CBSE Class 10 Mathematics Statistics VBQs Set D

SHORT ANSWER  Type Questions

Question. Two different dice are tossed together. Find the probability (i) of getting a doublet (ii) of getting a sum 10, of the numbers on the two dice.
Answer: (i) Two dice tossed together.
\(\Rightarrow\) Total outcomes = \( 36 \).
Doublet: \( \{(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)\} \Rightarrow 6 \) possibilities.
Probability = \(\frac{\text{Favourable outcome}}{\text{Total outcome}} = \frac{6}{36} = \frac{1}{6}\).
(ii) Sum of 10: \( \{(4,6), (6,4), (5,5)\} \Rightarrow 3 \) possibilities.
Probability = \(\frac{\text{Favourable outcome}}{\text{Total outcome}} = \frac{3}{36} = \frac{1}{12}\). 

Question. An integer is chosen at random between 1 and 100. Find the probability that it is: (i) divisible by 8. (ii) not divisible by 8.
Answer: Integers between 1 and 100.
\(\Rightarrow\) Total = \( 98 \) possible outcomes.
(i) Divisible by 8 \(\rightarrow 12\) numbers: \( \{8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96\} \).
\(\Rightarrow\) Probability = \(\frac{\text{Favourable outcome}}{\text{Total outcome}} = \frac{12}{98} = \frac{6}{49}\).
(ii) Not divisible by 8 \(\Rightarrow 98 - 12 = 86\) numbers.
\(\Rightarrow\) Probability = \(\frac{\text{Favourable outcome}}{\text{Total outcome}} = \frac{86}{98} = \frac{43}{49}\). 

Question. A die is thrown twice. Find the probability that 5 will not come up either time.
Answer: Total number of possible outcomes = \( 36 \).
Number of favourable outcomes = \( 25 \).
[Excluding \( (1, 5), (5, 1), (2, 5), (5, 2), (3, 5), (5, 3), (4, 5), (5, 4), (5, 5), (5, 6), (6, 5) \)]
So, required probability = \( \frac{25}{36} \).

Question. Cards marked with number 5 to 50 are placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that the number on the card taken out is (a) a prime number less than 10 (b) a number which is a perfect square. 
Answer: Total number of outcomes = \( 46 \).
(A) Let, the probability of getting a prime number less than 10 be \( E_1 \).
Then, favourable outcomes = \( 5, 7 \).
Total number of favourable outcomes = \( 2 \).
\( P(E_1) = \frac{2}{46} = \frac{1}{23} \).
(B) Let, the probability of getting a number which is a perfect square be \( E_2 \).
Then, favourable outcomes = \( 9, 16, 25, 36, 49 \).
Total number of favourable outcomes = \( 5 \).
\( P(E_2) = \frac{5}{46} \).

Question. Find the mean of the distribution:
Class: 1-3 | 3-5 | 5-7 | 7-10
Frequency: 9 | 22 | 27 | 17
[CBSE 2013]
Answer: We first find the class marks \( (x_i) \) of each class:
Class 1-3: \( x_i = 2, f_i = 9, f_i x_i = 18 \)
Class 3-5: \( x_i = 4, f_i = 22, f_i x_i = 88 \)
Class 5-7: \( x_i = 6, f_i = 27, f_i x_i = 162 \)
Class 7-10: \( x_i = 8.5, f_i = 17, f_i x_i = 144.5 \)
\( \sum f_i = 75 \), \( \sum f_i x_i = 412.5 \).
Mean, \( \bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{412.5}{75} = 5.5 \).
Hence, the mean of the given distribution is 5.5.

Question. Two dice are thrown at the same time. Find the probability of getting: (A) the same number on both dice. (B) different numbers on both dice.
Answer: (A) Let E be the event of getting the same number on both dice.
\( \therefore \) Total no. of all possible outcomes, \( T(E) = 36 \).
For getting the same no. on both dice, favourable outcomes are \( (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) \).
\( \therefore \) No. of favourable outcomes, \( F(E) = 6 \).
Required probability, \( P(E) = \frac{F(E)}{T(E)} = \frac{6}{36} = \frac{1}{6} \).
(B) Probability of getting different numbers on both dice = \( 1 - \text{Probability of getting same number} \).
\( P'(E) = 1 - P(E) = 1 - \frac{1}{6} = \frac{5}{6} \).

