NCERT Exemplar Class 10 Maths Solutions Chapter 13 Statistics and Probability

Question 1

In the formula $\bar{x}= a+\frac{\sum f_{i}d_{i}}{\sum f_{i}}$
For finding the mean of grouped data, $d_i$’s are deviations from
(A) lower limits of the classes
(B) upper limits of the classes
(C) midpoints of the classes
(D) frequencies of the class marks

Answer: [C]
Solution. Mean: It is the average of the given numbers/observations. It is easy to calculate the mean. First of all, add up all the observations and then divide
by the total number of observations.
We know that d_i = x_i – a
where x_i is data and a is the mean
So, $d_i$ is the deviation from the midpoint of the classes.

Question 2

While computing the mean of grouped data, we assume that the frequencies are
(A) evenly distributed over all the classes
(B) centred at the class marks of the classes
(C) centred at the upper limits of the classes
(D) centred at the lower limits of the classes

Answer: [B]
Solution. Mean: It is the average of the given numbers/observations. It is easy to calculate the mean. First of all, add up all the observations and then divide by the total number of observations.
Hence, while computing the mean of grouped data, we assume that the frequencies are centered at the class marks of the classes

Question 3

If x_i’s are the mid points of the class intervals of grouped data, fi’s are the corresponding frequencies and $\bar{x}$ is the mean, then $\Sigma\left(f_{i} x_{i}-\bar{x}\right)$ is equal to
(A) 0 (B) –1 (C) 1 (D) 2

Answer: [A]
Solution. Mean: It is the average of the given numbers/observations. It is easy to calculate the mean. First of all, add up all the observations and then divide by the total number of observations.
That is mean $\left ( \bar{x} \right )= \frac{\sum f_{i}x_{i}}{n}$
By cross multiplication, we get
$\sum f_{i}x_{i}= \bar{x}n\cdots (1)$
$\sum \left ( f_{i}x_{i} -\bar{x}\right )= \sum f_{i}x_{i} -\sum \bar{x}$
$=n\bar{x}-\sum \bar{x}$ (from equation (1))
$=n\bar{x}-n\bar{x}$
= 0

Question 4

In the formula $\bar{x}= a+h\left ( \frac{\sum f_{i}u_{i}}{\sum f_{i}} \right )$ for finding the mean of grouped frequency distribution, u_i =
(A) $\frac{x_{i}+a}{h}$ (B) h(xi – a) (C) $\frac{x_{i}-a}{h}$ (D) $\frac{a-x_{i}}{h}$

Answer:

Answer. [C]
Solution. Mean: It is the average of the given numbers/observations. It is easy to calculate the mean. First of all, add up all the observations and then divide by the total number of observations.
Also we know that d_i = x_i – a and $u_{i}= \frac{d_{i}}{h}$
$u_{i}= \frac{d_{i}}{h}$
put d_i = x_i – a
$u_{i}= \frac{x_{i}-a}{h}$
Hence, option C is correct.

Question 5

The abscissa of the point of intersection of the less than type and of the more than type cumulative frequency curves of a grouped data gives its
(A) mean
(B) median
(C) mode
(D) all three above

Answer: [B]
Solution: Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However mostly we use frequency distribution to summarize categorical variables.
If we make a graph of less than type and of more than type grouped data and find the intersection point, then the value at the abscissa is the median of the grouped data.
Hence, option (B) is correct.

Question 6

For the following distribution:

The sum of the lower limits of the median class and the modal class is
(A) 15
(B) 25
(C) 30
(D) 35

Answer: [B]
Solution: Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However mostly we use frequency distribution to summarize categorical variables.

$\frac{N}{2}=\frac{66}{2}=33$ which ies in the class 10-15.
Hence, the median class is 10 – 15
The class with the maximum frequency is the modal class, which is 15 – 20
The lower limit of the median class = 10
The lower limit of the modal class = 15
Sum = 10 + 15 = 25

Question 7

Consider the following frequency distribution:

The upper limit of the median class is
(A) 17 ) 17.5 (C) 18 (D) 18.5

Answer: [B]

Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However mostly we use frequency distribution to summarize categorical variables.
Class is not continuous. So we have to make 1t continuous first.

$\frac{N}{2}= \frac{57}{2}= 28\cdot 5$
Here, the median class is 15.5 – 17.5
The upper limit of the median class is 17.5.

Question 8

For the following distribution :

The modal class is
(A) 10-20 (B) 20-30 (C) 30-40 (D) 50-60

Answer:

Answer. [C]
Solution.
Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However, mostly we use frequency distribution to summarize categorical variables.

The class with the highest frequency is 30-40
Hence, 30 – 40 is the modal class.

Question 9

Consider the data:

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]

Solution: Frequency distribution: It tells how frequencies are distributed overvalues in a frequency distribution. However mostly we use frequency distribution to summarize categorical variables.

$\frac{N}{2}= \frac{67}{2}= 33\cdot 5$
Median class = 125 – 145
upper limit of median = 145
The class with the maximum frequency is the modal class, which is 125 – 145
lower limit of modal class = 125
Difference of the upper limit of median and lower limit of modal = 145 – 125 = 20

Question 10

The times, in seconds, taken by 150 athletes to run a 110 m hurdle race are tabulated below :

The number of athletes who completed the race in less than 14.6 seconds is :
(A) 11 (B) 71 (C) 82 (D) 130

Answer: [C]
Solution: Frequency:- The number of times an event occurs in a specific period is called frequency.
The number of athletes who are below 14.6 = frequency of class (13.8-14) + frequency of class (14- 14.2) +
frequency of class (14.2-14.4) + frequency of class (14.4-14.6)
= 2 + 4 + 5 + 71 = 82
Hence, the frequency of race completed in less than 14.6 = 82

Question 11

Consider the following distribution :

The frequency of the class 30-40 is
(A) 3 (B) 4 (C) 48 (D) 51

Answer: [A]
Solution: Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However mostly we use frequency distribution to summarize categorical variables.

So, the frequency of class 30 – 40 is 3.

