CBSE Class 10 Mathematics Statistics VBQs Set C

Very Short Answer Type Questions

Question. The mean and median of a distribution are both equal to 635.97. Find the mode.
Answer: Mean = Median = 635.97 [Given]. Mode = 3 median – 2 mean = 3 mean – 2 mean [Mean = Median] = mean = 635.97.

Question. Two dice are thrown simultaneously. What is the probability that the sum of the two numbers appearing on the top is 13?
Answer: No of possible outcomes = 36. No. of favourable outcomes is 0. (because the sum of number cannot be more than 12). So, required probability is \( \frac{0}{36} \), i.e., 0. 

Question. Find the class marks of the classes 15 – 35 and 45 – 60.
Answer: Class mark of 15 – 35 is \( \frac{15 + 35}{2} \), i.e., 25. Class mark of 45 – 60 is \( \frac{45 + 60}{2} \), i.e, 52.5. 

Question. A die is thrown once. What is the probability of getting a prime number.
Answer: Possible outcomes are 1, 2, 3, 4, 5, 6. Favourable outcomes are 2, 3, 5. So, required probability \( = \frac{3}{6} \) or \( \frac{1}{2} \). 

Question. A pair of dice is thrown once. What is the probability of getting a doublet?
Answer: Total number of possible outcomes = 36. Favourable cases { (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) }. \( \therefore \) Number of favourable outcomes = 6. \( \therefore \) Required probability \( = \frac{6}{36} = \frac{1}{6} \). 

Question. A die is thrown once. What is the probability of getting an even prime number?
Answer: Total number of possible outcomes = 6. Number of favourable outcomes = 1, as 2 is the only even prime number. So, the required probability is \( \frac{1}{6} \).

Question. A letter of the English alphabet is chosen at random. What is the probability that the chosen letter is a consonant?
Answer: Out of the 26 letters of english alphabets, 21 letters are consonants. So, required probability \( = \frac{21}{26} \). 

Question. A die is thrown once. What is the probability of getting a number less than 3?
Answer: All possible outcomes are 1, 2, 3, 4, 5, 6. Favourable outcomes are 1 and 2. So, required probability \( = \frac{2}{6} \), i.e., \( \frac{1}{3} \). 

Question. If the probability of winning a game is 0.07, what is the probability of losing it?
Answer: Probability of losing a game = 1 – p (winning a game) = 1 – 0.07 = 0.93. 

Question. The probability of selecting a rotten apple randomly from a heap of 900 apples is 0.18. What is the number of rotten apples in the heap?
Answer: A = getting a rotten apple. \( n(S) = 900 \) total apples. \( P(A) = 0.18 \). Let n(A) be number of rotten apples. Then, \( P(A) = \frac{n(A)}{n(S)} = \frac{n(A)}{900} \). \( 0.18 \times 900 = n(A) \). \( \therefore n(A) = 162 \). So, there are 162 rotten apples in the heap. 

Question. A card is drawn at random from a well shuffled pack of 52 playing cards. Find the probability of getting neither a red card nor a queen.
Answer: Total cards = 52. Total red cards & queens = 26 (red cards) + 2 (black queens) = 28. Probability of getting neither red card nor queen \( = \frac{52 - 28}{52} = \frac{24}{52} = \frac{12}{26} \) or \( \frac{6}{13} \).  

Question. A bag contains 3 red and 5 black balls. A ball is drawn at random from the bag. What is the probability that the drawn ball is not red?
Answer: Total number of balls in a bag = 3 + 5 = 8. Number of balls that are not red in colour = 5. Total number of favourable cases = 5. \( \therefore P(\text{getting a ball which is not red}) = \frac{5}{8} \). Hence, the required probability is \( \frac{5}{8} \). 

Question. Two different dice are tossed together. Find the probability that the product of the two numbers on the top of the dice is 6.
Answer: \( \frac{1}{9} \) 

Question. Write the empirical relationship among the three measures of central tendency mean, mode and median.
Answer: The required relationship is: \( 3 \text{Median} = \text{Mode} + 2 \text{Mean} \).

Question. Sarita buys a fish from a shop for her aquarium. The shopkeeper takes out a fish at random from a tank containing 10 male fish and 12 female fish. What is the probability that the fish taken out is a female fish?
Answer: Total number of fishes = 10 + 12 = 22. Probability (female fish) \( = \frac{\text{Total number of female fish}}{\text{Total number of fishes}} = \frac{12}{22} = \frac{6}{11} \).

