Solution

(c) We know that, $d_i=x_i-a$

i.e., $d_i$ ’s are the deviation from a of mid-points of the classes.

Solution

(b) In computing the mean of grouped data, the frequencies are centred at the class marks of the classes.

Solution

(a) $\because$

$ \bar{{}x}=\dfrac{\sum f_i x_i}{n} $

$ \begin{aligned} \Sigma(f_i x_i-\bar{{}x}) = \Sigma f_i x_i-\Sigma \bar{{}x} \\ & =n \bar{{}x}-n \bar{{}x} \\ & =0 \end{aligned} $

Solution

(c) Given, $\bar{{}x}=a+h \dfrac{\Sigma f_i u_i}{\Sigma f_i}$

Above formula is a step deviation formula.

$ u_i=\dfrac{x_i-a}{h} $

Solution

(b) Since, the intersection point of less than ogive and more than ogive gives the median on the abscissa.

Solution

(b) Here,

Now, $\dfrac{N}{2}=\dfrac{66}{2}=33$, which lies in the interval $10-15$. Therefore, lower limit of the median class is 10 .

The highest frequency is 20, which lies in the interval 15-20. Therefore, lower limit of modal class is 15 . Hence, required sum is $10+15=25$.

Solution

(b)

Given, classes are not continuous, so we make continuous by subtracting 0.5 from lower limit and adding 0.5 to upper limit of each class.

Here, $\dfrac{N}{2}=\dfrac{57}{2}=28.5$, which lies in the interval 11.5-17.5.

Hence, the upper limit is 17.5 .

Solution

(c)

Here, we see that the highest frequency is 30 , which lies in the interval $30-40$.

Solution

(c)

Here, $\dfrac{N}{2}=\dfrac{67}{2}=33.5$ which lies in the interval $125-145$.

Hence, upper limit of median class is 145 .

Here, we see that the highest frequency is 20 which lies in 125-145. Hence, the lower limit of modal class is 125 .

$\therefore$ Required difference $=$ Upper limit of median class -Lower limit of modal class

$=145-125=20$

Solution

(c) The number of atheletes who completed the race in less than 14.6 $=2+4+5+71=82$

Solution

(a)

Hence, frequency in the class interval $30-40$ is 3 .

Solution

(d) The event which cannot occur is said to be impossible event and probability of impossible event is zero.

Solution

(d) Since, probability of an event always lies between 0 and 1 .

Solution

(a) The probability of an event which is very unlikely to happen is closest to zero and from the given options 0.0001 is closest to zero.

Solution

(c) Since, probability of an event + probability of its complementry event $=1$

So, probability of its complementry event $=1$ - Probability of an event $=1-P$

Solution

(b) We know that, the probability expressed as a percentage always lie between 0 and 100. So, it cannot be less than 0 .

Solution

(c) Since, probability of an event always lies between 0 and 1 .

Solution

(a) In a deck of 52 cards, there are 12 face cards i.e., 6 red and 6 black cards. So, probability of getting a red face card $=\dfrac{6}{52}=\dfrac{3}{26}$

Solution

(a) A non-leap year has 365 days and therefore 52 weeks and 1 day. This 1 day may be Sunday or Monday or Tuesday or Wednesday or Thursday or Friday or Saturday. Thus, out of 7 possibilities, 1 favourable event is the event that the one day is Sunday. $\therefore$ Required probability $=\dfrac{1}{7}$

Solution

(a) When a die is thrown, then total number of outcomes $=6$

Odd number less than 3 is 1 only.

Number of possible outcomes $=1$

$\therefore \quad$ Required probability $=\dfrac{1}{6}$.

Solution

(d) In a deck of 52 cards, there are 13 cards of heart and 1 is ace of heart. Hence, the number of outcomes favourable to $E=51$

Solution

(b) Here, total number of eggs $=400$

Probability of getting a bad egg $=0.035$

$ \begin{matrix} \Rightarrow & \dfrac{\text{ Number of bad eggs }}{\text{ Total number of eggs }}=0.035 \\ \Rightarrow & \dfrac{\text{ Number of bad eggs }}{400}=0.035 \\ \therefore & \text{ Number of bad eggs }=0.035 \times 400=14 \end{matrix} $

Solution

(c) Given, total number of sold tickets $=6000$

Let she bought $x$ tickets.

Then, probability of her winning the first prize $=\dfrac{x}{6000}=0.08$ [given]

$\Rightarrow \quad x=0.08 \times 6000$

$\therefore \quad x=480$

Hence, she bought 480 tickets.

Solution

(a) Number of total outcomes $=40$

Multiples of 5 between 1 to $40=5,10,15,20,25,30,35,40$

$\therefore$ Total number of possible outcomes $=8$

$\therefore \quad$ Required probability $=\dfrac{8}{40}=\dfrac{1}{5}$

Solution

(c) Total numbers of outcomes $=100$

So, the prime numbers between 1 to 100 are $2,3,5,7,11,13,17,19,23,29,31,37,41$, $43,47,53,56,61,67,71,73,79,83,89$ and 97 .

$\therefore$ Total number of possible outcomes $=25$

$\therefore \quad$ Required probability $=\dfrac{25}{100}=\dfrac{1}{4}$

Solution

(b) Total number of students $=23$

Number of students in house $A, B$ and $C=4+8+5=17$

$\therefore \quad$ Remains students $=23-17=6$

So, probability that the selected student is not from $A, B$ and $C=\dfrac{6}{23}$

Solution

Not always, because for calculating median of a grouped data, the formula used is based on the assumption that the observations in the classes are uniformal distributed (or equally spaced).

Solution

No, it is not necessary that assumed mean consider as the mid-point of the class interval. It is considered as any value which is easy to simplify it.

Solution

No, the value of these three measures can be the same, it depends on the type of data.

Solution

Not always, It depends on the given data.

Solution

No, the probability of each is not $\dfrac{1}{4}$ because the probability of no girl in three children is zero and probability of three girls in three children is one.

Justification

So, these events are not equally likely as outcome one girl, means $g b b, b g b, b b g$ ’three girls’ means ’ $g 9 g$ ’ and so on.

Solution

No, the outcomes are not equally likely, because 3 contains half part of the total region, so it is more likely than 1 and 2 , since 1 and 2 , each contains half part of the remaining part of the region.

Solution

Apoorv throws two dice once.