Question 12

If an event cannot occur, then its probability is
(A) 1 (B)$\frac{3}{4}$ (C) $\frac{1}{2}$ (D) 0

Answer. [D]
Solution: Probability; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Here number of favorable cases is 0.
Probability = $\frac{Number \, of\, favourable\, case}{Total\, number\, of\, cases}$
$\Rightarrow$Probability = $\frac{0}{\left ( Total\, cases \right )}= 0$

Question 13

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:

Answer. [D]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
(A) 1/3
Here 0 < 1/3 < 1
Hence, it can be the probability of an event.
(B) 0.1
Here 0 < 0.1 < 1
Hence, it can be the probability of an event.
(C) 3% = 3/100 = 0.03
Here 0 < 0.03 < 1
Hence, it can be the probability of an event.
(D)17/16
Here $\frac{17}{16}> 1$
Hence $\frac{17}{16}$ is not a probability of event
Hence, option (D) is the correct answer.

Question 14

An event is very unlikely to happen. Its probability is closest to
(A) 0.0001 (B) 0.001 (C) 0.01 (D) 0.1

Answer:

Answer. [A]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
The descending order of option (A), (B), (C), (D) is
0.1 > 0.01 > 0.001 > 0.0001 that is (D) > (C) > (B) > (A)
We can also say that it is the order of happening of an event.
Here, 0.0001 is the smallest one.
Hence, 0.0001 is very unlikely to happen

Question 15

If the probability of an event is p, the probability of its complementary event will be
(A) p – 1 (B) p (C) 1 – p (D)$1-\frac{1}{p}$

Answer:

Answer. [C]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
The probability of an event = p
Let the probability of its complementary event = q
We know that total probability is equal to 1.
Hence, p + q = 1
$\Rightarrow$q = 1 – p

Question 16

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:

Answer. [B]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
The probability expressed as a percentage of an event A is between 0 to 100.
Hence, we can say that probability can never be less than 0.
Hence, option (B) is correct.

Question 17

If P(A) denotes the probability of an event A, then
(A) P(A) < 0 (B) P(A) > 1 (C) 0 ≤ P(A) ≤ 1 (D) –1 ≤ P(A) ≤ 1

Answer:

Answer. [C]
Solution. Probability: Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
(A) P(A) < 0
It does not represent the probability of event A because the probability of an event can never be less than 0.
(B) P(A) > 1
It does not represent the probability of event A because the probability of an event can never be greater than 1.
(C) 0 ≤ P(A) ≤ 1
It represents the probability of event A because the probability of an event always lies between 0 to 1.
(D) –1 ≤ P(A) ≤ 1
It does not represent the probability of event A because the probability of an event can never be equal to -1.
Hence, option (C) is correct.

Question 18

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:

Answer. [A]
Solution. Probability: Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total number of cases = 52
Red face cards = 6
Favorable cases = 6
Let event A is to selection of a card from 52 cards.
The probability that it is a red card is p(A)
$P\left ( A \right )= \frac{Number\, of\, favourable\, cases}{Total\, number\, of\, cases}$
$\Rightarrow$$P\left ( A \right )=\frac{6}{52}= \frac{3}{26}$

Question 19

The probability that a non-leap year selected at random will contain 53 Sundays is
(A) $\frac{1}{7}$ (B) $\frac{2}{7}$ (C) $\frac{3}{7}$ (D) $\frac{5}{7}$

Answer:

Answer. [A]
Solution. Probability: Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
In 365 days, there are 52 weeks and 1 day.
If it contains 53 Sundays, then the 1 day of the year must be Sunday.
But there are a total of 7 days.
Hence, the total number of favourable cases = 1
Hence probability of 53 sunday = $\frac{Number\, of\, favourable\, cases}{Total\, cases}$
$= \frac{1}{7}$

Question 20

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:

Answer. [A]
Solution. Probability: Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total no. of cases = 6
odd number less than 3 = 1
Number of favorable cases = 1
Probability = $= \frac{1}{6}$

Question 21

A card is drawn from a deck of 52 cards. The event E is that the card is not an ace of hearts. The number of outcomes favourable to E is
(A) 4 (B) 13 (C) 48 (D) 51

Answer:

Answer. [D]
Solution. Total number of cards = 52
Ace of hearts = 1
The card is not an ace of hearts = 52 – 1 = 51
The number of outcomes favourable to E = 51

Question 22

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) 28

Answer:

Answer. [B]
Solution. Probability: Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Let event A be to get a bad egg.
So, p (A) = 0.035 (given)
P(A) = $\frac{Number\, of\, favorable\ cases }{Total\, number\, of\, cases}$
$\Rightarrow$0.035 = $\frac{Number\, of\, favorable\ cases }{400}$
Number of favourable cases = $\frac{35}{1000}\times 400= \frac{140}{10}= 14$

Question 23

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:

Answer. [C]
Solution. Probability: Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total cases = 6000
Probability of getting first prize (p(A)) = 0.08
p(A) $= \frac{Number\, of\, tickets\, the\, bought}{Total\, tickets}$
$\Rightarrow$0.08 × 6000 = Number of tickets the bought
Number of tickets they bought = 480.

Question 24

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:

Answer. [A]
Solution. Probability: Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total tickets = 40
Number of tickets multiple of 5 = 5, 10, 15, 20, 25, 30, 35, 40
Total favourable cases = 8
Let A be the event of getting a ticket with a number multiple of 5.
p(A) = $\frac{Number\, of\, tickets\, the\, bought}{Total\, tickets}$
$\Rightarrow$$p\left ( A \right )= \frac{8}{40}= \frac{1}{5}$

Question 25

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:

Answer. [C]
Solution. Probability: Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total number of cases = 100
prime number from 1 to 100 = 2, 3, 5, 7, 9, 11, 13, 17, 19,
23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 83, 89, 97
Total prime numbers from 1 to 100 = 25
Probability of getting prime number = $\frac{prime\, no.\, from\, 1\, to\, 100}{Total\, number}$
$\\=\frac{25}{100}\\\\=\frac{1}{4}$

Question 26

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:

Answer. [B]
Solution. Probability: Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total students = 23
Students in A, B, C = 4 + 8 + 5 = 17
Students in C, D = 23 – 17 = 6
Number of favourable cases = 6
Let A be the event that the student is not from A, B, C
$p\left ( A \right )= \frac{Student\, from\, C,D}{Total\, students}$
$\Rightarrow$$p\left ( A \right )= \frac{6}{23}$

Question 1

The median of ungrouped data and the median calculated when the same data is grouped are always the same. Do you think that this is a correct statement? Give a reason.

