CBSE Class 10 Mathematics Statistics VBQs Set B

Question. A bag contains 3 red, 5 black and 7 white balls. A ball is drawn from the bag at random. The probability that the ball drawn is not black, is
(a) \( \frac{1}{3} \)
(b) \( \frac{9}{15} \)
(c) \( \frac{5}{10} \)
(d) \( \frac{2}{3} \)
Answer: (d)
Explanation: Out of the 15 (3 + 5 + 7) balls in the bag, 10 balls are not black. So, the required probability is \( \frac{10}{15} \), or \( \frac{2}{3} \). 

Question. The mean and median of a distribution are 14 and 15, respectively. The value of the mode is:
(a) 16
(b) 17
(c) 18
(d) 13
Answer: (b)
Explanation: The empirical formula.
\( 3 \text{ Median} = \text{Mode} + 2 \text{ Mean} \)
\( \Rightarrow \text{Mode} = 3 \text{ Median} - 2 \text{ Mean} \)
\( = 3(15) - 2(14) \)
\( = 45 - 28 \)
\( = 17 \). 

Question. While computing the mean of grouped data, we assume that the frequencies are:
(a) evenly distributed over all the classes.
(b) centred at the classmarks of the classes.
(c) centred at the upper limits of the classes.
(d) centred at the lower limits of the classes.
Answer: (b)
Explanation: In grouping the data, all the observations between the lower and upper limits of the class intervals are taken as one group; then the mid-value of class mark is taken for further calculation. In computing the mean of a grouped data, the frequencies are centered at the class marks of the classes. 

Question. If \( x_i \)’s are the mid-points of the class intervals of the grouped data, \( f_i \)’s are the corresponding frequencies and \( \bar{x} \) is the mean, then \( \sum (f_i x_i - \bar{x}) \) is equal to:
(a) 0
(b) –1
(c) 1
(d) 2
Answer: (a)
Explanation: Given that, \( x_i \)’s are the mid points of the class intervals, and \( f_i \)’s are the corresponding frequencies. The mean \( \bar{x} \) for the grouped data is
\( \bar{x} = \frac{\sum f_i x_i}{n} \)
\( \Rightarrow n\bar{x} = \sum f_i x_i \) ...(i)
Also \( \sum (f_i x_i - \bar{x}) = \sum f_i x_i - \sum \bar{x} \)
\( \sum (f_i x_i - \bar{x}) = n\bar{x} - n\bar{x} \) [Using eqn. (i)]
\( = 0 \). 

Question. Consider the data:
Class Frequency
65-85 : 4
85-105 : 5
105-125 : 13
125-145 : 20
145-165 : 14
165-185 : 7
185-205 : 4
The difference of the upper limit of the median class and the lower limit of the modal class is:
(a) 0
(b) 19
(c) 20
(d) 38
Answer: (c)
Explanation:
Class | Frequency | Cumulative Frequency
65-85 | 4 | 4
85-105 | 5 | 9
105-125 | 13 | 22
125-145 | 20 | 42
145-165 | 14 | 56
165-185 | 7 | 63
185-205 | 4 | 67
Now, the class whose cumulative frequency is greater than (and nearest to) \( \frac{n}{2} \) is called the median class.
Here, \( \frac{n}{2} = \frac{67}{2} = 33.5 \), which lies in the interval 125-145. i.e. median class. Hence, the upper limit of median class is 145.
Again, the class whose frequency is maximum is called the modal class.
Here, 20 is the highest frequency which lies in the interval 125-145. Hence, the lower limit of modal class is 125.
Thus, the required difference = Upper limit of median class – Lower limit of modal class \( = 145 - 125 = 20 \). 

Question. The times, in seconds, taken by 150 atheletes to run a 110 m hurdle race tabulated below:
Class | Frequency
13.8-14 | 2
14-14.2 | 4
14.2-14.4 | 5
14.4-14.6 | 71
14.6-14.8 | 48
14.8-15 | 20
The number of atheletes who completed the race in less then 14.6 seconds is:
(a) 11
(b) 71
(c) 82
(d) 130
Answer: (c)
Explanation: The number of athletes who completed the race in less than 14.6 seconds \( = 2 + 4 + 5 + 71 = 82 \). 

Question. Consider the following frequency distribution of the heights of 60 students of a class:
Height (in cm) | No of students
150-155 | 15
155-160 | 13
160-165 | 10
165-170 | 8
170-175 | 9
175-180 | 5
The upper limit of the median class in the given data is:
(a) 165
(b) 155
(c) 160
(d) 170
Answer: (a) 

Question. If an event cannot occur, then its probability is:
(a) 1
(b) \( \frac{3}{4} \)
(c) \( \frac{1}{2} \)
(d) 0
Answer: (d)
Explanation: The event which cannot occur is called the impossible event with a probability equal to zero. 