So total number of outcomes $=36$

Number of outcomes for getting product $36=1(6 \times 6)$

$\therefore \quad$ Probability for Apoorv $=\dfrac{1}{36}$

Also, Peehu throws one die,

So, total number of outcomes $=6$

Number of outcomes for getting square $36=1(6^{2}=36)$

$\therefore \quad$ Probability for Peehu $=\dfrac{1}{6}=\dfrac{6}{36}$

Hence, Peehu has better chance of getting the number 36 .

Solution

Yes, probability of each outcome is $\dfrac{1}{2}$ because head and tail both are equally likely events.

Solution

No, this is not correct.

Suppose we throw a die, then total number of outcomes $=6$

Possible outcomes $=1$ or 2 or 3 or 4 or 5 or 6

$\therefore \quad$ Probability of getting $1=\dfrac{1}{6}$

Now, probability of getting not $1=1-$ Probability of getting 1

$ =1-\dfrac{1}{6}=\dfrac{5}{6} $

Solution

I toss three coins together [given] So, total number of outcomes $=2^{3}=8$ and possible outcomes are (HHH), (HTT), (THT), (TTH), (HHT), (THH), (HTH) and (TTT) Now, probability of getting no head $=\dfrac{1}{8}$

Hence, the given conclusion is wrong because the probability of no head is $\dfrac{1}{8}$ not $\dfrac{1}{4}$.

Solution

No, if let we toss a coin, then we get head or tail, both are equally likely events. So, probability is $\dfrac{1}{2}$. If we toss a coin 6 times, then probability will be same in each case. So, the probability of getting a head is not 1 .

Solution

The outcome of next toss may or may not be tail, because on tossing a coin, we get head or tail so both are equally likely events.

Solution

No, let we toss a coin, then we get head or tail, both are equaly likely events. i.e., probability of each event is $\dfrac{1}{2}$. So, no question of expecting a tail to have a higher chance in 4th toss.

Solution

We know that, between 1 to 100 half numbers are even and half numbers are odd i.e., 50 numbers $(2,4,6,8, \ldots, 96,98,100)$ are even and 50 numbers $(1,3,5,7, \ldots, 97,99)$ are odd. So, both events are equally likely.

So, probability of getting even number $=\dfrac{50}{100}=\dfrac{1}{2}$

and probability of getting odd number $=\dfrac{50}{100}=\dfrac{1}{2}$

Hence, the probability of each is $\dfrac{1}{2}$.

Solution

We first, find the class mark $x_i$, of each class and then proceed as follows.

Therefore, mean $(\bar{{}x})=\dfrac{\Sigma f_i x_i}{\Sigma f_i}=\dfrac{412.5}{75}=5.5$

Hence, mean of the given distribution is 5.5 .

Solution

We first, find the class mark $x_i$ of each class and then proceed as follows

Therefore,

$ \text{ mean }(\bar{{}x})=\dfrac{\Sigma f_i x_i}{\Sigma f_i}=\dfrac{700}{20}=35 $

Hence, the mean of scores of 20 students in mathematics test is 35 .

Solution

Since, given data is not continuous, so we subtract 0.5 from the lower limit and add 0.5 in the upper limit of each class.

Now, we first find the class mark $x_i$ of each class and then proceed as follows

Therefore,

$ \bar{{}x}(\text{ mean })=\dfrac{\Sigma f_i x_i}{\Sigma f_i}=\dfrac{362}{28}=12.93 $

Hence, mean of the given data is 12.93 .

Solution

Since,

Since, given data is not continuous, so we subtract 0.5 from the lower limit and add 0.5 in the upper limit of each class.

$ \therefore \quad \text{ Mean }(\bar{{}x})=\dfrac{\Sigma f_i x_i}{\Sigma f_i}=\dfrac{780}{30}=26 $

Hence, the mean of pages written per day is 26 .

Solution

Since, given data is not continuous, so we subtract 0.5 from the lower limit and add 0.5 in the upper limit of each class.

Now we first, find the class mark $x_i$ of each class and then proceed as follows

$ \begin{matrix} \therefore \quad \text{ Assumed mean, } a=300.5 \\ \text{ Class width, } h=200 \end{matrix} $

and total observations, $N=50$

By step deviation method,

$ \begin{aligned} \text{ Mean } = a+h \times \dfrac{1}{N} \times \sum _{i=1}^{5} f_i u_i \\ & =300.5+200 \times \dfrac{1}{50} \times 14 \\ & =300.5+56=356.5 \end{aligned} $

Solution

We first, find the class mark $x_i$ of each class and then proceed as follows.

$ \begin{aligned} & \therefore \quad \text{ Assumed mean, } a=110, \\ & \text{ Class width, } h=4 \\ & \text{ and } \quad \text{ total observation, } N=100 \\ & \text{ By assumed mean method, } \\ & \text{ Mean }(\bar{{}x})=a+\dfrac{\sum f_i d_i}{\Sigma f_i} \\ &=110+\dfrac{-8}{100}=110-0.08=109.92 \end{aligned} $

Solution

We first find the class mark $x_i$, of each class and then proceed as follows

$\therefore$ Assumed mean $(a)=125$,

Class width $(h)=10$ and total observation $(N)=50$

By assumed mean method,

$ \begin{aligned} \text{ Mean }(\bar{{}x}) = a+\dfrac{\sum f_i d_i}{\sum f_i} \\ & =125+\dfrac{(-80)}{50} \\ & =125-1.6=123.4 kg \end{aligned} $

Solution

Here,

and

$\therefore$

$ \begin{aligned} \Sigma f_i = 50 \\ \Sigma f_i x_i = 724 \\ \text{ Mean } \bar{{}x} = \dfrac{\sum f_i x_i}{\Sigma f_i} \\ & =\dfrac{724}{50}=14.48 \end{aligned} $

Hence, mean mileage is $14.48 kmL^{-1}$.

$No$, the manufacturer is claiming mileage $1.52 kmh^{-1}$ more than average mileage.

Solution

The cumulative distribution (less than type) table is shown below

Solution

Here, we observe that 10 students have scored marks below 10 i.e., it lies between class interval 0-10. Similarly, 50 students have scored marks below 20. So, $50-10=40$ students lies in the interval $10-20$ and so on. The table of a frequency distribution for the given data is

Solution

Here, we observe that, all 34 students have scored marks more than or equal to 0 . Since, 32 students have scored marks more than or equal to 10 . So, $34-32=2$ students lies in the interval 0-10 and so on.