Answer:

Answer. [False]
Solution. Grouped data are data formed by aggregating individual observations of a variable into groups.
Ungrouped data:- The data that is not grouped is called ungrouped data.
The median is the middle number in the grouped data, but when the data is ungrouped, the median is also changed.
Hence, the median is not the same for grouped and ungrouped data

Question 2

In calculating the mean of grouped data, grouped in classes of equal width, we may use the formula
$\bar{x}= a+\frac{\sum f_{i}d_{i}}{\sum f_{i}}$where a is the assumed mean. a must be one of the mid-points of the classes. Is the last statement correct? Justify your answer.

Answer:

Answer. [False]
Solution. Grouped data are data formed by aggregating individual observations of a variable into groups.
Mean: It is the average of the given numbers. It is easy
to calculate the mean. First of all, add up all the numbers, then divide by how many numbers there are.
This last statement is not correct because a can be any point in the grouped data; it is not necessary that a must be the midpoint.
Hence, the statement is false.

Question 3

Is it true to say that the mean, mode, and median of grouped data will always be different? Justify your answer.

Answer:

Answer. [False]
Solution. Mean: It is the average of the given numbers. It is easy to calculate the mean. First of all, add up all the numbers, then divide by how many numbers are there.
Grouped data are data formed by aggregating individual observations of a variable into groups.
The mean, mode, and median of grouped data can be the same it will depend on what type of data is given.
Hence, the statement is false.

Question 4

Will the median class and modal class of grouped data always be different? Justify your answer.

Answer:

Answer. [False]
Solution. Grouped data are data formed by aggregating individual observations of a variable into groups.
The median is always the middle number, and the modal class is the class with the highest frequency it can happen that the median class is of the highest frequency.
So the given statement is false, the median class and mode class can be the same.

Question 5

In a family having three children, there may be no girls, one girl, two girls or three girls. So, the probability of each is $\frac{1}{4}$. Is this correct? Justify your answer.

Answer:

Answer. [False]
Solution. Probability: Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total children = 3
Cases – GGG, GGB, GBG, BGG, BBB, BBG, BGB, GBB were G is girl and B is boy.
Probability = $\frac{Number\, of\, favorable\ cases, }{Total\, number\, of\, cases}$
Probability of 0 girl = $\frac{1}{8}$
Probability of 1 girl = $\frac{3}{8}$
Probability of 2 girl = $\frac{3}{8}$
Probability of 3 girl = $\frac{1}{8}$
Here they are not equal to $\frac{1}{4}$

Question 6

A game consists of spinning an arrow which comes to rest pointing at one of the regions (1, 2 or 3) (Fig.). Are the outcomes 1, 2, and 3 equally likely to occur? Give reasons

Answer:

Answer. [False]
Solution. Probability: Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Here, 3 contains 50% of the region, and 1, 2 contain 25%, 15% of the region.
Probability= $\frac{Number\, of\, favorable\ cases }{Total\, number\, of\, cases}$
Probability of 3= $\frac{50}{100}= \frac{1}{2}$
Probability of 1= $\frac{25}{100}= \frac{1}{4}$
probability of 2= $\frac{25}{100}= \frac{1}{4}$
All probabilities are not equal. So the given statement is false.

Question 7

Apoorv throws two dice 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 the better chance of getting the number 36? Why

Answer:

Answer. [Peehu]
Solution. Probability: Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
As Apoorv throws two dice total cases = 36
Product is 36 when he get = (6, 6)
Number of favourable cases = 1
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability that Apoorv get 36 = $\frac{1}{36}$
Peehu throws are die total cases = 6
The square of 6 is 36
Hence case = 1
Probability that Peehu get 36 = $\frac{1}{6}$
Hence, Peehu has better cases to get 36.

Question 8

When we toss a coin, there are two possible outcomes - Head or Tail. Therefore, the probability of each outcome is $\frac{1}{2}$. Justify your answer.

Answer:

Answer. [True]
Solution. Probability: Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total cases when we toss a coin = 2(H, T)
Probability = $\frac{Number\, of\, favourable\ cases, }{Total\, number\, of\, cases}$
Probability of head = $\frac{1}{2}$
Probability of tail = $\frac{1}{2}$
Hence, the probability of each outcome is $\frac{1}{2}$.

Question 9

A student says that if you throw a die, it will show up 1 or not 1. Therefore, the probability of getting 1 and
the probability of getting ‘not 1’ each is equal to $\frac{1}{2}$. Is this correct? Give reasons.

Answer:

Answer. [False]
Solution. Probability: Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Here total cases = 6
Number of favourable cases in getting 1 = 1
Probability = $\frac{Number\, of\, favourable\ cases, }{Total\, number\, of\, cases}$
Probability of getting $1= \frac{1}{6}$
Number of favourable cases 'not 1' = 5 (2, 3, 4, 5, 6)
Probability of not 1 = $\frac{5}{6}$
Hence, they are not equal to $\frac{1}{2}$

Question 10

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:

Answer. [$\frac{1}{8}$ ]
Solution. Probability: Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total cases in tossing three coins = 8(HHH, HHT, HTH, THH, TTT, TTH, THT, HTT)
Number of cases with no head = TTT
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of no head =$\frac{1}{8}$
The conclusion that the probability of no head is $\frac{1}{4}$ is wrong because, as we calculated above, it comes out to $\frac{1}{8}$. Hence the probability of no head is $\frac{1}{8}$

Question 11

If you toss a coin 6 times and it comes down heads on each occasion. Can you say that the probability of getting a head is 1? Give reasons.

Answer:

Answer. [False]
Solution. Probability: Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
The probability of getting a head is 1, which means that we never get a tail. But this is not true because we have both heads and tails on a coin. Hence probability of getting a head is 1 is false.