Question. A bag contains 5 red, 8 green and 7 white balls. One ball is drawn at random from the bag. Find the probability of getting neither a green ball nor a red ball.
Answer: A ball which is neither a green ball nor a red ball is necessary a white ball. So, required probability \( = \frac{7}{5 + 8 + 7} = \frac{7}{20} \).

Question. Find the class marks of the classes 20-50 and 35-60.
Answer: Class-mark of 20 – 50 is \( \frac{20 + 50}{2} \), i.e., 35. Class-mark of 35 – 60 is \( \frac{35 + 60}{2} \), i.e, 47.5. 

Question. Two dice are thrown simultaneously. What is the probability that the product of the number appearing on the top is 1?
Answer: Total possible outcomes = 36. Only one outcome, say (1, 1) has the product of the two numbers as 1. So, the required probability is \( \frac{1}{36} \). 

Question. When we toss a coin, there are two possible outcomes – heads or tails. Therefore, the probability of each outcome is \( \frac{1}{2} \). Justify your answer.
Answer: Yes (True). Total no. of outcomes = 2. Probability of heads = Probability of tails \( = \frac{1}{2} \). As heads and tails both are equally likely events.

Question. The mean of 20 observations is 12. If each observation is increased by 5, then find the new mean.
Answer: Mean of 20 observations = 12. Here, each observation is increased by 5. So, new mean will also increased by 5. Thus, new mean = 12 + 5 = 17. 

Question. A letter is chosen from the letters of the word MAINTENANCE. What is the probability that it is N?
Answer: In the given word, there are in all eleven letters, of which three are N. So, the required probability is \( \frac{3}{11} \).

SHORT ANSWER (SA-I) Type Questions

Question. What is the probability that a randomly taken leap year has 52 Sundays ?
Answer: \( \therefore \) Favourable cases = (Sunday, Monday), (Saturday, Sunday). A leap year has 366 days. 364 days form 52 weeks. Remaining 2 days can be (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday). To have exactly 52 Sundays, the remaining 2 days must not include a Sunday. There are 5 such pairs out of 7. \( \therefore \) Probability \( = \frac{5}{7} \). [Note: The OCR provided text for the answer is truncated, but the logic follows the standard leap year probability question]. 

Question. A number is selected at random from natural numbers 1 to 20. Find the probability that the selected number is a prime number. 
Answer: Total number of outcomes = 20
Let, E be the probability that a number selected is a prime number.
Then, the prime number between 1 to 20 (or favourable cases) are 2, 3, 5, 7, 11, 13, 17, 19
\( \therefore \) Total of favourable outcomes = 8
\( P(E) = \frac{\text{Favourable outcomes}}{\text{Total number of outcomes}} = \frac{8}{20} = \frac{2}{5} \)
Hence, the required probability is \( \frac{2}{5} \).

Question. A number is chosen at random from the number –3, –2, –1, 0, 1, 2, 3. What will be the probability that square of this number is less than or equal to 1 ? 
Answer: The given number are = –3, –2, –1, 0, 1, 2, 3
The square of these numbere are = 9, 4, 1, 0, 1, 4, 9
\( \therefore \) Total number of outcomes = 7
The square of numbes less than or equal to 1 = 1, 0, 1
\( \therefore \) Total number of favourable cases = 3
P(square of getting a number is less than or equal to 1) = \( \frac{3}{7} \)
Hence, the required probability is \( \frac{3}{7} \).

Question. If two different coins are tossed together, then find the probability of getting two heads.
Answer: On tossing 2 coins, all possible outcomes are: (H, H), (H, T); (T, H); (T, T)
Total possible outcomes = 4
Let E be the probability of getting 2 heads.
Then, favourable outcomes = (H, H)
Total number of favourable outcomes = 1
\( \therefore P(E) = \frac{1}{4} \)
Hence, the required probability is \( \frac{1}{4} \).

Question. A letter of the English alphabets is chosen at random. Find the probability that the chosen letter is a letter of the word ‘Trignometry’. 
Answer: Total number of english alphabets are 26.
Total number of possible outcomes = 26
Number of distinct alphabets in TRIGNOMETRY = 9
\( \therefore \) Total number of favourable cases = 9
\( \therefore \) P(chosen letter is a letter from TRIGNOMETRY) = \( \frac{9}{26} \)
Hence, the required probability is \( \frac{9}{26} \).

