NCERT Solutions for Class 10 Maths Chapter 13 - Statistics And Probability

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Answer:

We know that, d_i = x_i$-$a, where, x_i are data and a is the assumed mean
i.e. d_i are the deviations from of mid$-$points of the classes.
Hence, the correct answer is option (C).

Answer:

In grouping the data all the observations between lower and upper limits of class marks are taken in one group then mid value or class mark is taken for further calculation.

In computing the mean of grouped data, the frequencies are centered at the class marks of the classes.

Hence, the correct answer is option (B).
 

Answer:

We know that,
$\sum _{i=1}^n \left(f_i\times x_i\right)$ =  $n\times \overline{x}$                                                                     .....(1)

$\sum _{i=1}^n \overline{x} = \overline{x} + \overline{x} + \overline{x} + \overline{x} ..... n times$   

$\Rightarrow \sum _{i=1}^n \overline{x} = n \overline{x}$                                                                       .....(2)

From (1) and (2)
$\sum _{i=1}^n \left(f_i\times x_i\right)$  =  $\sum _{i=1}^n \overline{x}$

$\Rightarrow$$\sum _{}$$\left(f_i{x}_i-\overline{x}\right)$ = 0

Hence, the correct answer is option (A).

Answer:

Above formula is a step deviation (shortcut) formula.
Where x_i are data values, a is assumed mean and h is class size, when class size is same we simplify the calculations of the mean by computing the coded mean of u₁, u₂, u₃…..where
$u_i=\frac{x_i-a}{h}$

Hence, the correct answer is option (C).

 

Answer:

The intersection point of less than ogive and more than ogive gives the median on the abscissa.
Hence, the correct answer is option (B).

Answer:

Here 
 

Answer:

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.
 

Answer:

Answer:

Answer:

The number of athletes who completed the race in less than 14.6 = 2 + 4 + 5 + 71 = 82

Hence, the correct answer is option (C).

Answer:

Answer:

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

Hence, the correct answer is option (D).

Answer:

Since, probability of an event always lies between 0 and 1, therefore probability of any event cannot be more than 1 i.e. it cannot be $\frac{17}{16}$.

Hence, the correct answer is option (D).  

Answer:

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.
Hence, the correct answer is option (A).

Answer:

Since, probability of an event + probability of its complementary event = 1 so,

Probability of its complementary event = (1 – Probability of an event) = 1 – p

Hence, the correct answer is option (C).

Answer:

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

Hence, the correct answer is option (B).
 

Answer:

Probability of an event always lies between 0 and 1.
Hence, the correct answer is option (C).

Answer:

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 = $\frac{6}{52}$ = $\frac{3}{26}$.

Hence, the correct answer is option (A).

Answer:

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 favorable event is the event that the one day is Sunday.
∴ Required probability = $\frac{1}{7}$
Hence, the correct answer is option (A).

Answer:

When a die is thrown, then total number of outcomes = 6
Odd number less than 3 is 1 only.
Number of possible outcomes = 1
∴ Required probability = $\frac{1}{6}$
Hence, the correct answer is option (A).

Answer:

In a deck of 52 cards, there are 13 cards of heart and 1 is ace of heart.
Therefore, the number of outcomes favorable to E = 52 $-$ 1 = 51

Hence, the correct answer is option (D).

Answer:

Here, total number of eggs = 400
Probability of getting a bad egg = 0.035
$\Rightarrow \mathrm{Probability} \mathrm{of} \mathrm{getting} \mathrm{bad} \mathrm{egg} = \frac{\mathrm{Number} \mathrm{of} \mathrm{bad} \mathrm{eggs}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{eggs}}$
$\Rightarrow 0.035 = \frac{\mathrm{Number} \mathrm{of} \mathrm{bad} \mathrm{eggs}}{400}$
∴ Number of bad eggs = 0.035 × 400 = 14
Hence, the correct answer is option (B).

Answer:

Given, total number of sold tickets = 6000
Let us assume that girl bought x tickets.
Then, probability of her winning the first prize is given as,
$\frac{x}{6000}=0.08$
∴ x = 480
Hence, the correct answer is option (C).

Answer:

Number of total outcomes = 40
Multiples of 5 between 1 to 40 = 5, 10, 15, 20, 25, 30, 35, 40
∴ Total number of possible outcomes = 8
∴ Required probability = $\frac{8}{40}$ = $\frac{1}{5}$
Hence, the correct answer is Option (A).

Answer:

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.
∴ Total number of possible outcomes = 25
∴ Required probability = $\frac{25}{100} = \frac{1}{4}$
Hence, the correct answer is option (C).