Question 12

Sushma tosses a coin 3 times and gets tails each time. Do you think that the outcome of the next toss will be a tail? Give reasons.

Answer:

Answer. [False]
Solution. Probability: Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
No, because when we toss a coin, we can get either tail or head, and the probability of each is $\frac{1}{2}$.
So, it is not necessary that she gets tails on the fourth toss. She can also get a head.

Question 13

If I toss a coin 3 times and get heads each time, should I expect a tail to have a higher chance in the 4th toss? Give a reason in support of your answer.

Answer:

Answer. [False]
Solution. Probability: Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
No, because we get head or tail after tossing a coin, the probability of both outcomes is $\frac{1}{2}$ .
Hence tail does not have a higher chance than the head.
Both have an equal chance.

Question 14

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:

Answer. [True]
Solution. Probability: Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total slips = 100
Slips with an even number = 50
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of slip with even number = $\frac{50}{100}= \frac{1}{2}$
Slips with odd number = 50
Probability of slip with odd number = $\frac{50}{100}= \frac{1}{2}$
Hence, the probability of each is $\frac{1}{2}$.

Question 1

Find the mean of the distribution:

Answer:

Answer. [5.5]
Solution. Here we calculate the mean by following the given steps:

1. Find the midpoint of each interval.

2. Multiply the frequency of each interval by its midpoint.

3. Get the sum of all the frequencies (f) and sum of all the (fx)

4. Now divide the sum of (fx) by the sum of (f)

   Class	Marks (xi)	Frequency(fi)	fixi
   1-3	2	9	18
   3-5	4	22	88
   5-7	6	27	162
   7-10	8.5	17	144.5
   		$\sum f_{i}= 75$	$\sum f_{i}x_{i}= 412\cdot 5$

   
   $mean\left ( \bar{x} \right )= \frac{\sum f_{i}x_{i}}{\sum f_{i}}= \frac{412\cdot 5}{75}= 5\cdot 5$

Question 2

Calculate the mean of the scores of 20 students in a mathematics test :

Answer:

Answer. [35]
Solution. Here we calculate the mean by following the given steps:

1. Find the midpoint of each interval.

2. Multiply the frequency of each interval by its midpoint.

3. Get the sum of all the frequencies (f) and the sum of all the (fx)
   4 Now divide the sum of (fx) by the sum of (f)

$\sum f_{i}= 20$ $\sum f_{i}x_{i}= 700$
$mean\left ( \bar{x} \right )= \frac{\sum f_{i}x_{i}}{\sum f_{i}}= \frac{700}{20}= 35$

Question 3

Calculate the mean of the following data

Answer:

Answer. [12.93]
Solution. Here we calculate the mean by following the given steps:

1. Find the midpoint of each interval.

2. Multiply the frequency of each interval by its midpoint.

3. Get the sum of all the frequencies (f) and the sum of all the (fx)

4. Now divide the sum of (fx) by the sum of (f)

$mean\left ( \bar{x} \right )= \frac{\sum f_{i}x_{i}}{\sum f_{i}}= \frac{362}{28}= 12\cdot 93$

Question 4

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

Find the mean number of pages written per day.

Answer:

Answer. [26]
Solution. Here we calculate the mean by following the given steps:

1. Find the midpoint of each interval.

2. Multiply the frequency of each interval by its midpoint.

3. Get the sum of all the frequencies (f) and sthe um of all the (fx)

4. Now divide the sum of (fx) by the sum of (f)

   No.of pages written per day	no.of days(fi)	(xi)	fixi
   16-18	1	17	17
   19-21	3	20	60
   22-24	4	23	92
   25-27	9	26	234
   28-30	13	29	377
   	$\sum f_{i}= 30$		$\sum f_{i}x_{i}= 780$

   
   $mean\left ( \bar{x} \right )= \frac{\sum f_{i}x_{i}}{\sum f_{i}}= \frac{780}{30}= 26$

Question 5

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

Find the mean daily income of employees.

Answer:

Answer. [356.5]
Solution. Here we calculate the mean by following the given steps:

1. Find the midpoint of each interval.

2. Multiply the frequency of each interval by its midpoint.

3. Get the sum of all the frequencies (f) and the sum of all the (fx)

4. Now divide the sum of (fx) by the sum of (f)

   Income (in Rs )	xi	No.of employees	fixi
   1-200	100.5	14	1407
   201-400	300.5	15	4507.5
   401-600	500.5	14	7007
   601-800	700.5	7	4903.5
   		$\sum f_{i}= 50$	$\sum f_{i}x_{i}= 17825$

   $mean\left ( \bar{x} \right )= \frac{\sum f_{i}x_{i}}{\sum f_{i}}= \frac{17825}{50}= 356.5$

Question 6

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 over the flights.

Answer:

Answer. [109]
Solution. Here we calculate the mean by following the given steps:

1. Find the midpoint of each interval.

2. Multiply the frequency of each interval by its midpoint.

3. Get the sum of all the frequencies (f) and the sum of all the (fx)

4. Now divide the sum of (fx) by the sum of (f)

$mean\left ( \bar{x} \right )= \frac{\sum f_{i}x_{i}}{\sum f_{i}}= \frac{10992}{100}= 109\cdot 92$
number of seats = 109

Question 7

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

Find the mean weight of the wrestlers.

Answer:

Answer. [123.4]
Solution. Here we calculate the mean by following the given steps:

1. Find the midpoint of each interval.

2. Multiply the frequency of each interval by its midpoint.

3. Get the sum of all the frequencies (f) and the sum of all the (fx)

4. Now divide the sum of (fx) by the sum of (f)

   Weight	fi	xi	fixi
   100-110	4	105	420
   110-120	14	115	1610
   120-130	21	125	2625
   130-140	8	135	1080
   140-150	3	145	435
   	$\sum f_{i}= 50$		$\sum f_{i}x_{i}=6170$

   
   $mean\left ( \bar{x} \right )= \frac{\sum f_{i}x_{i}}{\sum f_{i}}= \frac{6170}{50}= 123\cdot 4 \ \ kg$

Question 8

The mileage (km per litre) of 50 cars of the same model was tested by a manufacturer and details are tabulated as given below :

Find the mean mileage.
The manufacturer claimed that the mileage of the model was 16 km/litre. Do you agree with this claim?