Question. A coin is tossed two times. Find the probability of getting at most one head. 
Answer: Possible outcomes are \( \{(H, H), (H, T), (T, H), (T, T)\} \).
Total no. of outcomes, \( T(E) = 4 \).
Favourable outcomes of getting at most 1 head \( \{(T, T), (H, T), (T, H)\} \).
No. of favourable outcomes, \( F(E) = 3 \).
\( \therefore \) Required probability \( P(E) = \frac{F(E)}{T(E)} = \frac{3}{4} \).

Question. Two dices were rolled once. Find the probability of getting such numbers on the two dice, whose product is 12. 
Answer: When 2 dices are rolled, total possible outcomes are 36.
Favourable outcomes = \( \{(2, 6), (3, 4), (4, 3), (6, 2)\} = 4 \).
Therefore, probability = \( \frac{4}{36} = \frac{1}{9} \).

Question. Apoorv throws two dice at once and computes the product of the numbers appearing on the dice. Peehu throws one die and squares the number that appears on it. Who has a better chance of getting the number 36. Why?
Answer: Peehu has a better chance.
For Apoorv:
He throws two dice at once \(\therefore\) Total no. of outcomes \( T(E) = 36 \).
Number of outcomes for getting product 36 i.e., \( (6, 6) \).
Favourable outcome \( F(E) = 1 \).
\( P(E) \) or probability of getting the no. 36 for Apoorv = \( \frac{F(E)}{T(E)} = \frac{1}{36} \).
For Peehu:
She throws one dice \(\therefore\) Total no. of outcomes \( T'(E) = 6 \).
Number of outcomes for getting square 36 i.e., \( (6, 6) \) (as she squares the number on the die).
Favourable outcome \( F'(E) = 1 \).
\( \therefore \) Probability of getting square of the no. is \( P'(E) = \frac{F'(E)}{T'(E)} = \frac{1}{6} \).
Since \( P'(E) > P(E) \), Peehu has a better chance.

Question. From the following distribution, find the median:
Classes: 500 – 600 | 600 – 700 | 700 – 800 | 800 – 900 | 900 – 1000
Frequency: 36 | 32 | 32 | 20 | 30
Answer:
Classes | Frequency | Cum. Frequency
500 – 600 | 36 | 36
600 – 700 | 32 | 68
700 – 800 | 32 | 100
800 – 900 | 20 | 120
900 – 1000 | 30 | 150
Here, \( \frac{N}{2} = \frac{150}{2} = 75 \).
So, the median class is 700 – 800.
For this class, \( l = 700, h = 100, \frac{N}{2} = 75, cf = 68, f = 32 \).
So, \( \text{Median} = l + \frac{\frac{N}{2} - cf}{f} \times h \)
\( = 700 + \frac{75 - 68}{32} \times 100 = 700 + \frac{700}{32} = 721.875 \).

Question. The probability of selecting a blue marble at random from a jar that contains only blue, black and green marbles is \( \frac{1}{5} \). The probability of selecting a black marble at random from the same jar is \( \frac{1}{4} \). If the jar contains 11 green marbles, find the total number of marbles in the jar. 
Answer: Let the total number of marbles in jar be \( x \).
Given: \( P(\text{blue marble}) = \frac{1}{5} \), \( P(\text{black marble}) = \frac{1}{4} \).
Number of green marbles = \( 11 \).
\( P(\text{green marbles}) = 1 - [P(\text{blue marble}) + P(\text{black marbles})] \)
\( = 1 - \left(\frac{1}{5} + \frac{1}{4}\right) = 1 - \frac{9}{20} = \frac{11}{20} \).
But, \( P(\text{green marbles}) = \frac{11}{x} \).
\( \frac{11}{x} = \frac{11}{20} \) or \( x = 20 \).