Answer:

Total number of students = 23

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

∴ Remaining students = 23 $-$ 17 = 6

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

Hence, the correct answer is option (B).

Answer:

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 uniformly distributed (or equally spaced).

Answer:

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.

Answer:

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

Answer:

Not always, it depends on the given data.

Answer:

No, the probability of each is not $\frac{1}{4}$.
Let boys be B and girls be G.
Outcomes can be {BBB, GGG, BBG, BGB, GBB, GGB, GBG, BGG}.
Total number of outcomes = 8
So,
P(No girl) = $\frac{1}{8}$

P(1 girl) = $\frac{3}{8}$

P(2 girls) = $\frac{3}{8}$

P(3 girls) =  $\frac{1}{8}$
 

Answer:

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.

Answer:

Apoorv throw two dice once.
So, total number of outcomes = 6² = 36
Number of outcomes for getting product 36 = 1 (6 × 6)
∴ Probability for Apoorv = $\frac{1}{36}$

Peehu throws one die.
So, total number of outcomes = 6
Number of outcomes for getting square 36 = 1 (6²)
∴ Probability for Peehu = $\frac{1}{6}$

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

Answer:

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

Answer:

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
∴ Probability of getting 1 = $\frac{1}{6}$
Now, probability of getting not 1 = 1 – Probability of getting 1
                                                      = $1-\frac{1}{6}$ 
                                                      = $\frac{5}{6}$

Answer:

Total number of outcomes = 2³ = 8
Possible outcomes are (HHH), (HTT), (THT), (TTH), (HHT), (THH), (HTH) and (TTT).
Now, probability of getting no head = $\frac{1}{8}$
Hence, the given conclusion is wrong because the probability of no head is $\frac{1}{8}$.

Answer:

No, if  we toss a coin, then possible outcomes are head or tail.
Both are equally likely events.
So, probability is $\frac{1}{2}$.
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.

Answer:

On tossing a coin, possible outcomes can be head or tail.
Both are equally likely events.
So, the outcome of next toss may or may not be tail. 

Answer:

Possible outcomes at each toss are Head or Tail.
Both are equally likely at each toss.
Hence, at the 4th toss, Tail does not have a higher chance, Head and Tail have equal chances.

Answer:

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

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


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


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

Answer:

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

Answer:

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

Answer:

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
 

Answer:

Since, the table is not continuous, we subtract 0.5 and 0.5 to the lower limit and upper limit respectively.
 

Answer:

No need to convert discontinuous classes into continuous for class mark because class mark of C.I. are same and gives same result of $\overline{x}$.

Answer:

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

Answer:

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

Answer:

Answer:

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

Answer:

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 lie in the interval 10 – 20 and so on. The table of a frequency distribution for the given data is:
 

Answer:

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 lie in the interval 0 $-$10 and so on.

Now, we construct the frequency distribution table.
 

Answer:

Answer:

(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.
Less than type cumulative frequency distribution is as follows:
 

Answer:

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 continuing in the same manner, we get the complete frequency distribution table for given data.
 

Answer:

First we construct a cumulative frequency table.

Answer:

First we construct the cumulative frequency table.

Answer:

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

Here, l = 10000, f_m = 41,  f₁ = 26,  f₂ = 16 and h = 5000.

$\therefore \mathrm{Mode}=l+\left(\frac{f_m-f_1}{2f_m-f_1-f_2}\right)\times h$
              = 10000 + $\frac{\left(41-26\right)}{(2\times 41-26-16)}\times 5000$
              = 10000 + $\left(\frac{15}{40}\right)\times 5000$
              = 10000 + 15 × 125
              = 10000 + 1875
              = 11875

Hence, the modal income is ₹11875 per month.

Answer:

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

Here, l = 201, f_m = 26, f₁ = 12, f₂ = 20 and (class width) h = 1.

$\therefore \mathrm{Mode}=l+\left(\frac{f_m-f_1}{2f_m-f_1-f_2}\right)\times h$

             = 201 + $\frac{\left(26-12\right)}{2\times 26-12-20}\times 1$
             = 201 + $\frac{14}{20}$

             = 201 + 0.7
             = 201.7 g

Hence, the modal weight is 201.7 g.

Answer:

Total number of possible outcomes = 6² = 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).
∴ Number of possible outcomes = 6
Now, required probability = $\frac{6}{36}$ = $\frac{1}{6}$


(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
∴ Required probability = $\frac{30}{36}$ = $\frac{5}{6}$

Answer:

Two dice are thrown simultaneously.  [given]
So, that 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
∴ Required probability = $\frac{6}{36}$ = $\frac{1}{6}$


(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
∴ Required probability = $\frac{15}{36}$ = $\frac{5}{12}$


(iii) Sum of the numbers appearing on the dice is 1.
It is not possible, so its probability is 0.