Answer:

Answer. [14.48]
Solution. Here we calculate the mean by following
1. Find the midpoint of each interval.
2. Multiply the frequency of each interval by its midpoint.
3. Get the sum of all the frequencies (f) and the sum of all the (fx)
4. Now divide the sum of (fx) by the sum of (f)

$mean\left ( \bar{x} \right )= \frac{\sum f_{i}x_{i}}{\sum f_{i}}= \frac{724}{50}= 14\cdot 48 \ \ km/L$

Question 9

The following is the distribution of weights (in kg) of 40 persons :

Construct a cumulative frequency distribution (of the less than type) table for the data above.

Answer:

Solution. Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However, mostly we use frequency distribution to summarize categorical variables.

Question 10

The following table shows the cumulative frequency distribution of marks of 800 students in an examination:

Construct a frequency distribution table for the data above.

Answer:

Solution. Frequency distribution: It tells how frequencies are distributed over values in a
frequency distribution. However mostly we use frequency distribution to summarize categorical variables.

Question 11

Form the frequency distribution table from the following data :

Answer:

Solution. Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However mostly we use frequency distribution to summarize categorical variables.

Question 12

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

Answer:

Solution. Frequency distribution: It tells how frequencies are distributed overvalues in a frequency distribution. However mostly we use frequency distribution to
summarize categorical variables.
$a= 12$ ( because first term of frequency and cumulative frequency is same )
$\Rightarrow$12 + b = 25
$\Rightarrow$b = 25 – 12
$\Rightarrow$b = 13
$\Rightarrow$25 + 10 = c
$\Rightarrow$35= c
$\Rightarrow$c + d = 43
$\Rightarrow$35 + d = 43
$\Rightarrow$d = 43 – 35
$\Rightarrow$d=8
$\Rightarrow$43 + e = 48
$\Rightarrow$e = 48 – 43
$\Rightarrow$e =5
$\Rightarrow$48+2 = f
$\Rightarrow$50 = f
Ans. a = 12, b = 13, c = 35, d = 8, e = 5, f = 50

Question 13

The following are the ages of 300 patients getting medical treatment in a hospital on a particular day :

(i) Less than type cumulative frequency distribution.
(ii) More than type cumulative frequency distribution

Answer:

Solution. Frequency distribution: It tells how frequencies are distributed overvalues in a frequency distribution. However mostly we use frequency distribution to summarize categorical variables.

Question 15

Weekly income of 600 families is tabulated below :

Compute the median income.

Answer:

Answer. [1263.15]
Solution. n = 600
$\frac{n}{2}= \frac{600}{2}= 300$
$\iota$= 1000, l = 1000, cf = 250, f = 190
median = $\iota +\left ( \frac{\frac{n}{2}-cf}{f} \right )\times h$
$= 1000+\frac{\left ( 300-250 \right )}{190}\times 1000$
$= 1000+\frac{50}{190}\times 1000$
$= 1000+\frac{5000}{19}$
$= \frac{19000+5000}{19}= \frac{24000}{19}= 1263\cdot 15$
Median = 1263.15

Question 16

The maximum bowling speeds, in km per hour, of 33 players at a cricket coaching centre are given as follows :

Calculate the median bowling speed.

Answer:

Answer. [109.16]
Solution. Here n = 33
$\frac{n}{2}= \frac{33}{2}= 16\cdot 5$
$\iota = 100$
h = 15, f = 9 , cf = 11
Median $= \iota +\left ( \frac{\frac{n}{2}-cf}{f} \right )\times h$
$=100+\frac{\left ( 16\cdot 5-11 \right )\times 15}{9}$
$=100+\left ( \frac{55}{10\times 9} \right )\times 15$
$=100+\frac{55\times 5}{30}$
$=100+\frac{275}{30}$
=100 + 9.16 $\Rightarrow$ 109.16

Question 17

The monthly income of 100 families are given as below :

Calculate the modal income.

Answer:

Answer. [11875]
Solution. Here l = 10000, f₁ = 41, f₀ = 26, f₂ = 16, h = 5000
Mode = $\iota +\left ( \frac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}} \right )\times h$
=$1000+\left ( \frac{41-26}{2\times 41-26-16} \right )\times 5000$
=$1000+\frac{15}{40}\times 5000$
= 10000 + 1875 = 11875
Modal income is 11875 Rs.

Question 18

The weight of coffee in 70 packets are shown in the following table :

Determine the modal weight.

Answer:

Answer. [201.7 g]
Solution.
Here l = 201, f1 = 26, f0 = 12, f2 = 20, h = 1
$\Rightarrow 201+\left ( \frac{26-12}{2\times 26-12-20} \right )\times 1$
$\Rightarrow 201+\left ( \frac{14}{52-32} \right )$
$\Rightarrow 201+\frac{14}{20}$
$\Rightarrow 201+0\cdot 7= 201\cdot 7 g$

Question 19

Two dice are thrown at the same time.
(i) Find the probability of getting same number on both dice.
(ii) Find the probability of getting different numbers on both dice.

Answer:

Solution. Probability ; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total number of cases after thrown of two dice = 36
(i) Same number = (1, 1), (2, 2), (3, 3), (4. 4), (5, 5), (6, 6)
Same number cases = 6
Let A be the event of getting same number.
Probability [p(A)] = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
$p\left ( A \right )= \frac{6}{36}= \frac{1}{6}$
(ii) Different number cases = 36 – same number case
= 36 –6 = 30
Let A be the event of getting different number
Probability [p(A)]= $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
$p\left ( A \right )= \frac{30}{36}= \frac{5}{6}$

Question 20

Two dice are thrown simultaneously. What is the probability that the sum of the numbers appearing on the dice is
(i)7?
(ii)a prime number?
(iii)1?