Question. Calculate the mode of the following distribution:
Class: 10-15 | 15-20 | 20-25 | 25-30 | 30-35
Frequency: 4 | 7 | 20 | 8 | 1
Answer: In the given distribution, the modal class is 20 – 25, with the maximum frequency 20.
\(\therefore\) Lower limit of modal class, \( x_k = 20 \).
Frequency of modal class, \( f_k = 20 \).
Frequency of class preceding the modal class, \( f_{k-1} = 7 \).
Frequency of class succeeding the modal class, \( f_{k+1} = 8 \).
Height of the class marks, \( h = 5 \).
\(\therefore \text{Mode, } M = x_k + \left\{ \frac{f_k - f_{k-1}}{2f_k - f_{k-1} - f_{k+1}} \right\} \times h \)
\( = 20 + \left\{ \frac{20 - 7}{40 - 7 - 8} \right\} \times 5 = 20 + \frac{13 \times 5}{25} = 20 + 2.6 = 22.6 \).
Hence, the mode of the given distribution is 22.6.

Question. Calculate the mean of the following data:
Class: 4-7 | 8-11 | 12-15 | 16-19
Frequency: 5 | 4 | 9 | 10
Answer: Given frequency table is in inclusive form, so we need to convert the data into exclusive form or continuous by subtracting 0.5 from the lower limit and adding 0.5 in the upper limit of each class. Also, we find class marks \( (x_i) \) of each class and then proceed as follows:
Class 3.5-7.5: \( x_i = 5.5, f_i = 5, f_i x_i = 27.5 \)
Class 7.5-11.5: \( x_i = 9.5, f_i = 4, f_i x_i = 38 \)
Class 11.5-15.5: \( x_i = 13.5, f_i = 9, f_i x_i = 121.5 \)
Class 15.5-19.5: \( x_i = 17.5, f_i = 10, f_i x_i = 175 \)
\( \sum f_i = 28 \), \( \sum f_i x_i = 362 \).
Mean \( \bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{362}{28} = 12.93 \).

Question. The following table gives the number of pages written by Sarika for completing her own book for 30 days:

Calculate the average number of pages written in 30 days. 
Answer: We need to convert the data into continuous classes by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class.

\( \therefore \text{Mean, } \bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{780}{30} = 26 \).
[Alternate Method]

\( a = \text{assumed mean} = 23 \)
\( \text{Mean, } \bar{x} = a + \frac{\sum f_i d_i}{\sum f_i} = 23 + \frac{90}{30} = 23 + 3 \)
\( \therefore \bar{x} = 26 \)
Hence, the mean the number of the pages written per day is 26.

Question. Find the mode of the following frequency distribution. 

Answer: In the given frequency distribution the modal class is 30-40, because it has the maximum frequency of 16.
Here, \( x_k = 30, f_k = 16, f_{k-1} = 10, f_{k+1} = 12 \) and \( h = 10 \)
\( \therefore \text{Mode, } M_0 = x_k + \left\{ h \times \frac{(f_k - f_{k-1})}{(2f_k - f_{k-1} - f_{k+1})} \right\} \)
\( = 30 + \left\{ 10 \times \frac{(16 - 10)}{32 - 10 - 12} \right\} \)
\( = 30 + \left\{ \frac{10 \times 6}{10} \right\} \)
\( = 30 + 6 \)
\( = 36 \)
Hence, the mode of the given frequency distribution is 36.

Question. All kings, jacks and diamonds have been removed from a pack of playing cards and the remaining cards are well-shuffled. A card is then drawn at random. Find the probability that the drawn card is a
(A) face card.
(B) black card. 
Answer: Number of cards removed = 4 kings + 4 jacks + 11 diamond (since, one jack & king of diamond is removed) = 19
\( \therefore \) Total number of remaining cards = 52 - 19 = 33
Now, there will be 3 queens left as face cards.
(A) Number of face cards (since, face cards are only jacks, queens and kings) = 3
\( \therefore P(\text{a face card}) = \frac{3}{33} = \frac{1}{11} \)
Hence, the required probability is \( \frac{1}{11} \).
(B) Number of black cards = 26 - 2 kings - 2 jacks = 26 - 4 = 22
\( \therefore P(\text{black card}) = \frac{22}{33} = \frac{2}{3} \)
Hence, the required probability is \( \frac{2}{3} \).