Answer:

Number of total outcomes = 36

(i) 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
∴ Required probability = $\frac{4}{36}$ = $\frac{1}{9}$

(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
∴ Required probability = $\frac{4}{36}$= $\frac{1}{9}$

(iii) Product of the numbers on the top of the dice cannot be 7. So, its probability is 0.

Answer:

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, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (4, 2), (4, 1), (5, 1), (6, 1).

Number of possible ways = 16

Probability = $\frac{\mathrm{No}. \mathrm{of} \mathrm{favourable} \mathrm{outcomes}}{\mathrm{Total} \mathrm{no}. \mathrm{of} \mathrm{outcomes}}$

∴ Required probability = $\frac{16}{36}$  
                                     = $\frac{4}{9}$

Answer:

Number of total outcome = n(S) = 36

(i) Let E₁ = Event of getting sum 2 = {(1, 1), (1, 1)}
 n(E₁) = 2
$\therefore \mathrm{P}\left({\mathrm{E}}_1\right)=\frac{n\left({\mathrm{E}}_1\right)}{n\left(\mathrm{S}\right)}=\frac{2}{36}=\frac{1}{18}$


(ii) Let E₂ = Event of getting sum 3 = {(1, 2), (1, 2), (2, 1), (2, 1)}
n(E₂) = 4
$\therefore \mathrm{P}\left({\mathrm{E}}_2\right)=\frac{n\left({\mathrm{E}}_2\right)}{n\left(\mathrm{S}\right)}=\frac{4}{36}=\frac{1}{9}$


(iii) Let E₃ = Event of getting sum 4 = {(2, 2), (2, 2), (3, 1), (3, 1), (1, 3), (1, 3)}
n(E₃) = 6
$\therefore \mathrm{P}\left({\mathrm{E}}_3\right)=\frac{n\left({\mathrm{E}}_3\right)}{n\left(\mathrm{S}\right)}=\frac{6}{36}=\frac{1}{6}$

(iv) Let E₄ = Event of getting sum 5 = {(2, 3), (2, 3), (4, 1), (4, 1), (3, 2), (3, 2)}
n(E₄) = 6
$\therefore \mathrm{P}\left({\mathrm{E}}_4\right)=\frac{n\left({\mathrm{E}}_4\right)}{n\left(\mathrm{S}\right)}=\frac{6}{36}=\frac{1}{6}$

(v) Let E₅ = Event of getting sum 6 = {(3, 3), (3, 3), (4, 2), (4, 2), (5, 1), (5, 1)}
n(E₅) = 6
$\therefore \mathrm{P}\left({\mathrm{E}}_5\right)=\frac{n\left({\mathrm{E}}_5\right)}{n\left(\mathrm{S}\right)}=\frac{6}{36}=\frac{1}{6}$

(vi) Let E₆ = Event of getting sum 7 = {(4, 3), (4, 3), (5, 2), (5, 2), (6, 1), (6, 1)}
n(E₆) = 6
$\therefore \mathrm{P}\left({\mathrm{E}}_6\right)=\frac{n\left({\mathrm{E}}_6\right)}{n\left(\mathrm{S}\right)}=\frac{6}{36}=\frac{1}{6}$

(vii) Let E₇ = Event of getting sum 8 = {(5, 3), (5, 3), (6, 2), (6, 2)}
n(E₇) = 4
$\therefore \mathrm{P}\left({\mathrm{E}}_7\right)=\frac{n\left({\mathrm{E}}_7\right)}{n\left(\mathrm{S}\right)}=\frac{4}{36}=\frac{1}{9}$

(viii) Let E₈ = Event of getting sum 9 = {(6, 3), (6, 3)}
n(E₈) = 2
$\therefore \mathrm{P}\left({\mathrm{E}}_8\right)=\frac{n\left({\mathrm{E}}_8\right)}{n\left(\mathrm{S}\right)}=\frac{2}{36}=\frac{1}{18}$

Answer:

The possible outcomes, if a coin is tossed 2 times are S = {(HH), (TT), (HT), (TH)}
n(S) = 4
Let E = Event of getting at most one head = {(TT), (HT), (TH)}
n(E) = 3
Hence, required probability = $\frac{n\left(\mathrm{E}\right)}{n\left(\mathrm{S}\right)}=\frac{3}{4}$.