Now, we construct the frequency distribution table.

Solution

On comparing last two tables, we get

$ \begin{aligned} & a=12 \\ & \therefore \quad 12+b=25 \\ & \Rightarrow \quad b=25-12=13 \\ & 22+b=c \\ & c=22+13=35 \\ & 22+b+d=43 \\ & \Rightarrow \quad 22+13+d=43 \\ & \Rightarrow \quad d=43-35=8 \\ & \text{ and } 22+b+d+e=48 \\ & \Rightarrow \quad 22+13+8+e=48 \\ & \Rightarrow \quad e=48-43=5 \\ & \text{ and } \quad 24+b+d+e=f \\ & \Rightarrow \quad 24+13+8+5=f \\ & \therefore \quad f=50 \end{aligned} $

Solution

(i) We observe that the number of patients which take medical treatment in a hospital on a particular day less than 10 is 0 . Similarly, less than 20 include the number of patients which take medical treatment from 0-10 as well as the number of patients which take medical treatment from 10-20.

So, the total number of patients less than 20 is $0+60=60$, we say that the cumulative frequency of the class $10-20$ is 60 . Similarly, for other class.

(ii) Also, we observe that all 300 patients which take medical treatment more than or equal to 10. Since, there are 60 patients which take medical treatment in the interval 10-20, this means that there are $300-60=240$ patients which take medical treatment more than or equal to 20. Continuing in the same manner.

Solution

Here, we observe that, 17 students have scored marks below 20 i.e., it lies between class interval 0-20 and 22 students have scored marks below 40 , so $22-17=5$ students lies in the class interval 20-40 continuting in the same manner, we get the complete frequency distribution table for given data.

Solution

First we construct a cumulative frequency table.

It is given that, $n=600$

$ \therefore \quad \dfrac{n}{2}=\dfrac{600}{2}=300 $

Since, cumulative frequency 440 lies in the interval $1000-2000$.

Here, (lower median class) $l=1000$,

$ f=190, c f=250, \text{ (class width) } h=1000 $

and (total observation) $n=600$

$ \begin{aligned} \therefore \quad \text{ Median } = l+\dfrac{\dfrac{n}{2}-c f}{f} \times h \\ & =1000+\dfrac{(300-250)}{190} \times 1000 \\ & =1000+\dfrac{50}{190} \times 1000 \\ & =1000+\dfrac{5000}{19} \\ & =1000+263.15=1263.15 \end{aligned} $

Hence, the median income is ₹ 1263.15 .

Solution

First we construct the cumulative frequency table

It is given that, $n=33$

$ \therefore \quad \dfrac{n}{2}=\dfrac{33}{2}=16.5 $

So, the median class is $100-115$.

where,

$ \begin{aligned} & \text{ lower limit }(l)=100, \\ & \text{ frequency }(f)=9 \end{aligned} $

$ \text{ cumulative frequency }(c f)=11 $

and class width $(h)=15$

$ \begin{aligned} \text{ Median } = l+\dfrac{\dfrac{n}{2}-c f}{f} \times h \\ & =100+\dfrac{(16.5-11)}{9} \times 15 \\ & =100+\dfrac{5.5 \times 15}{9}=100+\dfrac{82.5}{9}=100+9.17 \\ & =109.17 \end{aligned} $

Hence, the median bowling speed is $109.17 km / h$.

Solution

In a given data, the highest frequency is 41, which lies in the interval 10000-15000.

Here, $l=10000, f_m=41, f_1=26, f_2=16$ and $h=5000$

$ \begin{aligned} \therefore \quad \text{ Mode } = l+\dfrac{f_m-f_1}{2 f_m-f_1-f_2} \times h \\ & =10000+\dfrac{41-26}{2 \times 41-26-16} \times 5000 \\ & =10000+\dfrac{15}{82-42} \times 5000 \\ & =10000+\dfrac{15}{40} \times 5000 \\ & =10000+15 \times 125=10000+1875=\text{₹} 11875 \end{aligned} $

Hence, the modal income is ₹ 11875

Solution

In the given data, the highest frequency is 26 , which lies in the interval 201-202 Here,

$ l=201, f_m=26, f_1=12, f_2=20 \text{ and (class width) } h=1 $

$ \begin{aligned} \therefore \quad \text{ Mode } = l+\dfrac{f_m-f_1}{2 f_m-f_1-f_2} \quad \times h=201+\dfrac{26-12}{2 \times 26-12-20} \times 1 \\ =201+\dfrac{14}{52-32}=201+\dfrac{14}{20}=201+0.7=201.7 g \end{aligned} $

Hence, the modal weight is $201.7 g$.

Solution

Two dice are thrown at the same time.

[given]

So, total number of possible outcomes $=36$

(i) We have, same number on both dice.

So, possible outcomes are $(1,1),(2,2),(3,3),(4,4),(5,5)$ and $(6,6)$.

$\therefore$ Number of possible outcomes $=6$

Now, required probability $=\dfrac{6}{36}=\dfrac{1}{6}$ (ii) We have, different number on both dice.

So, number of possible outcomes

$=36-$ Number of possible outcomes for same number on both dice

$ =36-6=30 $

$\therefore$ Required probability $=\dfrac{30}{36}=\dfrac{5}{6}$

Solution

Two dice are thrown simultaneously.

[given]

So, total number of possible outcomes $=36$

(i) Sum of the numbers appearing on the dice is 7 .

So, the possible ways are $(1,6),(2,5),(3,4),(4,3),(5,2)$ and $(6,1)$.

Number of possible ways $=6$

$\therefore$ Required probability $=\dfrac{6}{36}=\dfrac{1}{6}$

(ii) Sum of the numbers appearing on the dice is a prime number i.e., 2, 3, 5, 7 and 11.

So, the possible ways are $(1,1),(1,2),(2,1),(1,4),(2,3),(3,2),(4,1),(1,6),(2,5)$, $(3,4),(4,3),(5,2),(6,1),(5,6)$ and $(6,5)$.

Number of possible ways $=15$

$\therefore$ Required probability $=\dfrac{15}{36}=\dfrac{5}{12}$

(iii) Sum of the numbers appearing on the dice is 1.