Question. The daily income of a sample of 50 employees are tabulated as follows:

Find the mean daily income of the employees. 
Answer: The given data is not continuous, so we need to convert the data into continuous by subtracting 0.5 from the lower limit and adding 0.5 in the upper limit of each class:

\( a = \text{assumed mean} = 300.5 \)
\( d_i = x_i - a \).
\( \text{Mean } \bar{x} = a + \frac{\sum f_i d_i}{\sum f_i} = 300.5 + \frac{2800}{50} \)
\( = 300.5 + 56 \)
\( \bar{x} = 356.5 \)
Thus, the mean daily income of the employees is ₹ 356.50.

Question. A bag contains 12 balls out of which some are white and the others are red. If the probability of drawing a white ball at random from the bag is \( \frac{2}{3} \), then find how many red balls are there in the bag. 
Answer: Let the number of red balls in bag be 'x'. The total number of balls in bag = 12
Then, number of white balls = 12 - x
\( P(\text{drawing a white ball}) = \frac{\text{number of white balls}}{\text{Total number of balls}} \)
\( = \frac{12 - x}{12} \)
But, \( P(\text{drawing a white ball}) = \frac{2}{3} \)
\( \therefore \frac{12 - x}{12} = \frac{2}{3} \)
\( \Rightarrow 12 - x = 8 \)
\( \Rightarrow x = 4 \)

Question. An aircraft has 120 passenger seats. The number of seats occupied during 100 flights is given in the following table:

Determine the mean number of seats occupied during the flights. [NCERT]
Answer:

Here, assumed mean, a = 110
\( \therefore \text{Mean, } \bar{x} = a + \frac{\sum f_i d_i}{\sum f_i} \)
\( = 110 + \frac{(-8)}{100} = 110 - 0.08 \)
\( = 109.92 \)
Thus, the mean number of seats occupied over the flights is 109.92 or 110 (approx).

Question. A game consists of tossing a one-rupee coin 3 times and noting the outcome each time. Ramesh wins the game if all the tosses give the same result (i.e. three heads or three tails) and loses otherwise. Find the probability of Ramesh losing the game. 
Answer: When 3 coins are tossed simultaneously all possible outcomes are:
HHH, HHT, HTH, THH, HTT, THT, TTH, TTT
Total number of possible outcomes = 8
Let \( E_1 \) be the event of not getting 3 heads or 3 tails i.e. losing.
Then, the outcomes for losing the game are:
HHT, HTH, THH, HTT, THT, TTH.
Number of total outcomes for losing games = 6
\( \therefore P(E_1) = \frac{6}{8} = \frac{3}{4} \)

Question. Two different dice are thrown together. Find the probability that the numbers obtained.
(A) have a sum less than 7
(B) have a product less than 16
(C) is a doublet of odd numbers. 
Answer: The outcomes when 2 dice are thrown together 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), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
Total number of outcomes = 36
(A) Let E be the event of getting the numbers whose sum is less than 7.
The favourable cases are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2) and (5, 1)
Number of favourable outcomes = 15
\( \therefore p(E_1) = \frac{15}{36} = \frac{5}{12} \)
Hence, the required probability is \( \frac{5}{12} \).
(B) Let \( E_2 \) be the event of getting the numbers whose product is less than 16.
The outcomes in favour of event \( E_2 \) 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), (5, 1), (5, 2), (5, 3), (6, 1) and (6, 2).
Number of favourable outcomes = 25
\( \therefore p(E_2) = \frac{25}{36} \)
Hence, the required probability is \( \frac{25}{36} \).
(C) Let \( E_3 \) be the event of getting the numbers which are doublets of odd numbers.
The outcomes in favour of event \( E_3 \) are (1, 1), (3, 3) and (5, 5)
Number of favourable outcomes = 3
\( \therefore P(E_3) = \frac{3}{36} = \frac{1}{12} \)
Hence the required probability is \( \frac{1}{12} \).

