 5 (1 option).
• The Tens digit can be any digit from 0 to 9 (10 options).

The numbers will look like: 805, 815, 825 \dots, 895
Total number of such favorable numbers = 1 \times 10 \times 1 = 10.

Next, we find the total number of all possible 3-digit numbers.
The smallest 3-digit number is 100 and the largest is 999.
\text{Total 3-digit numbers} = 999 - 100 + 1 = 900.

Now, calculate the probability:
\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}

\text{Probability} = \frac{10}{900} = \frac{1}{90}

Final Answer: There are \boxed{ 10 } such integers, and the probability is \boxed{ \frac{1}{90}}.

Q5: For a moderately skewed distribution, the mode and median are respectively 30.1 and 27.9. What is the mean of the distribution?

Solution:
There is a well-known empirical relationship between Mean, Median, and Mode given by Karl Pearson’s formula:
\text{Mode} = 3 \text{ Median} - 2 \text{ Mean}

We are given:
\text{Mode} = 30.1
\text{Median} = 27.9

Substitute these values into the empirical formula:
30.1 = 3 (27.9) - 2 \text{ Mean}
30.1 = 83.7 - 2 \text{ Mean}
2 \text{ Mean} = 83.7 - 30.1
2 \text{ Mean} = 53.6

\text{ Mean} = \frac{53.6}{2}

\text{ Mean} = 26.8

Now, solve for the Mean. Rearrange the terms:

Final Answer: The mean of the distribution is \boxed{26.8}

3 Marks Questions

Q6: The mode of the following data is 67. Find the missing frequency x.

Solution:
We are given that the Mode is 67.
Since 67 lies in the class interval 60-70, our modal class is 60-70.

From the table, we identify the following variables for the mode formula:

• Lower limit of the modal class, l = 60
• Class size, h = 10
• Frequency of the modal class, f_1 = 15
• Frequency of the class preceding the modal class, f_0 = x
• Frequency of the class succeeding the modal class, f_2 = 12

The formula for the mode of grouped data is: \text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h

Substitute the known values into the formula:
67 = 60 +\left( \frac{15 - x}{2(15) - x -12} \right) \times 10

67 - 60 = \left( \frac{15 - x}{30 - x - 12} \right) \times 10 7 = \left( \frac{15 - x}{18 - x} \right) \times 10

Cross-multiply to solve for x:
7(18 - x) = 10(15 - x)

126 - 7x = 150 - 10x

3x = 24

x = 8

Final Answer: The missing frequency \boxed{ x = 8 }.

5 marks Questions

Q7: The monthly expenditure of milk in 200 families of a housing society is given below… Find the value of x and also, find the median and mean of expenditure.

Solution:
Step 1: Finding the value of x
We are given that the total number of families is 200.

Summing the given frequencies gives us the total:
24 + 40 + 33 + x + 30 + 22 +16 + 7 = 200
172 + x = 200
x = 200 - 172 = 28

Step 2: Finding the Median First, construct the Cumulative Frequency (cf) column:

We calculate \frac{N}{2} = \frac{200}{2} = 100.

The cumulative frequency just greater than 100 is 125, which belongs to the class 2500-3000.

• Lower limit, l = 2500
• Cumulative frequency of preceding class, cf = 97
• Frequency of median class, f = 28
• Class size, h = 500

\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h

\text{Median} = 2500 + \left( \frac{100 - 97}{28} \right) \times 500

\text{Median} = 2500 + \left( \frac{3}{28} \right) \times 500

\text{Median} = 2500 + \frac{1500}{28}

\text{Median} = 2500 + 53.57

\text{Median} = 2553.57

Step 3: Finding the Mean (using Step-Deviation Method)
Let assumed mean A = 2750
Class size h = 500.