It is not possible, so its probability is zero.

Solution

Number of total outcomes $=36$

(i) When product of the numbers on the top of the dice is 6 .

So, the possible ways are $(1,6),(2,3),(3,2)$ and $(6,1)$.

Number of possible ways $=4$

$\therefore \quad$ Required probability $=\dfrac{4}{36}=\dfrac{1}{9}$

(ii) When product of the numbers on the top of the dice is 12 .

So, the possible ways are $(2,6),(3,4),(4,3)$ and $(6,2)$.

Number of possible ways $=4$

$\therefore \quad$ Required probability $=\dfrac{4}{36}=\dfrac{1}{9}$

(iii) Product of the numbers on the top of the dice cannot be 7 . So, its probability is zero.

Solution

Number of total outcomes $=36$

When product of numbers appearing on them is less than 9 , then possible ways are $(1,6)$, $(1,5)(1,4),(1,3),(1,2),(1,1),(2,2),(2,3),(2,4),(3,2),(4,2),(4,1),(3,1),(5,1),(6,1)$ and $(2,1)$.

Number of possible ways $=16$

$\therefore$ Required probability $=\dfrac{16}{36}=\dfrac{4}{9}$

Solution

Number of total outcomes $=n(s)=36$

(i) Let $E_1=$ Event of getting sum $2={(1,1),(1,1)}$

$ \begin{matrix} & n(E_1)=2 \\ \therefore & P(E_1)=\dfrac{n(E_1)}{n(S)}=\dfrac{2}{36}=\dfrac{1}{18} \end{matrix} $

(ii) Let $E_2=$ Event of getting sum $3={(1,2),(1,2),(2,1),(2,1)}$

$ \begin{aligned} & n(E_2)=4 \\ & \therefore \quad P(E_2)=\dfrac{n(E_2)}{n(S)}=\dfrac{4}{36}=\dfrac{1}{9} \end{aligned} $

(iii) Let $E_3=$ Event of getting sum $4={(2,2),(2,2),(3,1),(3,1),(1,3),(1,3)}$

$ \begin{matrix} \therefore & n(E_3)=6 \\ \therefore & P(E_3)=\dfrac{n(E_3)}{n(S)}=\dfrac{6}{36}=\dfrac{1}{6} \end{matrix} $

(iv) Let $E_4=$ Event of getting sum $5={(2,3),(2,3),(4,1),(4,1),(3,2),(3,2)}$

$ \begin{matrix} \therefore & n(E_4)=6 \\ \therefore & P(E_4)=\dfrac{n(E_4)}{n(S)}=\dfrac{6}{36}=\dfrac{1}{6} \end{matrix} $

(v) Let $E_5=$ Event of getting sum $6={(3,3),(3,3),(4,2),(4,2),(5,1),(5,1)}$

$ \begin{matrix} n(E_5) = 6 \\ \therefore & P(E_5)=\dfrac{n(E_5)}{n(S)}=\dfrac{6}{36}=\dfrac{1}{6} \end{matrix} $

(vi) Let $E_6=$ Event of getting sum $7={(4,3),(4,3),(5,2),(5,2),(6,1),(6,1)}$

$ \begin{matrix} \therefore & n(E_6)=6 \\ \therefore & P(E_6)=\dfrac{n(E_6)}{n(S)}=\dfrac{6}{36}=\dfrac{1}{6} \end{matrix} $

(vii) Let $E_7=$ Event of getting sum $8={(5,3),(5,3),(6,2),(6,2)}$

$ \begin{matrix} \therefore & n(E_7)=4 \\ \therefore & P(E_7)=\dfrac{n(E_7)}{n(S)}=\dfrac{4}{36}=\dfrac{1}{9} \end{matrix} $

(viii) Let $E_8=$ Event of getting sum $9={(6,3),(6,3)}$

$ \begin{matrix} \therefore & n(E_8)=2 \\ \therefore & P(E_8)=\dfrac{n(E_8)}{n(S)}=\dfrac{2}{36}=\dfrac{1}{18} \end{matrix} $

Solution

The possible outcomes, if a coin is tossed 2 times is

$ \begin{aligned} & \therefore = {(H H),(T T),(H T),(T H)} \\ & n(S) = 4 \end{aligned} $

Let $E=$ Event of getting atmost one head

$ \begin{matrix} ={(T T),(H T),(T H)} \\ \therefore \quad n(E) = 3 \\ \text{ Hence, } \quad \text{ required probability } = \dfrac{n(E)}{n(S)}=\dfrac{3}{4} \end{matrix} $

Solution

The possible outcomes if a coin is tossed 3 times is

$ \begin{aligned} & S={(H H H),(T T T),(H T T),(T H T),(T T H),(T H H),(H T H),(H H T)} \\ & \therefore \quad n(S)=8 \end{aligned} $

(i) Let $E_1=$ Event of getting all heads $={(H H H)}$

$ \begin{matrix} \therefore & n(E_1)=1 \\ \therefore & P(E_1)=\dfrac{n(E_1)}{n(S)}=\dfrac{1}{8} \end{matrix} $

(ii) Let $E_2=$ Event of getting atleast 2 heads $={(H H T),(H T H),(T H H),(H H H)}$

$ \begin{matrix} \therefore & n(E_2)=4 \\ \therefore & P(E_2)=\dfrac{n(E_2)}{n(S)}=\dfrac{4}{8}=\dfrac{1}{2} \end{matrix} $

Solution

The total number of sample space in two dice, $n(S)=6 \times 6=36$

$ \begin{aligned} & \text{ Let } \quad E=\text{ Event of getting the numbers whose difference is } 2 \\ & ={(1,3),(2,4),(3,5),(4,6),(3,1),(4,2),(5,3),(6,4)} \\ & \therefore \quad n(E)=8 \\ & \therefore \quad P(E)=\dfrac{n(E)}{n(S)}=\dfrac{8}{36}=\dfrac{2}{9} \end{aligned} $

Solution

If a ball is drawn out of 22 balls ( 5 blue +7 green +10 red), then the total number of outcomes are

$ n(S)=22 $

(i) Let $E_1=$ Event of getting a red ball

$ \begin{matrix} n(E_1)=10 \\ \therefore \quad \text{ Required probability }=\dfrac{n(E_1)}{n(S)}=\dfrac{10}{22}=\dfrac{5}{11} \end{matrix} $

(ii) Let $E_2=$ Event of getting a green ball

$ \begin{aligned} n(E_2) = 7 \\ \therefore \quad \text{ Required probability }= & \dfrac{n(E_2)}{n(S)}=\dfrac{7}{22} \end{aligned} $

(iii) Let $E_3=$ Event getting a red ball or a green ball i.e., not a blue ball.