Question. A lot consists of 144 ball pens of which 20 are defective. The customers will buy a ball pen if it is good, but will not buy a defective ball pen. The shopkeeper draws one pen at random from the lot and gives it to the customers. What is the probability that
(A) customer will buy the ball pen
(B) customer will not buy the ball pen 
Answer: Total number of ball pens = 144
\( \therefore \) Total number of outcomes is 144.
Defective ball pens = 20
Non-defective ball pens = 144 - 20 = 124
(A) Let \( E_1 \) be the probability that customer will buy a pen i.e., ball-pen is non-defective.
Total number of non-defective pens = 124
\( P(E_1) = \frac{124}{144} = \frac{31}{36} \)
Hence, the probability of buying a pen is \( \frac{31}{36} \).
(B) Probability of not buying pen
= 1 - probability of buying a pen
= 1 - \( P(E_1) \)
= 1 - \( \frac{31}{36} = \frac{5}{36} \)
Hence, the probability of not-buying a pen is \( \frac{5}{36} \).

Question. The weights (in kg) of 50 wrestlers are recorded in the following table:

Find the mean weight of the wrestlers. 
Answer:

Assumed mean, a = 125
\( \therefore \text{Mean } \bar{x} = a + \frac{\sum f_i d_i}{\sum f_i} = 125 - \frac{80}{50} \)
\( = 125 - 1.6 \)
\( = 123.4 \)
Hence, the mean weight of the wrestler is 123.4 kg.

Question. A coin is tossed 3 times. Write all the possible outcomes. Find the probability of getting at least 2 heads. 
Answer: When 3 coins are tossed simultaneously all possible outcomes are: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.
Total number of possible outcomes = 8
Let E be the probability of getting at least 2 heads.
Then, favourable outcomes are: HHH, HHT, HTH, THH.
Number of favourable outcomes = 4
\( \therefore P(E) = \frac{4}{8} = \frac{1}{2} \)
Hence, the required probability is \( \frac{1}{2} \).

Question. Black aces and black queens are removed from a pack of 52 cards. The remaining cards are reshuffled and then a card is drawn. Find the probability of getting:
(A) a black card
(B) an ace. 
Answer: Number of cards removed = 2 + 2 = 4
Total number of remaining cards = 52 - 4 = 48
Now, there are 2 aces and 2 queens of red colour left.
(A) Number of black cards left = 26 - 4 = 22
\( P(\text{getting a black card}) = \frac{22}{48} = \frac{11}{24} \)
Hence, the required probability is \( \frac{11}{24} \).
(B) Number of aces = 4 - 2 = 2
\( P(\text{getting an ace}) = \frac{2}{48} = \frac{1}{24} \)
Hence, the required probability is \( \frac{1}{24} \).

Question. In a single throw of a pair of different dice, what is the probability of getting (A) a prime number on each dice (B) a total of 9 or 11 ? 
Answer: When 2 dice are thrown, the total possible outcomes are 36.
(A) Let \( E_1 \) be the event of getting a prime number on both dice.
Then the favourable outcomes = (2, 2), (2, 3), (2, 5), (3, 2), (3, 3), (3, 5), (5, 2), (5, 3), (5, 5). [Note: text in OCR lists 8, but (2,2) is a prime-prime pair. Favourable outcomes = 9]
Number of favourable outcomes = 9
\( P(E_1) = \frac{9}{36} = \frac{1}{4} \)
Hence, the required probability is \( \frac{1}{4} \).
(B) Let \( E_2 \) be the event of getting a total of 9 or 11 on both dice.
Then, the favourable outcomes = (3, 6), (4, 5), (5, 4), (6, 3), (5, 6), (6, 5)
Number of favourable outcomes = 6
\( P(E_2) = \frac{6}{36} = \frac{1}{6} \)
Hence, the required probability is \( \frac{1}{6} \).

Question. A carton of 24 bulbs contain 6 defective bulbs. One bulb is drawn at random. What is the probability that the bulb is not defective? If the bulb selected is defective and it is not replaced and a second bulb is selected at random from the rest, what is the probability that the second bulb is defective? 
Answer: Total no. of bulbs, \( T(E) = 24 \)
Good bulbs = 18
Defective bulbs = 6
Let \( E_1 \) be the event of selecting a non-defective bulbs i.e., selecting a good bulb.
Number of favourable outcomes, \( F(E) = 18 \).
\( \therefore \) Probability that bulb is not defective, \( P(E) = \frac{F(E)}{T(E)} = \frac{18}{24} = \frac{3}{4} \)
Suppose, the selected bulb is defective and not replaced, then total no. of bulbs remaining in the carton = 23.
i.e., total remaining bulb \( T(E') = 23 \).
Out of them, 18 are good bulbs and 5 are defective.
i.e., \( F(E') = 5 \)
\( \therefore P \text{ (selecting second defective bulb)} = \frac{F(E')}{T(E')} = \frac{5}{23} \).