Question. 20 tickets, on which numbers 1 to 20 are written, are mixed thoroughly and then a ticket is drawn at random out of them. Find the probability that the number on the drawn ticket is a multiple of 3 or 7. 
Answer: Total number of possible outcomes = 20
Let E be the probability of getting a ticket number which is a multiple of 3 or 7
Then, the favourable outcomes are : 3, 6, 7, 9, 12, 14, 15, 18
Number of favourable outcomes = 8
\( P(E) = \frac{\text{Favourable outcomes}}{\text{Possible outcomes}} = \frac{8}{20} = \frac{2}{5} \)
Hence, the required probability is \( \frac{2}{5} \).

Question. Cards marked with number 3, 4, 5, ...., 50 are placed in a box and mixed thoroughly. A card is drawn at random from the box. Find the probability that the selected card bears a perfect square number. 
Answer: Given cards are numbered from 3, 4, 5, ...., 50
Total possible outcomes (Total number of cards) = 48
Let E be the event of drawing a card having a perfect square.
Then, the favourable outcomes = 4, 9, 16, 25, 36, 49
Number of favourable outcomes = 6
\( \therefore P(E) = \frac{6}{48} = \frac{1}{8} \)

Question. Will the median class and modal class of grouped data always be different? justify your answer. 
Answer: Not always. Median gives the middle value of data whereas mode gives the maximum occuring frequency. So it really depends on the data given. The median and modal class may be the same in cases where the distribution of data is perfectly symmetrical.

Question. In a family with three children, there may be no girl, one girl, two girls or three girls. So, the probability of each is \( \frac{1}{4} \). Is this correct? Justify your answer.
Answer: No (False).
In a family with 3 children, events are (b, b, b), (g, b, b), (b, g, b), (b, b, g), (g, g, g), (g, g, b), (g, b, g), (g, b, b)
T(E) = 8
Required probability = \( \frac{\text{No. of favourable outcomes}}{\text{Total no. of outcomes}} \)
Total no. of outcomes = 8
Probability of having no girl = \( \frac{1}{8} \)
Probability of having 1 girl = \( \frac{3}{8} \)
Probability of having 2 girls = \( \frac{3}{8} \)
Probability of having 3 girls = \( \frac{1}{8} \)
Hence, the probability of each is not \( \frac{1}{4} \).

Question. Find the mode of the following distribution:
Marks: 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60
Number of Students: 4 | 6 | 7 | 12 | 5 | 6
Answer: Here, the modal class is 30 – 40; as this class marks has the maximum frequency 12.
So, for this class.
\( l = 30, f_1 = 12, f_0 = 7, f_2 = 5, h = 10 \)
So, \( \text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h \)
\( = 30 + \frac{12 - 7}{24 - 7 - 5} \times 10 = 30 + \frac{50}{12} = 34\frac{1}{6} \)

Question. Find the mode of the following distribution :
Classes: 0-20 | 20-40 | 40-60 | 60-80 | 80-100
Frequency: 10 | 8 | 12 | 16 | 4
Answer: Here, the modal class is 60-80. as it has maximum frequency.
For this class,
\( l = 60, h = 20, f_1 = 16, f_0 = 12 \text{ and } f_2 = 4 \)
So, \( \text{Mode} = 60 + \frac{16 - 12}{32 - 12 - 4} \times 20 = 60 + \frac{4}{16} \times 20 = 60 + 5 = 65 \)

Question. I toss three coins together. The possible outcomes are no heads, 1 head, 2 heads and 3 heads. So, I say that the probability of no heads is \( \frac{1}{4} \). What is wrong with this conclusion? 
Answer: No. When three coins are tossed together: Possible outcomes S = {HHH, HTT, THT, TTH, HHT, HTH, THH, and TTT}. Total outcomes, (T) = \( 2^3 = 8 \). No. of favourable outcomes (getting no heads) F(E) = 1. \( \therefore \) Probability of getting no heads = \( \frac{F(E)}{T(E)} = \frac{1}{8} \). Outcome ‘no heads’ means {TTT} \( \therefore P(\text{no heads}) = \frac{1}{8} \). Outcome ‘one head’ means {THT, HTT, TTH} \( \therefore P(\text{one head}) = \frac{3}{8} \). Outcome ‘two head’ means {HHT, HTH, THH} \( P(\text{Two head}) = \frac{3}{8} \). Hence, the probability of getting no head is \( \frac{1}{8} \) and not \( \frac{1}{4} \).