$ \begin{aligned} & n(E_3)=(10+7)=17 \\ & \therefore \quad \text{ Required probability }=\dfrac{n(E_3)}{n(S)}=\dfrac{17}{22} \end{aligned} $

Solution

If we remove one king, one queen and one jack of clubs from 52 cards, then the remaining cards left, $n(S)=49$

(i) Let $E_1=$ Event of getting a heart

$ \begin{aligned} & n(E_1)=13 \\ \therefore & P(E_1)=\dfrac{n(E_1)}{n(S)}=\dfrac{13}{49} \end{aligned} $

(ii) Let $E_2=$ Event of getting a king

$n(E_2)=3$ [since, out of 4 king, one club cards is already removed]

$\therefore \quad P(E_2)=\dfrac{n(E_2)}{n(S)}=\dfrac{3}{49}$

Solution

(i) $L$ et $E_3=$ Event of getting a club

$n(E_3)=(13-3)=10$

$\therefore$ Required probability $=\dfrac{n(E_3)}{n(S)}=\dfrac{10}{49}$

(ii) Let $E_4=$ Event of getting 10 of hearts

$ n(E_4)=1 $

[because in 52 playing cards only 13 are the heart cards and only one 10 in 13 heart cards]

$\therefore \quad$ Required probability $=\dfrac{n(E_4)}{n(S)}=\dfrac{1}{49}$

Solution

In out of 52 playing cards, 4 jacks, 4 queens and 4 kings are removed, then the remaining cards are left, $n(S)=52-3 \times 4=40$.

(i) Let $E_1=$ Event of getting a card whose value is 7

$E=$ Card value 7 may be of a spade, a diamond, a club or a heart

$ \begin{matrix} \therefore & n(E_1)=4 \\ \therefore & P(E_1)=\dfrac{n(E_1)}{n(S)}=\dfrac{4}{40}=\dfrac{1}{10} \end{matrix} $

(ii) Let $E_2=$ Event of getting a card whose value is greater than 7

$=$ Event of getting a card whose value is 8,9 or 10

$ \begin{matrix} \therefore & n(E_2)=3 \times 4=12 \\ \therefore & P(E_2)=\dfrac{n(E_2)}{n(S)}=\dfrac{12}{40}=\dfrac{3}{10} \end{matrix} $

(iii) Let $E_3=$ Event of getting a card whose value is less than 7 $=$ Event of getting a card whose value is 1, 2, 3, 4, 5 or 6

$ \begin{matrix} \therefore & n(E_3)=6 \times 4=24 \\ \therefore & P(E_3)=\dfrac{n(E_3)}{n(S)}=\dfrac{24}{40}=\dfrac{3}{5} \end{matrix} $

Solution

The number of integers between 0 and 100 is

$ n(S)=99 $

(i) Let $E_1=$ Event of choosing an integer which is divisible by 7

$=$ Event of choosing an integer which is multiple of 7

$={7,14,21,28,35,42,49,56,63,70,77,84,91,98}$

$ \begin{matrix} \therefore & n(E_1)=14 \\ \therefore & P(E_1)=\dfrac{n(E_1)}{n(S)}=\dfrac{14}{99} \end{matrix} $

(ii) Let $E_2=$ Event of choosing an integer which is not divisible by 7

$ \begin{matrix} \therefore & n(E_2) = n(S)-n(E_1) \\ & =99-14=85 \\ \therefore & P(E_2) = \dfrac{n(E_2)}{n(S)}=\dfrac{85}{99} \end{matrix} $

Solution

Total number of out comes with numbers 2 to $101, n(s)=100$

(i) Let $E_1=$ Event of selecting a card which is an even number $={2,4,6, \ldots 100}$

$ \begin{array}{ll} [\text{ in an AP, } l=a+(n-1) d, \text{ here } l = 100, a=2 \text{ and } d=2 \Rightarrow 100=2+(n-1) 2 \\ \Rightarrow (n-1)=49 \Rightarrow n=50] \\ \therefore n(E_1)=50 \\ \therefore \text{Requied probability} = \dfrac{n(E_1)}{n(S)}=\dfrac{50}{100}=\dfrac{1}{2} \end{array} $

(ii) Let $E_2=$ Event of selecting a card which is a square number

$ \begin{aligned} & \begin{matrix} ={4,9,16,25,36,49,64,81,100} \\ ={(2)^{2},(3)^{2},(4)^{2},(5)^{2},(6)^{2},(7)^{2},(8)^{2},(9)^{2},(10)^{2}} \\ \end{matrix} \\ \therefore & \quad \quad n(E_2)=9 \\ & \text{ Hence, required probability }=\dfrac{n(E_2)}{n(S)}=\dfrac{9}{100} \end{aligned} $

Solution

We know that, in english alphabets, there are ( 5 vowels +21 consonants) $=26$ letters. So, total number of outcomes in english alphabets are,

$ \begin{aligned} & n(S)=26 \\ & \text{ Let } \quad E=\text{ Event of choosing a english alphabet, which is a consonent } \\ & ={b, c, d, f, g, h, j, k, l, m, n, p, q, r, s, t, v, w, x, y, z} \\ & \therefore \quad n(E)=21 \end{aligned} $

Hence, required probability $=\dfrac{n(E)}{n(S)}=\dfrac{21}{26}$

Solution

Total number of sealed envelopes in a box, $n(S)=1000$

Number of envelopes containing cash prize $=10+100+200=310$

Number of envelopes containing no cash prize,

$ \begin{matrix} & n(E)=1000-310=690 \\ \therefore & P(E)=\dfrac{n(E)}{n(S)}=\dfrac{690}{1000}=\dfrac{69}{100}=0.69 \end{matrix} $

Solution

Total number of slips in a box, $n(S)=25+50=75$

From the chart it is clear that, there are 11 slips which are marked other than ₹ 1 .