Question. At a fete, cards bearing numbers 1 to 1000, (one number on one card, are put in a box. Each player selects one card at random and that card is not replaced. If the selected card has a perfect square greater than 500, the player wins a prize. What is the probability that:
(A) the first player wins a prize?
(B) the second player wins a prize, if the first has won? 
Answer: Total no. of outcomes, \( T(E) = 1000 \).
(A) Let \( E_1 \) be the event that the first player wins the prize i.e. selects a card which is a perfect square greater than 500.
Favourable outcomes \( \{529, 576, 625, 676, 729, 784, 841, 900, 961\} \).
\( \therefore \) No. of favourable outcomes, \( F(E_1) = 9 \)
Required probability, \( P(E_1) = \frac{F(E_1)}{T(E)} = \frac{9}{1000} \).
(B) Let \( E_2 \) be the event that the second player wins the prize i.e., the remaining cards which have a perfect square greater than 500 i.e., 8.
No. of favourable outcomes, \( F(E_2) = 8 \).
Total no. of outcomes as 1 card was already selected by first player \( T(E_2) = 999 \).
\( \therefore \) Required probability, \( P(E_2) = \frac{F(E_2)}{T(E_2)} = \frac{8}{999} \).
Hence, \( P(E_1) = \frac{9}{1000} \) and \( P(E_2) = \frac{8}{999} \).

Question. The table below shows the salaries of 280 persons:

Calculate the median salary of the data. 
Answer: Distribution of frequencies:
No. of people = 280.
\( \Rightarrow \frac{n}{2} = 140 \), the 140th term lies in class interval 10-15.
\( \Rightarrow \) median class = 10-15.
\( l=10, h=5, f=133, \frac{n}{2}=140, cf=49 \).
We know, \( \text{median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h \).
\( \Rightarrow \text{median} = 10 + \frac{140 - 49}{133} \times 5 \)
\( = 10 + \frac{91}{133} \times 5 \)
\( = 10 + \frac{13}{19} \times 5 \)
\( = 10 + \frac{65}{19} \)
\( = 10 + 3.421 \)
\( = 13.421 \).
The median salary is 13.421 thousand rupees.

Question. A bag contains 15 white and some black balls. If the probability of drawing a black ball from the bag is thrice that of drawing a white ball, find the number of black balls in the bag. 
Answer: Let there be \( x \) black balls and 15 white balls.
Total balls \( = n(S) = 15 + x \)
\( P(\text{drawing black ball}) = 3 \times P(\text{drawing white ball}) \)
\( \frac{x}{(15 + x)} = 3 \times \frac{15}{(15 + x)} \)
\( x = 3 \times 15 \)
\( x = 45 \)
\( \therefore \) There are 45 black balls in the bag. 

Question. From a pack of 52 playing cards, Jacks, Queens and Kings of red colour are removed. From the remaining, a card is drawn at random. Find the probability that drawn card is:
(i) a black King
(ii) a card of red colour
(iii) a card of black colour 
Answer: Total cards = 52. Red Jacks, Queens, Kings removed = 2 + 2 + 2 = 6.
Remaining cards = 52 - 6 = 46.
(i) Number of black Kings = 2. \( P(\text{black King}) = \frac{2}{46} = \frac{1}{23} \).
(ii) Number of red cards left = 26 - 6 = 20. \( P(\text{red card}) = \frac{20}{46} = \frac{10}{23} \).
(iii) Number of black cards left = 26. \( P(\text{black card}) = \frac{26}{46} = \frac{13}{23} \).

The mean of the following frequency distribution is 62.8 and the sum of all the frequencies is 50. Compute the missing frequencies \(f_1\) and \(f_2\).