Question. Which of the following cannot be the probability of an event?
(a) \( \frac{1}{3} \)
(b) 0.1
(c) 3%
(d) \( \frac{17}{16} \)
Answer: (d)
Explanation: Here, \( 0.1 = \frac{1}{10} \), \( 3\% = \frac{3}{100} \). The probability of an event always lies from 0 to 1. As \( \frac{17}{16} > 1 \), \( \therefore \) Option (d) is correct.

Question. If P(E) = 0.005, then the probability of “not E” is:
(a) 0.05
(b) 0.5
(c) 0.995
(d) 0.95
Answer: (c)
Explanation: \( P(\text{not E}) = 1 - P(E) = 1 - 0.005 = 0.995 \).

Question. The probability expressed as a percentage of a particular occurrence can never be:
(a) less than 100
(b) less than 0
(c) greater than 1
(d) anything but a whole number
Answer: (b)
Explanation: We know that: Probability of an event E is \( 0 \le P(E) \le 1 \). Thus, when probability is expressed in terms of percentage, it always lies from 0 to 100. Thus it cannot be less than 0. 

Question. If P(a) denotes the probability of an event a, then:
(a) \( P(a) < 0 \)
(b) \( P(a) > 1 \)
(c) \( 0 \le P(a) \le 1 \)
(d) \( -1 \le P(a) \le 0 \)
Answer: (c)
Explanation: If P(a) denotes the probability of an event a, then \( 0 \le P(a) \le 1 \) i.e., probability can be any number that lies from 0 to 1. 

Question. A number from numbers 1 to 100 was chosen at random. What is the probability that this number is a prime number that lies between 75 and 85?
(a) \( \frac{1}{10} \)
(b) \( \frac{1}{50} \)
(c) \( \frac{1}{25} \)
(d) \( \frac{7}{100} \)
Answer: (b)
Explanation: Prime numbers between 75 and 85 are 79 and 83. So, required probability is \( \frac{2}{100} = \frac{1}{50} \).

Question. A card is selected from a deck of 52 cards. The probability of its being a red face card is:
(a) \( \frac{3}{26} \)
(b) \( \frac{3}{13} \)
(c) \( \frac{2}{13} \)
(d) \( \frac{1}{2} \)
Answer: (a)
Explanation: In a deck of 52 cards: Total no. of cards = 52. No. of face cards = 12. No. of black face cards = 6 (3 spades and 3 clubs). No. of red face cards = 6 (3 hearts & 3 diamonds). Probability of an event E is \( P(E) = \frac{\text{No. of favourable outcomes}}{\text{No. of all possible outcomes of experiment}} = \frac{6}{52} = \frac{3}{26} \). 

Question. When a die is thrown, the probability of getting an odd number less than 3 is:
(a) \( \frac{1}{6} \)
(b) \( \frac{1}{3} \)
(c) \( \frac{1}{2} \)
(d) 0
Answer: (a)
Explanation: When a dice is thrown: Total no. of outcomes = 6 = {1, 2, 3, 4, 5, 6}. Odd no. less than 3 = 1 = {1}. \( \therefore \) No. of favourable outcome = 1. \( \therefore \) Required probability \( = \frac{\text{No. of favourable outcome}}{\text{Total no. of outcomes}} = \frac{1}{6} \). [NCERT]

Question. The probability of getting a bad egg in a lot of 400 is 0.035. The number of bad eggs in the lot is:
(a) 7
(b) 14
(c) 21
(d) 58
Answer: (b)

Question. A girl calculates that the probability of her winning the first prize in a lottery is 0.08. If 6000 tickets are sold, how many tickets has she bought?
(a) 40
(b) 240
(c) 480
(d) 750
Answer: (c)

Question. One ticket is drawn at random from a bag containing tickets numbered 1 to 40. The probability that the selected ticket has a number which is a multiple of 5 is:
(a) \( \frac{1}{5} \)
(b) \( \frac{3}{5} \)
(c) \( \frac{4}{5} \)
(d) \( \frac{1}{3} \)
Answer: (a)

Question. Someone is asked to take a number from 1 to 100. The probability that it is a prime is:
(a) \( \frac{1}{5} \)
(b) \( \frac{6}{25} \)
(c) \( \frac{1}{4} \)
(d) \( \frac{13}{50} \)
Answer: (c)
Explanation: Total no. of outcomes = 100. Prime numbers from 1 to 100 are {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}. \( \therefore \) No. of favourable outcomes = 25. Required probability \( = \frac{\text{No. of favourable outcomes}}{\text{Total no. of outcomes}} \). \( \therefore \text{Required probability} = \frac{25}{100} = \frac{1}{4} \). 

Question. A school has five houses A, B, C, D and E. A class has 23 students, 4 from house A, 8 from house B, 5 from house C, 2 from house D and the rest from house E. A single student is selected at random to be the class monitor. The probability that the selected student is not from A, B and C is:
(a) \( \frac{4}{23} \)
(b) \( \frac{6}{23} \)
(c) \( \frac{8}{23} \)
(d) \( \frac{17}{23} \)
Answer: (b)
Explanation: Total no. of outcomes = 23. No. of students in house A, B & C = 4 + 8 + 5 = 17. \( \therefore \) Remaining students = 23 – 17 = 6. Required probability \( = \frac{\text{No. of favourable outcomes}}{\text{Total no. of outcomes}} \). Probability that the selected students is not from A, B and C = \( \frac{6}{23} \).