Question. A bag contains slips numbered from 1 to 100. If Fatima chooses a slip at random from the bag, it will either be an odd number or an even number. Since this situation has only two possible outcomes, so. the probability of each is \( \frac{1}{2} \). Justify. 
Answer: From 1 to 100, there are 50 even and 50 odd numbers. Total number of outcomes T(E) = 100, Number of favourable outcomes (even no.), F(E) = 50. So, P(E) i.e., probability of getting an even no. = \( \frac{50}{100} = \frac{1}{2} \). Similarly, probability of getting an odd no. = \( \frac{1}{2} \). Hence, probability of getting an odd no = probability of getting even no = \( \frac{1}{2} \).

Question. A Group Housing ociety has 600 members, who have their houses in the campus and decided to hold a Tree lantation rive on the occation of New Year. Each household was given the choice of planting a samplings of its choice. The number of different types of saplings planted were.
(i) Neem – 125
(ii) Peepal – 165
(iii) Creepers – 50
(iv) Fruit plants – 150
(v) Flowering plants – 110
On the opening ceremony. one of the plants is selected randomly for a prize. after reading the above passage, answer the following questions.
What is the probability that the selected plant is:
(A) A fruit plant or a flowering plant ?
(B) Either a neem plant or a peepal plant ? 
Answer: (A) Of the 600 plants, there are 150 fruit plants and 110 flowering plants. So, required probability = \( \frac{150 + 110}{600} = \frac{260}{600} = \frac{13}{30} \).
(B) Of the 600 plants, there are 290 (125 + 165) plants which are either neem plants or peepal plants. So, required probability = \( \frac{290}{600} = \frac{29}{60} \).

Question. Find the mode of the following frequency distribution:
Class: 15-20 | 20-25 | 25-30 | 30-35 | 35-40 | 40-45
Frequency: 3 | 8 | 9 | 10 | 3 | 2
Answer: The modal class is 30-35, as this class has the maximum frequency of 10. For this class, \( l = 30, h = 5, f_1 = 10, f_0 = 9 \text{ and } f_2 = 3 \). So, \( \text{mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h = 30 + \frac{10 - 9}{20 - 9 - 3} \times 5 = 30 + \frac{5}{8} = 30.625 \).

Question. If a number x is chosen at random from the number –3, –2, –1, 0, 1, 2, 3, What is probability that \( x^2 < 4 \)? 
Answer: All possible outcomes are –3, –2, –1, 0, 1, 2, 3. Favourable outcomes are – 2, – 1, 0, 1, 2 (As \( x^2 \ge 4 \)). So, required probability = \( \frac{5}{7} \).

Question. If X, M and Z are denoting mean, median and mode of a data and X : M = 9 : 8, then the ratio M : Z is? 
Answer: Mode = 3 median – 2 mean \( \Rightarrow Z = 3M - 2X \). Given \( X : M = 9 : 8 \Rightarrow \frac{X}{M} = \frac{9}{8} \Rightarrow X = \frac{9M}{8} \). On putting the value of X in equation: \( Z = 3M - 2 \times \frac{9M}{8} = 3M - \frac{9M}{4} = \frac{12M - 9M}{4} = \frac{3M}{4} \). \( \frac{M}{Z} = \frac{4}{3} \). \( M : Z = 4 : 3 \).

Question. Find the mean of the following distribution:
Class: 3-5 | 5-7 | 7-9 | 9-11 | 11-13
Frequency: 5 | 10 | 10 | 7 | 8
[CBSE 2020]
Answer:
Class | Frequency | Class mark (x) | fx
3-5 | 5 | 4 | 20
5-7 | 10 | 6 | 60
7-9 | 10 | 8 | 80
9-11 | 7 | 10 | 70
11-13 | 8 | 12 | 96
\( \sum f = 40, \sum fx = 326 \). So, mean = \( \frac{\sum fx}{\sum f} = \frac{326}{40} = 8.15 \).

Question. Find the mode of the following data:
Class: 0-20 | 20-40 | 40-60 | 60-80 | 80-100 | 100-120 | 120-140
Frequency: 6 | 8 | 10 | 12 | 6 | 5 | 3
Answer: As the maximum frequency is 12 for the class 60-80. So, the modal class is 60-80. For this class, \( l = 60, h = 20, f_1 = 12, f_0 = 10 \text{ and } f_2 = 6 \). So, \( \text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h = 60 + \frac{12 - 10}{24 - 10 - 6} \times 20 = 60 + \frac{40}{8} = 60 + 5 = 65 \). Thus, mode is 65.