Answer: We know that,
Mean \( = A + \frac{\sum f_i u_i}{\sum f_i} \times h\)
\( 62.8 = 50 + \frac{28 - f_1 + f_2}{30 + f_1 + f_2} \times 20 \)
\( \frac{28 - f_1 + f_2}{30 + f_1 + f_2} = \frac{12.8}{20} \)
\( 560 + 20 f_2 - 20 f_1 = 384 + 12.8 f_1 + 12.8 f_2 \)
\( 32.8 f_1 - 7.2 f_2 - 176 = 0 \)
\( 4.1 f_1 - 0.9 f_2 - 22 = 0 \) .....(i)
Also, we are given that \( f_1 + f_2 + 30 = 50 \)
i.e. \( f_1 + f_2 = 20 \) .....(ii)
From eqs. (i) and (ii), we get
\( f_1 = 8 \) and \( f_2 = 12 \)

Question. The distribution given below show the number of wickets taken by bowlers in one-day cricket matches, Find the mean and the median for the numbers of wickets taken. 

Answer: Calculation of Mean

So, Mean \( = A + \frac{\sum f_i u_i}{\sum f_i} = 120 + \frac{240}{45} = 120 + 5.33 = 125.33 \)

Calculation of Median

Here, \(N = 45, \text{ i.e., } \frac{N}{2} = 22.5\)
So, median class is 100 – 140
For this class, \(l = 100, h = 40, cf = 12, \frac{N}{2} = 22.5, f = 16\)
So, Median \( = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h = 100 + \left( \frac{22.5 - 12}{16} \right) \times 40 = 100 + 26.25 = 126.25 \)

Question. The median of the following data is 525. Find the values of \(x\) and \(y\), if total frequency is 100:

Answer:

It is given that \(\sum f = 100\). So, \(76 + x + y = 100 \Rightarrow x + y = 24 \) ...(i)
Also, it is given that median is 525. So, the median class is 500 – 600.
\(l = 500, h = 100, \frac{N}{2} = 50, cf = 36 + x \text{ and } f = 20\)
So, Median \( = l + \frac{\frac{N}{2} - cf}{f} \times h \)
\( 525 = 500 + \frac{50 - (36 + x)}{20} \times 100 \)
\( 25 = 5(14 - x) \)
\( 14 - x = 5 \Rightarrow x = 9 \)
From (i), \( y = 15 \).
Thus, \(x = 9, y = 15 \).

Question. Find the mean marks of the students for the following distribution:

Answer:

\( \therefore \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} = \frac{4140}{80} = 51.75 \).

Alternate Method: Deviation method (Assumed mean \(a = 55\))
Mean \( \bar{x} = a + \frac{\sum f_i d_i}{\sum f_i} = 55 + \frac{-260}{80} = 55 - 3.25 = 51.75 \).

Question. Find the values of frequency ‘x’ and ‘y’ in the following frequency distribution table, if \(N = 100\) and median is 32. 

Answer:

the median class is 30-40:
Now, \(75 + x + y = 100 \Rightarrow x + y = 25 \) ...(i)
Median, \(M_e = l + \frac{\frac{N}{2} - cf}{f} \times h \)
Here, \(l = 30, N = 100, cf = 35 + x, f = 30, h = 10\)
\( 32 = 30 + \left( \frac{50 - (35 + x)}{30} \right) \times 10 \)
\( 2 = \frac{15 - x}{3} \Rightarrow 6 = 15 - x \Rightarrow x = 9 \)
If we put the value of ‘x’ in equation (i)
\( y = 25 - 9 = 16 \)
Hence, the values of ‘x’ and ‘y’ are 9 and 16 respectively.

Question. The weights of tea in 70 packets are shown in the following table:

Find the mean weight of the packets. [CBSE 2014, 12]
Answer:

Here, assumed mean, \(a = 203.5\), Class width, \(h = 1\)
Using assumed mean method, Mean \( \bar{x} = a + \frac{\sum f_i d_i}{\sum f_i} \times h = 203.5 + \frac{(-108)}{70} \times 1 = 203.5 - 1.54 = 201.96 \)
Hence, required mean weight is 201.96 g.