$\therefore \quad$ Required probability $=\dfrac{\text{ Number of slips other than ₹ } 1}{\text{ Total number of slips }}=\dfrac{11}{75}$

Solution

$\because$ Total number of bulbs, $n(S)=24$

Let $E_1=$ Event of selecting not defective bulb $=$ Event of selecting good bulbs

$ \begin{aligned} & n(E_1)=18 \\ \therefore & P(E_1)=\dfrac{n(E_1)}{n(S)}=\dfrac{18}{24}=\dfrac{3}{4} \end{aligned} $

Suppose, the selected bulb is defective and not replaced, then total number of bulbs remains in a carton, $n(S)=23$.

In them, 18 are good bulbs and 5 are defective bulbs.

$\therefore P($ selecting second defective bulb $)=\dfrac{5}{23}$

Solution

Total number of figures

$ n(S)=8 \text{ triangles }+10 \text{ squares }=18 $

(i) $P$ (lost piece is a triangle) $=\dfrac{8}{18}=\dfrac{4}{9}$

(ii) $P$ (lost piece is a square) $=\dfrac{10}{18}=\dfrac{5}{9}$

(iii) $P$ (square of blue colour) $=\dfrac{6}{18}=\dfrac{1}{3}$

(iv) $P$ (triangle of red colour) $=\dfrac{5}{18}$

Solution

Total possible outcomes of tossing a coin 3 times, $S={(H H H),(T T T),(H T T),(T H T),(T T H),(T H H),(H T H),(H H T)}$ $n(S)=8$

(i) Let $E_1=$ Event that Sweta losses the entry fee

$=$ She tosses tail on three times

$n(E_1)={(T T T)}$

$P(E_1)=\dfrac{n(E_1)}{n(S)}=\dfrac{1}{8}$

(ii) Let $E_2=$ Event that Sweta gets double entry fee $=$ She tosses heads on three times $={(H H H)}$

$ \begin{aligned} & n(E_2)=1 \\ \therefore & P(E_2)=\dfrac{n(E_2)}{n(S)}=\dfrac{1}{8} \end{aligned} $

(iii) Let $E_3=$ Event that Sweta gets her entry fee back

$ \begin{aligned} & =\text{ Sweta gets heads one or two times } \\ & ={(H T T),(T H T),(T T H),(H H T),(H T H),(T H H)} \\ & \therefore \quad n(E_3)=6 \\ & \therefore \quad P(E_3)=\dfrac{n(E_3)}{n(S)}=\dfrac{6}{8}=\dfrac{3}{4} \end{aligned} $

Solution

Given, a die has its six faces marked ${0,1,1,1,6,6}$

$\therefore$ Total sample space, $n(S)=6^{2}=36$

(i) The different score which are possible are 6 scores i.e., 0,1,2,6,7 and12.

(ii) Let $E=$ Event of getting a sum 7

$ \begin{aligned}& =\{(1,6),(1,6),(1,6),(1,6),(1,6),(1,6),(6,1),(6,1),(6,1),(6,1),(6,1),(6,1)\} \\ & \therefore \quad n(E)=12 \\ & \therefore \quad P(E)=\dfrac{n(E)}{n(S)}=\dfrac{12}{36}=\dfrac{1}{3} \end{aligned} $

Solution

Given, total number of mobile phones

$ n(S)=48 $

(i) Let $E_1=$ Event that Varnika will buy a mobile phone

$=$ Varnika buy only, if it is good mobile

$\therefore n(E_1)=42$

$\therefore P(E_1)=\dfrac{n(E_1)}{n(S)}=\dfrac{42}{48}=\dfrac{7}{8}$

(ii) Let $E_2=$ Event that trader will buy only when it has no major defects

$=$ Trader will buy only 45 mobiles

$ \begin{matrix} \therefore & n(E_2)=45 \\ \therefore & P(E_2)=\dfrac{n(E_2)}{n(S)}=\dfrac{45}{48}=\dfrac{15}{16} \end{matrix} $

Solution

Given that, A bag contains total number of balls $=24$

A bag contains number of red balls $=x$

A bag contains number of white balls $=2 x$

and a bag contains number of blue balls $=3 x$

By condition, $\quad x+2 x+3 x=24$

$\Rightarrow \quad 6 x=24$

$\therefore \quad x=4$ $\therefore$ Number of red balls $=x=4$

Number of white balls $=2 x=2 \times 4=8$

and number of blue balls $=3 x=3 \times 4=12$

So, total number of outcomes for a ball is selected at random in a bag contains 24 balls.

$\Rightarrow \quad n(S)=24$

(i) Let $E_1=$ Event of selecting a ball which is not red i.e., can be white or blue.

$\therefore n(E_1)=$ Number of white balls + Number of blue balls

$ \Rightarrow \hspace{10 mm} n(E_1) =8+12=20 \\ \therefore \hspace{10 mm} \text{Requied probability} = \dfrac{n(E_1)}{n(S)}=\dfrac{20}{24}=\dfrac{5}{6} $

(ii) Let $E_2=$ Event of selecting a ball which is white

$\therefore n(E_2)=$ Number of white balls $=8$

So, required probability $=\dfrac{n(E_2)}{n(S)}=\dfrac{8}{24}=\dfrac{1}{3}$

Solution

Given that, 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 so, the total number of outcomes are $n(S)=1000$

If the selected card has a perfect square greater than 500 , then player wins a prize.

(i) Let $E_1=$ Event first player wins a prize $=$ Player select a card which is a perfect square greater than 500

$={529,576,625,676,729,784,841,900,961}$

$={(23)^{2},(24)^{2},(25)^{2},(26)^{2},(27)^{2},(28)^{2},(29)^{2},(30)^{2},(31)^{2}}$

$\therefore \quad n(E_1)=9$

So, required probability $=\dfrac{n(E_1)}{n(S)}=\dfrac{9}{1000}=0.009$

(ii) First, has won i.e., one card is already selected, greater than 500 , has a perfect square. Since, repeatition is not allowed. So, one card is removed out of 1000 cards. So, number of remaining cards is 999 .

$\therefore$ Total number of remaining outcomes, $n(S^{\prime})=999$

Let $E_2=$ Event the second player wins a prize, if the first has won.

$=$ Remaining cards has a perfect square greater than 500 are 8 .