Answer:

(i) Answer. [1/6]
Solution.Probability ; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases after throwing of two dice = 36
Cases when total is 7 = (1, 6), (6, 1), (3, 4), (4, 3), (2, 5), (5, 2)
Total cases = 6
Let A be the event of getting total 7
Probability [p(A)]= $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting sum 7 = $\frac{6}{36}= \frac{1}{6}$

(ii) Answer.[5/12]
Solution. Probability ; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases = 36
Prime number as a sum = (1, 1), (1, 2), (2, 1), (1, 4),
(4, 1), (2, 3), (3, 2), (1, 6), (6, 1), (3, 4), (4, 3), (2, 5), (5, 2), (6, 5), (5, 6)
Cases = 15
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability that sum is a prime number = $\frac{15}{36}= \frac{5}{12}$
(iii) Answer.[0]
Solution.Probability ; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases = 36
pairs from which we get sum 1 = 0
Cases = 0
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting sum 1 = $\frac{0}{36}= 0$

Question 21

Two dice are thrown together. Find the probability that the product of the numbers on the top of the dice is
(i)6
(ii) 12
(iii) 7

Answer:

(i) Answer. [1/9]
Solution.Probability; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases = 36
For getting product 6 = (1, 6,), (6, 1), (2, 3), (3, 2)
Cases = 4
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting product 6 =$\frac{4}{36}= \frac{1}{9}$

(ii) Answer.[1/9]
Solution.Probability; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases = 36
product 12 = (2, 6), (6, 2), (3, 4), (4, 3)
Cases = 4
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting product 12 =$\frac{4}{36}= \frac{1}{9}$

(iii) Answer.[0]
Solution.Probability; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases = 36
Product 7 = 0 (case)
Cases = 0
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting product 7 =$\frac{0}{36}= 0$

Question 22

Two dice are thrown at the same time and the product of numbers appearing on them is noted. Find the probability that the product is less than 9.

Answer:

Answer.[4/9]
Solution.Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases in throwing two dice = 36
Product less than 9 cases = (1, 1). (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (4, 1), (4, 2), (5, 1), (6, 1)
Number of favourable cases = 16
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting product less than 9 = $\frac{16}{36}= \frac{4}{9}$

Question 23

Two dice are numbered 1, 2, 3, 4, 5, 6 and 1, 1, 2, 2, 3, 3, respectively. They are thrown and the sum of the numbers on them is noted. Find the probability of getting each sum from 2 to 9 separately.

Answer:

Solution.Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total number of cases = 36
case of getting sum 2 = ( 1 , 1 ) ( 1 , 1 )
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting sum 2 = $\frac{2}{36}= \frac{1}{18}$
case of getting sum 3 = (1, 2), (1, 2), (2, 1), (2, 1)
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting sum 3= $\frac{4}{36}= \frac{1}{9}$
case of getting sum 4 = (1, 3), (1, 3), (2, 2), (2, 2), (3, 1), (3, 1)
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting sum 4= $\frac{6}{36}= \frac{1}{6}$
case of getting sum 5 = (2, 3), (2, 3), (4, 1),(4,1) (3, 2), (3, 2)
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting sum 5 = $\frac{6}{36}= \frac{1}{6}$
case of getting sum 6 = (3, 3), (3, 3), (4, 2), (4, 2), (5, 1), (5, 1)
probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting sum 6= $\frac{6}{36}= \frac{1}{6}$
case of getting sum 7 = (4, 3), (4, 3), (5, 2), (5, 2), (6, 1), (6, 1)
probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting sum 7= $\frac{6}{36}= \frac{1}{6}$
case of getting sum 8 = (5, 3), (5, 3), (6, 2), (6, 2)
probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting sum 8= $\frac{4}{36}= \frac{1}{9}$
case of getting sum 9 = (6, 3), (6, 3)
probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting sum 9= $\frac{2}{36}= \frac{1}{18}$

Question 24

A coin is tossed two times. Find the probability of getting at most one head.

Answer:

Answer. [3/4]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases = 4 (HH, TT, HT, TH)
Cases of at most 1 head = HT, TH, TT
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting at most 1 head = $\frac{3}{4}$

Question 25

A coin is tossed 3 times. List the possible outcomes. Find the probability of getting
(i) all heads
(ii)at least 2 heads

Answer:

(i) Answer. [1/8]
Solution.Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Possible outcomes = (HHH, HHT, HTH, THH, TTT, TTH, THT, HTT)
Total cases = 8
Cases of getting all heads = (HHH)
Number of favourable cases = 1
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting all heads = $\frac{1}{8}$

(ii) Answer.[1/2]
Solution.Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Possible outcomes = 8 (HHH, HHT, HTH, THH, TTT, TTH, THT, HTT)
Cases of getting at least 2 heads = (HHH, HHT, HTH, THH)
Favorable cases = 4
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting at least 2 heads = $\frac{4}{8}= \frac{1}{2}$

Question 26

Two dice are thrown at the same time. Determine the probability that the difference of the numbers on the two dice is 2.

Answer:

Answer. [2/9]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Cases of getting difference 2 = (1, 3), (3, 1), (2, 4), (4, 2), (3, 5), (5, 3), (4, 6), (6, 4)
Favourable cases = 8
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting difference 2 = $\frac{8}{36}= \frac{2}{9}$

Question 27

A bag contains 10 red, 5 blue and 7 green balls. A ball is drawn at random. Find the probability of this ball being a
(i) red ball
(ii)green ball
(iii) not a blue ball

Answer:

(i) Answer.[5/11]
Solution.Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total balls = 10 + 5 + 7 = 22
Red balls = 10
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting red ball = $\frac{10}{22}= \frac{5}{11}$

(ii) Answer.[7/22]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total balls = 10 + 5 + 7 = 22
Green balls = 7
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting green ball = $\frac{7}{22}$

(iii) Answer.[17/22]
Solution.Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total balls = 10 + 5 + 7 = 22
Not a blue ball = 22 – (blue ball)
= 22 – 5 = 17
robability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting not a blue ball = $\frac{17}{22}$