Question. If the median of the following frequency distribution is 32.5. Find the values of \(f_1\) and \(f_2\). 

Answer:

Given: Median \( = 32.5 \). Then, the median class is 30–40.
Median \( M_e = l + \left\{ \frac{\frac{N}{2} - cf}{f} \right\} \times h \)
\( 32.5 = 30 + \left\{ \frac{20 - (14 + f_1)}{12} \right\} \times 10 \)
\( 2.5 = \frac{10}{12} (6 - f_1) \)
\( 3 = 6 - f_1 \Rightarrow f_1 = 3 \)
Also, \( 31 + f_1 + f_2 = 40 \Rightarrow 31 + 3 + f_2 = 40 \Rightarrow f_2 = 6 \)
Hence, the values of \(f_1\) and \(f_2\) are 3 and 6, respectively.

Question. The mean of the following distribution is 18. Find the frequency of the class 19-21. 

Answer:

Given, mean \( \bar{x} = 18 \)
\( 18 = \frac{704 + 20 f}{40 + f} \)
\( 18(40 + f) = 704 + 20 f \)
\( 720 + 18 f = 704 + 20 f \)
\( 2f = 16 \Rightarrow f = 8 \)
Hence, the value of the frequency ‘f’ for class 19-21 is 8.

Question. From a pack of 52 playing cards. Jacks and Kings of red colour and Queens and Aces of black colour are removed. The remaining cards are mixed and a card is drawn at random. Find the probability that the drawn card is.
(A) a black Queen
(B) a card of red colour
(C) a Jack of black colour
(D) a face card 
Answer: Number of cards removed = (2 + 2 + 2 + 2) = 8
Total number of remaining cards = (52 – 8) = 44
Now, there are 2 jacks, 2 kings of black colour and 2 queens, 2 aces of red colour left.
(A) Number of black queens = 0. \( \therefore P(\text{getting a black queen}) = \frac{0}{44} = 0 \)
(B) Number of red cards = (26 – 4) = 22. \( \therefore P(\text{getting a red card}) = \frac{22}{44} = \frac{1}{2} \)
(C) Number of jacks of black colour = 2. \( \therefore P(\text{getting a black jack}) = \frac{2}{44} = \frac{1}{22} \)
(D) We know that jacks, queen and kings are face cards. Remaining face cards = (2 + 2 + 2) = 6. \( \therefore P(\text{getting a face card}) = \frac{6}{44} = \frac{3}{22} \)

Question. Daily wages of 110 workers, obtained in a survey, are tabulated below:

Compute the mean daily wages and modal daily wages of these workers.
Answer:

Mean daily wages \( = 170 + \frac{1}{110} \times 20 = Rs 170.19 \) approx.
Mode \( = 160 + \frac{22 - 20}{44 - 20 - 18} \times 20 = Rs 166.67 \) approx.

Question. Find the unknown entries a, b, c, d, e, f in the following distribution of the heights of the students in a class: 

Answer:

Comparing the columns:
\( a = 12 \)
\( 12 + b = 25 \Rightarrow b = 13 \)
\( c = 22 + b = 22 + 13 = 35 \)
\( 22 + b + d = 43 \Rightarrow 35 + d = 43 \Rightarrow d = 8 \)
\( 22 + b + d + e = 48 \Rightarrow 43 + e = 48 \Rightarrow e = 5 \)
\( f = 50 \)
Hence, \( a = 12, b = 13, c = 35, d = 8, e = 5, f = 50 \).

Question. A card is drawn at random from a well shuffled deck of playing cards. Find the probability that the card drawn is:
(A) a card of spade or an ace.
(B) a black king.
(C) neither a jack nor a king.
(D) either a king or a queen. 
Answer: (A) \( P(\text{spade or an ace}) = \frac{13+3}{52} = \frac{16}{52} = \frac{4}{13} \)
(B) \( P(\text{a black king}) = \frac{2}{52} = \frac{1}{26} \)
(C) \( P(\text{neither a jack nor a king}) = \frac{52 - 8}{52} = \frac{44}{52} = \frac{11}{13} \)
(D) \( P(\text{either a king or a queen}) = \frac{4+4}{52} = \frac{8}{52} = \frac{2}{13} \)