Question. A child has a die whose 6 faces show the letters given below: A B C A A B. The die is thrown once. What is the probability of getting (i) A (ii) B ?
Answer: Total number of outcomes = 6.
(i) Let the probability of getting A be \( E_1 \). Then, favourable number of cases = 3. \( P(E_1) = \frac{3}{6} = \frac{1}{2} \).
(ii) Let the probability of getting B be \( E_2 \). Then, favourable number of cases = 2. \( P(E_2) = \frac{2}{6} = \frac{1}{3} \).

Question. A card is drawn at random from a pack of 52 playing cards. Find the probability of drawing a card which is neither a spade nor a king. 
Answer: Total number of possible outcomes = Total number of playing cards = 52. Total numbers of spade cards + 3 kings (as one king is already in spade) = 13 + 3 = 16. \( \therefore \) Cards which are neither spade nor kings: = 52 – 16 = 36. \( P(\text{getting a card neither spade nor king}) = \frac{36}{52} = \frac{9}{13} \). Hence, the required probability is \( \frac{9}{13} \).

Question. A die is thrown once. Find the probability of getting a number which (A) is a prime number (B) lies between 2 and 6. 
Answer: In a single throw of a die, all possible outcomes are 1, 2, 3, 4, 5, 6. Total number of possible outcomes = 6.
(A) Let \( E_1 \) be the event of getting a prime number. Then, the favourable outcomes are 2, 3, 5. No. of favourable outcomes = 3. \( \therefore P(E_1) = \frac{3}{6} = \frac{1}{2} \).
(B) Let, \( E_2 \) be the event of getting a number between 2 and 6. Then, the favourable outcomes are 3, 4, 5. No. of favourable outcomes = 3. \( \therefore P(E_2) = \frac{3}{6} = \frac{1}{2} \).
Hence, the probability of getting a prime number and getting a number between 2 and 6 is \( \frac{1}{2} \) each.

Question. 20 cards from 11 to 30, are put in a box and mixed thoroughly. A card is then drawn from the box at random. Find the probability that the number on the drawn card is a prime number.
Answer: The total number of outcomes = 20. Let, E be the event of getting a prime number on the drawn card. Then, favourable outcomes = 11, 13, 17, 19, 23, 29. \( \therefore \) Number of favourable outcomes = 6. \( P(E) = \frac{\text{Favourable outcomes}}{\text{Total outcomes}} = \frac{6}{20} = \frac{3}{10} \). Hence, the required probability is \( \frac{3}{10} \).

Question. Find the probability that in the leap year there will be 53 Tuesdays. 
Answer: In a leap year there are 366 days. 366 days = 52 weeks + 2 days. These 2 days can be : (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday), (Sunday, Monday). Total number of outcomes = 7. Number of favourable outcomes = 2. P(53 Tuesdays in a leap year) = \( \frac{2}{7} \). Hence, the required probability is \( \frac{2}{7} \).

Question. Two different dice are thrown together. Find the probability that the product of the numbers appeared is less than 18. 
Answer: When 2 dice are thrown, then all possible outcomes are: (1, 1) to (6, 6). Number total possible outcomes = 36. Let, E be the probability that the product of the numbers appeared is less than 18. Then, favourable cases are: (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (4, 1), (4, 2), (4, 3), (4, 4), (5, 1), (5, 2), (5, 3), (6, 1), (6, 2). Number of favourable cases = 26. \( P(E) = \frac{26}{36} = \frac{13}{18} \). Hence, the required probability is \( \frac{13}{18} \).

Question. 15 cards numbered from 1 to 15 are put in a box and mixed thoroughly. Then, a card is drawn at random from the box. Find the probability that the number on the drawn card is divisible by 2 or 3. 
Answer: Total number of tickets = 15. Total possible outcomes = 15. Let E be the probability that the number on the drawn card is divisible by 2 or 3. Then, numbers divisible by 2 or 3 are = 2, 3, 4, 6, 8, 9, 10, 12, 14, 15. Total number of favourable cases = 10. \( P(E) = \frac{10}{15} = \frac{2}{3} \).

Question. An integer is chosen between 70 and 100, Find the probability that it is (A) a prime number (B) divisible by 7 
Answer: Total number of integers = 29. (A) Prob. (prime number) = \( \frac{6}{29} \). (B) Prob. (number divisible by 7) = \( \frac{4}{29} \).