Question 28

The king, queen and jack of clubs are removed from a deck of 52 playing cards and then well shuffled. Now one card is drawn at random from the remaining cards. Determine the probability that the card is
(i) a heart
(ii)a king

Answer:

(i) Answer.[13/49]
Solution.Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cards = 52 – 3 ( three cards are removed)
= 49
Total hearts = 13
Favourable cases = 13
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting a heart = $\frac{13}{49}$

(ii) Answer.[3/49]
Solution.Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cards = 52 – 3 ( three cards are removed)
= 49
Total king = 4 – 1 = 3 ( 1 king is removed)
favourable cases = 3
bability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting a King= $\frac{3}{49}$

Question 29

The king, queen and jack of clubs are removed from a deck of 52 playing cards and then well shuffled. Now one card is drawn at random from the remaining cards. What is the probability that the card is
(i)a club
(ii) 10 of hearts

Answer:

(i) Answer.[10/49]
Solution.Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cards = 52 – 3 = 49 ( three cards are removed)
Total club = 13 – 3 = 10 ( 3 club cards are removed)
favourable cases = 10
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting a club = $\frac{10}{49}$

(ii) Answer.[1/49]
Solution.Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cards = 52 – 3 = 49 ( three cards are removed)
10 of heart = 1
favourable cases = 1
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting a heart = $\frac{1}{49}$

Question 30

All the jacks, queens and kings are removed from a deck of 52 playing cards. The remaining cards are well shuffled and then one card is drawn at random. Giving ace a value 1 similar value for other cards, find the probability that the card has a value
(i)7
(ii) greater than 7
(iii) less than 7

Answer:

(i) Answer.[1/10]
Solution.Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cards = 52 – 12 = 40 ( 12 cards are removed)
card with number 7 = 4
favourable cases = 4
probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting card 7= $\frac{4}{10}= \frac{1}{10}$

(ii) Answer. [3/10]
Solution.Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cards = 52 – 12 = 40 ($\mathbb{Q}$ 12 cards are removed)
Cards greater than 7 =8,9,10 (3 × 4 = 12)
favourable cases = 12
probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting card 7= $\frac{12}{40}= \frac{3}{10}$
(iii) Answer. [3/5]
Solution.Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cards = 52 – 12 = 40 ($\because$ 12 cards are removed)
Cards less than 7 = 1, 2, 3, 4, 5, 6 (6 × 4 = 24)
favourable cases = 24
probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting card 7= $\frac{24}{40}= \frac{6}{10}= \frac{3}{5}$

Question 31

An integer is chosen between 0 and 100. What is the probability that it is
(i) divisible by 7 ?
(ii) not divisible by 7?

Answer:

(i) Answer.[14/99]
Solution.Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total number = 99 (between 0 to 100)
Number divisible by 7 = (7, 14, 21, 28,35, 42, 49, 56, 63, 70, 77, 84, 91, 98)
Favourable cases = 14
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting number divisible by 7 = $\frac{14}{99}$

(ii) Answer.[85/99]
Solution.Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total number between 0 to 100 = 99
Number divisible by 7 = (7, 14, 21, 28,35, 42, 49, 56, 63, 70, 77, 84, 91, 98)
Favourable cases = 14
robability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting number divisible by 7 = $\frac{14}{99}$
$= 1-\frac{14}{99}$
$= \frac{99-14}{99}= \frac{85}{99}$

Question 32

Cards with numbers 2 to 101 are placed in a box. A card is selected at random. Find the probability that the card has
(i) an even number
(ii)a square number

Answer:

(i) Answer.[1/2]
Solution.Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total numbers from 2 to 101 = 100
Total even numbers from 2 to 101 = 50
Favourable cases = 50
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability that card is with even number = $\frac{50}{100}= \frac{1}{2}$

(ii) Answer.[9/100]
Solution.Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total number from 2 to 101 = 100
Square numbers from 2 to 101 = (4, 9, 16, 25, 36, 49, 64, 81, 100)
Favourable cases = 9
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability that the card is with a square number = $\frac{9}{100}$

Question 33

A letter of English alphabets is chosen at random. Determine the probability that the letter is a consonant.

Answer:

Answer.[21/26]
Solution.Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total alphabets = 26
Total consonant = 21
Favourable cases = 21
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability that alphabet is consonant = $\frac{21}{26}$

Question 34

There are 1000 sealed envelopes in a box, 10 of them contain a cash prize of Rs 100 each, 100 of them contain a cash prize of Rs 50 each and 200 of them contain a cash prize of Rs 10 each and rest do not contain any cash prize. If they are well shuffled and an envelope is picked up out, what is the probability that it contains no cash prize ?

Answer:

Answer.[0.69]
Solution.Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total envelopes = 1000
Envelopes with no cash prize = Total envelopes – envelopes with cash prize
= 1000 – 10 – 100 – 200 = 690
Favourable cases = 690
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability that the envelope is no cash prize = $\frac{690}{1000}= \frac{69}{100}= 0\cdot 69$

Question 35

Box A contains 25 slips of which 19 are marked Rs 1 and other are marked Rs 5 each. Box B contains 50 slips of which 45 are marked Rs 1 each and others are marked Rs 13 each. Slips of both boxes are poured into a third box and reshuffled. A slip is drawn at random. What is the probability that it is marked other than Rs 1 ?

Answer:

Answer.[11/75]
Solution.Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total slips = 25 + 50 = 75
Slips marked other than 1 = Rs. 5 slips + Rs. 13 slips
= 6 + 5 = 11
Favourable cases = 11
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability that slips is not marked 1 $= \frac{11}{75}$

Question 36

A carton of 24 bulbs contain 6 defective bulbs. One bulbs 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:

Answer.[5/23]
Solution.Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total bulbs = 24
Defective = 6
not defective = 18
Probability that the bulb is not defective = $\frac{18}{24}= \frac{3}{4}$
Let the bulbs is defective and it is removed from 24 bulb.
Now bulbs remain = 23
In 23 bulbs, non-defective bulbs = 18
defective = 5
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Now probability that the bulb is defective = $\frac{5}{23}$ .