Question. A set of numbers consists of three 4s, two 5s, six 6s, eight 8s and seven 10s. What is the mode of this collection of numbers?
(a) 10
(b) 7.5
(c) 7
(d) 8
Answer: (d)
Explanation: Here, set of numbers are : 4, 4, 4, 5, 5, 6, 6, 6, 6, 6, 6, 8, 8, 8, 8, 8, 8, 8, 8, 10, 10, 10, 10, 10, 10, 10. So, mode of this collection of numbers is 8. As it occurs maximum number of times. 

Question. If a letter is chosen at random from the letter of English alphabet, then the probability that it is a letter of the word ‘DELHI’ is:
(a) \( \frac{1}{5} \)
(b) \( \frac{1}{26} \)
(c) \( \frac{5}{26} \)
(d) \( \frac{21}{26} \)
Answer: (c)
Explanation: Total number of alphabets in English = 26 letters. Number of letters in a word ‘DELHI’ = 5 letters. So, the number of favourable outcomes = 5. Probability \( = \frac{5}{26} \).

Question. A dice is thrown twice. The probability of getting 4, 5 or 6 in the first throw and 1, 2, 3 or 4 in the second throw is:
(a) \( \frac{1}{3} \)
(b) \( \frac{2}{3} \)
(c) \( \frac{1}{2} \)
(d) \( \frac{1}{4} \)
Answer: (a)
Explanation: Here, let P(a) and P(b) be the probability of the events. Then, P(A and B) = P(a) \(\cdot\) P(b) \( = \frac{3}{6} \times \frac{4}{6} = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3} \). 

Question. The median of a set of 9 distinct observations is 20.5. If each of the largest 4 observation of the set is increased by 2, then the median of the new set:
(a) is increased by 2.
(b) is decreased by 2.
(c) is two times the original median.
(d) remains the same as that of the original set.
Answer: (d)
Explanation: Remains the same as that of the original set. Since, n = 9 then, median term \( = \frac{9 + 1}{2} = 5\text{th term} \). Now, last four observations are increased by 2. But the median is \( 5\text{th} \) observation, which remains unchanged. There will be no change in median. 

Fill in the Blanks

Fill in the below blanks/tables with suitable information:

Question. The probability of an event that is sure to happen, is .................... .
Answer: 1
Explanation: Probability of a sure event is 1.

Question. If the probability of an event E happening is 0.023, then \( P(\bar{E}) = .................... . \)
Answer: 0.977
Explanation: We know that \( P(E) + P(\bar{E}) = 1 \Rightarrow P(\bar{E}) = 1 - P(E) = 1 - 0.023 = 0.977 \). 

Question. A number is chosen at random from the numbers –5, –4, –3, –2, –1, 0, 1, 2, 3, 4, 5. Then the probability that square of this number is less than or equal to 1 is .................... .
Answer: \( \frac{3}{11} \) 

Question. Mode of observations 4, 3, 1, 2, 3, 4, 4 is .................... .
Answer: 4
Explanation: The most frequent observation in the given data is 4.

Question. Number of face cards in a pack of 52 cards is .................... .
Answer: 12
Explanation: There are 12 face cards in the pack of 52 cards.

Question. When a digit is choosen at random from the digits, 1 to 9, then the probability of this chosen digit to be a prime number is .................... .
Answer: \( \frac{4}{9} \)
Explanation: Between 1 to 9, there are 4 prime numbers (i.e. 2, 3, 5, 7). \( \therefore P(\text{a prime number}) = \frac{4}{9} \).

Question. The upper limit of the median class of the following frequency distribution is .................. .
Class | 0-5 | 6-11 | 12-17 | 18-23 | 24-29
Frequency | 13 | 10 | 15 | 8 | 11
Answer: 17.5
Explanation: The classes in exclusive form are: (–0.5)-5.5; 5.5-11.5; 11.5-17.5; 17.5-23.5; 23.5-24.5 with cumulative frequencies of 13, 23, 38, 46 and 57. Here, N = 57. So, \( \frac{N}{2} = 28.5 \). Thus, median class is 11.5 – 17.5 whose upper limit is 17.5.

Question. .................. is calculated using the formula: \( l + \frac{\frac{N}{2} - cf}{f} \times h \).
Answer: Median

Question. The probability of getting a number which is neither prime nor composite in single throw of a dice is .................. .
Answer: \( \frac{1}{6} \)
Explanation: Of the six numbers only one number, i.e. 1 is neither prime nor composite. So, P(neither prime nor composite) \( = \frac{1}{6} \).

Question. Total number of outcomes in a single throw of three coins is .................. .
Answer: 8
Explanation: Total number of outcomes in a single throw of three coins is eight.