$\therefore \quad n(E_2)=9-1=8$

So, required probability $=\dfrac{n(E_2)}{n(S^{\prime})}=\dfrac{8}{999}$

Solution

Solution

Here, we observe that, 5 students have scored marks below 10, i.e. it lies between class interval 0-10 and 9 students have scored marks below 20.

So, $(9-5)=4$ students lies in the class interval $10-20$. Continuing in the same manner, we get the complete frequency distribution table for given data.

Here, (assumed mean) $a=45$

and (class width) $h=10$

By step deviation method,

$ \begin{aligned} \text{ Mean }(\bar{{}x}) = a+\dfrac{\sum f_i u_i}{\sum f_i} \times h=45+\dfrac{29}{85} \times 10=45+\dfrac{58}{17} \\ & =45+3.41=48.41 \end{aligned} $

Solution

Here, we observe that, all 100 residents of a town have age equal and above 0. Since, 90 residents of a town have age equal and above 10.

So, $100-90=10$ residents lies in the interval 0-10 and so on. Continue in this manner, we get frequency of all class intervals. Now, we construct the frequency distribution table.

Here, (assumed mean) $a=35$

and (class width) $h=10$

By step deviation method,

$ \begin{aligned} Mean(\bar{{}x}) = a+\dfrac{\sum f_i u_i}{\sum f_i} \times h \\ & =35+\dfrac{(-40)}{100} \times 10 \\ & =35-4=31 . \end{aligned} $

Hence, the required mean age is $31 yr$.

Solution

First, we find the class marks of the given data as follows.

Here, (assume mean) a $=203.5$

and (class width) $\quad h=1$

By assumed mean method,

$ \begin{aligned} Mean(\bar{{}x}) = a+\dfrac{\sum f_i d_i}{\sum f_i} \\ & =203.5-\dfrac{108}{70} \\ & =203.5-1.54=201.96 \end{aligned} $

Hence, the required mean weight is $201.96 g$.

Solution

We observe that, the number of packets less than 200 is 0 . Similarly, less than 201 include the number of packets from 0-200 as well as the number of packets from 200-201.

So, the total number of packets less than 201 is $0+13=13$. We say that, the cumulative frequency of the class $200-201$ is 13 . Similarly, for other class.

To draw the less than type ogive, we plot the points $(200,0),(201,13),(202,40)(203,58)$, $(204,68),(205,69)$ and $(206,70)$ on the paper and join by free hand.

$\because$ Total number of packets $(n)=70$

Now,

Firstly, we plot a point $(0,35)$ on $Y$-axis and draw a line $y=35$ parallel to $X$-axis. The line cuts the less than ogive curve at a point. We draw a line on that point which is perpendicular to $X$-axis. The foot of the line perpendicular to $X$-axis is the required median.

$\therefore$ Median weight $=201.8 g$

Solution

For less than type table we follow the Q.5.

Here, we observe that, the weight of all 70 packets is more than or equal to 200 . Since, 13 packets lie in the interval 200-201. So, the weight of $70-13=57$ packets is more than or equal to 201. Continuing in this manner we will get remaining more than or equal to 202, 203, 204, 205 and 206.

To draw the less than type ogive, we plot the points $(200,0),(201,13),(202,40),(203,58)$, $(204,68),(205,69),(206,70)$ on the paper and join them by free hand.

To draw the more than type ogive plot the points $(200,70),(201,57),(202,30),(203,12)$, $(204,2),(205,1),(206,0)$ on the the graph paper and join them by free hand.

Hence, required median weight $=I$ ntersection point of $X$-axis $=201.8 g$.

Solution

First, we construct a cumulative frequency table

(i) Here, median class is $10-15$, because 140 lies in it.

Lower limit $(l)=10$, Frequency $(f)=133$,

Cumulative frequency $(c f)=49$ and class width $(h)=5$

$ \therefore \quad \begin{aligned} \text{ Median } = l+\dfrac{\dfrac{N}{2}-c f}{f} \times h \\ & =10+\dfrac{(140-49)}{133} \times 5 \\ & =10+\dfrac{91 \times 5}{133} \\ & =10+\dfrac{455}{133}=10+3.421 \\ & =\text{₹} 13.421(\text{ in thousand) } \\ & =13.421 \times 1000 \\ & =\text{₹} 13421 \end{aligned} $

(ii) Here, the highest frequency is 133, which lies in the interval 10-15, called modal class. Lower $limit(l)=10$, class width $(h)=5, f_m=133, f_1=49$, and $f_2=63$.

$ \begin{array}{rl} \therefore \quad \text{ Mode } & = l+\dfrac{f_m-f_1}{2 f_m-f_1-f_2} \times h \\ & =10+\dfrac{133-49}{2 \times 133-49-63} \times 5 \\ & =10+\dfrac{84 \times 5}{266-112}=10+\dfrac{84 \times 5}{154}=10+2.727 \\ & =\text{₹} 12.727 \text{ (in thousand) } \\ & =12.727 \times 1000=\text{₹} 12727 \end{array} $

Hence, the median and modal salary are ₹ 13421 and ₹ 12727, respectively.

Solution

First we calculate the class mark of given data

Given that, sum of all frequencies $=120$

$ \begin{matrix} \Rightarrow & \sum f_i=68+f_1+f_2=120 \\ \Rightarrow & f_1+f_2=52 \\ \text{ Here, } & \text{ (assumed mean) } a=50 \\ \text{ and } & \text{ (class width) } h=20 \end{matrix} $

By step deviation method,

$$ \begin{matrix} \text{ Mean } = a+\dfrac{\sum f_i u_i}{\sum f_i} \times h \\ \Rightarrow & 50 = 50+\dfrac{(4+f_2-f _{1)}.}{120} \times 20 \\ \Rightarrow & 4+f_2-f_1 = 0 \\ \Rightarrow & -f_2+f_1 = 4 \tag{ii} \end{matrix} $$

On adding Eqs. (i) and (ii), we get

$ \begin{matrix} 2 & 2 f_1 = 56 \\ \Rightarrow & f_1 = 28 \end{matrix} $

Put the value of $f_1$ in Eq. (i), we get

$ \begin{aligned} & f_2=52-28 \\ & f_2=24 \end{aligned} $

Hence, $f_1=28$ and $f_2=24$.

Solution

Given,

$ N=90 $

$ \therefore \quad \dfrac{N}{2}=\dfrac{90}{2}=45 $

which lies in the interval 50-60.