Question 37

A child’s game has 8 triangles of which 3 are blue and rest are red, and 10 squares of which 6 are blue and rest are red. One piece is lost at random. Find the probability that it is a
(i)triangle
(ii) square
(iii) square of blue colour
(iv) triangle of red colour

Answer:

(i) Answer.[4/9]
Solution.Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total piece = 8 + 10 = 18
Total triangles = 8
Favourable cases = 8
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability that piece is a triangle $= \frac{8}{18}= \frac{4}{9}$

(ii) Answer.[5/9]
Solution.Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total piece = 8 + 10 = 18
Total square = 10
Favourable cases = 10
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability that the piece is a square = $\frac{10}{18}= \frac{5}{9}$

(iii) Answer.[1/3]
Solution.Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total piece = 10 + 8 = 18
Square of blue color = 6
favourable cases = 6
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability that piece is a square of blue color = $\frac{6}{18}= \frac{1}{3}$

(iv) Answer.[5/18]
Solution.Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total piece = 10 + 8 = 18
triangle of red color = 8 – 3 = 5
favourable cases = 5
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability that piece is a triangle of red colour $= \frac{5}{18}$

Question 38

In a game, the entry fee is Rs 5. The game consists of a tossing a coin 3 times. If one or two heads show, Shweta gets her entry fee back. If she throws 3 heads, she receives double the entry fees. Otherwise she will lose. For tossing a coin three times, find the probability that she
(i) loses the entry fee.
(ii) gets double entry fee.
(iii) just gets her entry fee.

Answer:

(i) Answer.[1/8]
Solution.Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases = 8(HHH, HHT, HTH, THH, TTT, TTH, THT, HTT)
case in which the lose entry = 8 – (in which she gets entry book + in which she gets double)
= 8 – 6 (HHT, HTH, THH, TTH, THT, HTT) – 1(HHH)
= 8 – 7 = 1
Favourable cases = 1
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability that she will lose money = $\frac{1}{8}$

(ii) Answer.[1/8]
Solution.Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases = 8(HHH, HHT, HTH, THH, TTT, TTH, THT, HTT)
case in which she gets double entry = HHH
favourable cases = 1
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability that she gets double entry fee = $\frac{1}{8}$

(iii) Answer.[3/4]
Solution.Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases = 8(HHH, HHT, HTH, THH, TTT, TTH, THT, HTT)
case in which she gets entry book = 6(HHT, HTH, THH, TTH, THT, HTT)
favourable cases = 6
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability that she gets entry fees = $\frac{6}{8}= \frac{3}{4}$

Question 39

A die has its six faces marked 0, 1, 1, 1, 6, 6. Two such dice are thrown together and the total score is recorded.
(i) How many different scores are possible?
(ii) What is the probability of getting a total of 7?

Answer:

(i) Answer.[6]
Solution. Count the number of sums we can notice by using two dice of (0, 1, 1, 1, 6, 6) type.
We can get a sum of 0 = (0,0)
We can get a sum of 1 = (0,1) , (1,0)
We can get a sum of 2 = (1,1)
We can get a sum of 6 = (0,6) , (6,0)
We can get a sum of 7 = (6,1) , (1,6)
We can get a sum of 12 = (6,6)
We can get a score of 0, 1, 2, 6, 7, 12
Hence we can get 6 different scores.

(ii) Answer.[4/9]
Solution.Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases = 36
Case of getting sum 7 = (1, 6), (1, 6), (1, 6), (1, 6), (1, 6), (1, 6), (6,1), ( 6,1), (6,1), ( 6,1), (6,1), (6,1),
Number of favourable cases = 12
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting a total 7 = $\frac{12}{36}= \frac{1}{3}$

Question 40

A lot consists of 48 mobile phones of which 42 are good, 3 have only minor defects and 3 have major defects. Varnika will buy a phone if it is good but the trader will only buy a mobile if it has no major defect. One phone is selected at random from the lot. What is the probability that it is
(i) acceptable to Varnika?
(ii) acceptable to the trader?
Answer:

(i) Answer.[7/8]
Solution.Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total mobiles = 48
Minor defective = 3
major defective = 3
good = 42
Varnika buy only good so favourable cases = 42
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability that acceptable to Varnika = $\frac{42}{48}= \frac{7}{8}$
(ii) Answer.[15/16]
Solution.Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
total mobiles = 48
good = 42
minor defect = 3
major defect = 3
trader accept only good and minor defect.
So favourable cases = 42 + 3 = 45
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability that trader accept $\frac{45}{48}= \frac{15}{16}$

Question 41

A bag contains 24 balls of which x are red, 2x are white and 3x are blue. A ball is selected at random. What is the probability that it is
(i) not red?
(ii) white?

Answer:

(i) Answer.[5/6]
Solution.Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total balls = red + white + blue
24 = x + 2x + 3x
6x = 24
x = 4
Red balls = x = 4
White balls = 2x = 2 × 4 = 8
Blue balls = 3x = 3 × 4 = 12
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability that ball is not red = $\frac{blue+white}{24}$
$= \frac{8+12}{24}= \frac{20}{24}= \frac{5}{6}$

(ii) Answer.[1/3]
Solution.Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total balls = red + white + blue
24 = 6x
x = 4
white balls = 2x = 2 × 4 = 8
Favourable cases = 8
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability of getting on white ball = $\frac{8}{24}= \frac{1}{3}$

Question 42

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
(i) the first player wins a prize?
(ii) the second player wins a prize, if the first has won?

Answer:

(i) Answer.[0.009]
Solution.Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cards = 1000
Player wins prize with cards = (529, 576, 625, 676, 729, 784, 841, 900, 961)
Favourable cases = 9
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability that player wins = $\frac{9}{1000}= 0\cdot 009$

(ii) Answer.[0.008]
Solution.Probability: Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Now the total cards are = 1000 – 1 = 999
Now the total winning cards = 9 – 1 = 8
Probability = $\frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}$
Probability that second player wins after first = $\frac{8}{999}\approx 0\cdot 008$