Using the formula d_i = x_i - A:

Using the formula u_i = \frac{d_i}{h}:

\Sigma f_i u_i = (-72 - 80 - 33) + (30 + 44 + 48 + 28) = -185 + 150 = -35

\text{Mean} = A + \left( \frac{\Sigma f_i u_i}{N} \right) \times h

\text{Mean} = 2750 + \left( \frac{-35}{200} \right) \times 500

\text{Mean} = 2750 - \left( \frac{35}{2} \right) \times 5

\text{Mean} = 2750 - \frac{175}{2}

\text{Mean} = 2750 - 87.5

\text{Mean} = 2662.5

Final Answer: x = 28, \text{Median} = 2553.57 \text{Mean} = 2662.5

Q8: Two different dice are thrown together. Find the probability that the number obtained have
(i) even sum
(ii) even product

Solution:
When two dice are thrown together, the total number of possible outcomes is 6 \times 6 = 36.

(i) Probability of getting an even sum:
For the sum of two numbers to be even, both numbers must be even OR both numbers must be odd.

• Possibilities for (Even, Even): 3 evens on die 1 and 3 evens on die 2 \rightarrow 3 \times 3 = 9 outcomes.
• Possibilities for (Odd, Odd): 3 odds on die 1 and 3 odds on die 2 \rightarrow 3 \times 3 = 9 outcomes. Total favorable outcomes = 9 + 9 = 18.

P(\text{Even Sum}) = \frac{18}{36} = \boxed{ \frac{1}{2} }

(ii) Probability of getting an even product:
For a product of two numbers to be even, at least one of the numbers must be even.
The only way to get an odd product is if BOTH numbers are odd.
We already know there are 9 outcomes where both numbers are odd.
Therefore, the number of outcomes with an even product is the total outcomes minus the odd product outcomes:
Total favorable outcomes = 36 - 9 = 27.

P(\text{Even Product}) = \frac{27}{36} = \boxed{ \frac{3}{4}}

Case Study: 5 marks Question

Q9: Cards on which numbers 1, 2, 3 ………. 100 are written (one number on one card and no number is repeated), put in a bag and are mixed thoroughly. A card is drawn at random from the bag.

Solution:
Total number of cards in the bag = 100.

(i) What is the probability that card taken out has a odd number?
There are exactly half odd and half even numbers from 1 to 100.
Favorable outcomes = 50 (1, 3, 5, …, 99).

P(\text{Odd number}) = \frac{50}{100} = \boxed{ \frac{1}{2} }

(ii) What is the probability that card taken out has a two digit odd number?
There are 50 odd numbers in total.
The single-digit odd numbers are 1, 3, 5, 7, 9 (which are 5 numbers).
Favorable outcomes = Total odd numbers – Single-digit odd numbers Favorable outcomes = 50 - 5 = 45.

P(\text{Two-digit odd number}) = \frac{45}{100} = \boxed{ \frac{9}{20}}

(iii) What is the probability that card taken out has a odd number which is multiple of 11?
The odd multiples of 11 between 1 and 100 are: 11, 33, 55, 77, 99.
Favorable outcomes = 5.

P(\text{Odd multiple of 11}) = \frac{5}{100} = \boxed{ \frac{1}{20}}

(iv) What is the probability that card taken out has an odd number which is not less than 70?
“Not less than 70” means numbers from 70 to 100.
The odd numbers in this range are: 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99.
By counting them, we find there are exactly 15 odd numbers.
Favorable outcomes = 15.

P(\text{Odd number } \geq 70) = \frac{15}{100} = \boxed{ \frac{3}{20}}

(v) What is the probability that card taken out has an odd number which is not multiple of 11?
We know from part (i) that there are 50 odd numbers in total.
We know from part (iii) that 5 of these are multiples of 11.
Favorable outcomes = Total odd numbers – Odd numbers that are multiples of 11
Favorable outcomes = 50 - 5 = 45.

P(\text{Odd number NOT multiple of 11}) = \frac{45}{100} = \boxed{ \frac{9}{20}}