Lower limit, $l=50, f=20, c f=40+p, h=10$

$ \begin{matrix} \therefore & \text{ Median } = l+\dfrac{\dfrac{N}{2}-c f}{f} \times h \\ & = 50+\dfrac{(45-40-p)}{20} \times 10 \\ \Rightarrow & 50 = 50+\dfrac{5-p}{2} \\ \Rightarrow & 0 = \dfrac{5-p}{2} \\ \therefore & p = 5 \\ \text{ Also, } & 78+p+q = 90 \\ \Rightarrow & 78+5+ & q = 90 \\ \Rightarrow & q = 90-83 \\ \therefore & q = 7 \end{matrix} $

Solution

To draw the less than type ogive, we plot the points $(124,0),(128,5),(132,13),(136,30)$, $(140,54),(144,70),(148,82),(152,88),(156,92),(160,95),(164,96)$ and join all these point by free hand.

Here,

$ \dfrac{N}{2}=\dfrac{96}{2}=48 $

We take, $y=48$ in $y$-coordinate and draw a line parallel to $X$-axis, meets the curve at $A$ and draw a perpendicular line from point $A$ to the $X$-axis and this line meets the $X$-axis at the point which is the median i.e., median $=141.17$.

Solution

(i) Here, $N=200$

Now, $\dfrac{N}{2}=\dfrac{200}{2}=100$, which lies in the interval 15-20.

Lower limit, $l=15, h=5, f=80$ and $c f=55$

$ \begin{aligned} \therefore \quad \text{ Median } = l+\dfrac{\dfrac{N}{2}-c f}{f} \times h=15+\dfrac{100-55}{80} \times 5 \\ & =15+\dfrac{45}{16}=15+2.81=17.81 hec \end{aligned} $

(ii) In a given table 80 is the highest frequency.

So, the modal class is $15-20$.

$ \begin{aligned} & \text{ Here, } l=15, f_m=80, f_1=30, f_2=40 \text{ and } h=5 \\ & \qquad \begin{aligned} \text{ Mode } = l+\dfrac{f_m-f_1}{2 f_m-f_1-f_2} \times h \\ & =15+\dfrac{80-30}{2 \times 80-30-40} \times 5 \\ & =15+\dfrac{50}{160-70} \times 5 \\ & =15+\dfrac{50}{90} \times 5=15+\dfrac{25}{9} \\ & =15+2.77=17.77 hec \end{aligned} \end{aligned} $

Solution

We observe that, the annual rainfall record of a city less than 0 is 0 . Similarly, less than 10 include the annual rainfall record of a city from 0 as well as the annual rainfall record of a city from 0-10.

So, the total annual rainfall record of a city for less than $10 cm$ is $0+22=22$ days. Continuing in this manner, we will get remaining less than 20,30,40,50, and 60 .

Also, we observe that annual rainfall record of a city for 66 days is more than or equal to $0 cm$. Since, 22 days lies in the interval $0-10$. So, annual rainfall record for $66-22=44$ days is more than or equal to $10 cm$. Continuing in this manner we will get remaining more than or equal to 20, 30, 40, 50 and 60.

Now, we construct a table for less than and more than type.

To draw less than type ogive we plot the points $(0,0),(10,22),(20,32),(30,40),(40,55)$, $(50,60),(60,66)$ on the paper and join them by free hand.

To draw the more than type ogive we plot the points $(0,66),(10,44),(20,34),(30,26),(40$, $11),(50,6)$ and $(60,0)$ on the graph paper and join them by free hand.

$\because$ Total number of days $(n)=66$

Now, $\quad \dfrac{n}{2}=33$

Firstly, we plot a line parallel to $X$-axis at intersection point of both ogives, which further intersect at $(0,33)$ on $Y$-axis. Now, we draw a line perpendicular to $X$-axis at intersection point of both ogives, which further intersect at $(21.25,0)$ on $X$-axis. Which is the required median using ogives.

Hence, median rainfall $=21.25 cm$.

Solution

First, we calculate class marks as follows

Here, (assumed mean) $a=170$,

and (class width) $h=30$

By step deviation method,

$ \text{ Average } \begin{aligned} (\bar{{}x}) = a+\dfrac{\sum f_i u_i}{\sum f_i} \times h=170+\dfrac{1}{100} \times 30 \\ & =170+0.3=170.3 \end{aligned} $

Hence, average duration is $170.3 s$.

For calculating median from a cumulative frequency curve

We prepare less than type or more than type of give

We observe that, number of calls in less than $95 s$ is 0 . Similarly, in less than $125 s$ include the number of calls in less than $95 s$ as well as the number of calls from 95-125.s So, the total number of calls less than $125 s$ is $0+14=14$. Continuing in this manner, we will get remaining in less than 155, 185, 215 and $245 s$.

Now, we construct a table for less than ogive (cumulative frequency curve).

Less than type

To draw less than type ogive we plot them the points $(95,0),(125,14)(155,36),(185,64)$, $(215,85),(245,100)$ on the paper and join them by free hand.

$ \therefore \quad \dfrac{n}{2}=\dfrac{100}{2}=50 $

Now, point 50 taking on $Y$-axis draw a line parallel to $X$-axis meet at a point $P$ and draw a perpendicular line from $P$ to the $X$-axis, the intersection point of $X$-axis is the median.

Hence, required median is 170

Solution

(i)

(ii)

To draw less than type ogive, we plot the points $(0,0),(20,6),(40,17),(60,34),(80,46)$, $(100,50)$, join all these points by free hand.

Now,

Taking $Y=25$ on $Y$-axis and draw a line parallel to $X$-axis, which meets the curve at point $A$. From point $A$, we draw a line perpendicular to $X$-axis, where this meets that point is the required median i.e., 49.4.

(iii) Now,

$ \dfrac{N}{2}=\dfrac{50}{2}=25 $

which lies is the interval $40-60$.

$ \begin{matrix} \therefore \quad l = 40, h=20, c f=17 \text{ and } f=17 \\ \therefore \quad \text{ Median } = l+\dfrac{\dfrac{N}{2}-c f}{f} \times h \\ & =40+\dfrac{(25-17)}{17} \times 20 \\ & =40+\dfrac{8 \times 20}{17} \\ & =40+9.41 \\ & =49.41 \end{matrix} $

(iv) Yes, median distance calculated by parts (ii) and (iii) are same.