Answer:

The possible outcomes if a coin is tossed 3 times is

S = {(HHH), (TTT), (HTT), (THT), (TTH), (THH), (HTH), (HHT)}
n(S) = 8

(i) Let E₁ = Event of getting all heads = {(HHH)}
∴ n(E₁) = 1
$\therefore \mathrm{P}\left({\mathrm{E}}_1\right)=\frac{n\left({\mathrm{E}}_1\right)}{n\left(\mathrm{S}\right)}=\frac{1}{8}$

(ii) Let E₂ = Event of getting at least 2 heads = {(HHT), (HTH), (THH), (HHH)}
∴ n(E₂) = 4
$\therefore \mathrm{P}\left({\mathrm{E}}_2\right)=\frac{n\left({\mathrm{E}}_2\right)}{n\left(\mathrm{S}\right)}=\frac{4}{8}=\frac{1}{2}$

Answer:

The total number of sample space in two dice, n(S) = $6^2=36$.
Let E be the event of getting the numbers whose difference is 2.

$\Rightarrow$ E = {(1, 3), (2, 4), (3, 5), (4, 6), (3, 1), (4, 2), (5, 3), (6, 4)}
∴ n(E) = 8
$\therefore \mathrm{P}\left(\mathrm{E}\right)=\frac{n\left(\mathrm{E}\right)}{n\left(\mathrm{S}\right)}=\frac{8}{36}=\frac{2}{9}$

Answer:

No. of red ball = 10
No. of blue ball = 5
No. of green balls = 7
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 E₁ = Event of getting a red ball
∴ n(E₁) = 10
$\therefore \mathrm{Required} \mathrm{probability}=\frac{n\left({\mathrm{E}}_1\right)}{n\left(\mathrm{S}\right)}=\frac{10}{22}=\frac{5}{11}$

(ii) Let E₂ = Event of getting a green ball n(E₂) = 7.
$\therefore \mathrm{Required} \mathrm{probability}=\frac{n\left({\mathrm{E}}_2\right)}{n\left(\mathrm{S}\right)}=\frac{7}{22}$

(iii) Let E₃ = Event getting a red ball or a green ball i.e. not a blue ball.
n(E₃) = (10 + 7) = 17

$\therefore \mathrm{Required} \mathrm{probability}=\frac{n\left({\mathrm{E}}_3\right)}{n\left(\mathrm{S}\right)}=\frac{17}{22}$

Answer:

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 E₁ = Event of getting a heart
n(E₁) = 13
$\therefore \mathrm{Required} \mathrm{probability}=\frac{n\left({\mathrm{E}}_1\right)}{n\left(\mathrm{S}\right)}=\frac{13}{49}$

(ii) Let E₂ = Event of getting a king
n(E₂) = 3                            [Since, out of 4 kings, one club cards is already removed]
$\therefore \mathrm{Required} \mathrm{probability}=\frac{n\left({\mathrm{E}}_2\right)}{n\left(\mathrm{S}\right)}=\frac{3}{49}$

Answer:

i)
Total number of cards = 52 $-$ 3 = 49
Let E₁ = Event of getting a club
n(E₁) = (13 $-$ 3) = 10
∴ Required probability = $\frac{n\left({\mathrm{E}}_1\right)}{n\left(\mathrm{S}\right)}$= $\frac{10}{49}$

(ii) Let E₂ = Event of getting 10 of hearts
n(E₂) = 1                     [In 52 playing cards only 13 are the heart cards and only one 10 in 13 heart cards]
∴ Required probability = $\frac{n\left({\mathrm{E}}_2\right)}{n\left(\mathrm{S}\right)}$ = $\frac{1}{49}$
 

Answer:

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

(i) Let E₁ = Event of getting a card whose value is 7
E₁ = Card of value 7 may be of a spade, a diamond, a club or a heart
∴ n(E₁) = 4
$\therefore \mathrm{P}\left({\mathrm{E}}_1\right)=\frac{n\left({\mathrm{E}}_1\right)}{n\left(\mathrm{S}\right)}=\frac{4}{40}=\frac{1}{10}$

(ii) Let E₂ = Event of getting a card whose value is greater than 7
= Event of getting a card whose value is 8, 9 or 10
∴ n(E₂) = 3 × 4 = 12
$\therefore \mathrm{P}\left({\mathrm{E}}_2\right)=\frac{n\left({\mathrm{E}}_2\right)}{n\left(\mathrm{S}\right)}=\frac{12}{40}=\frac{3}{10}$

(iii) Let E₃ = 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
∴ n(E₃) = 6 × 4 = 24
$\therefore \mathrm{P}\left({\mathrm{E}}_3\right)=\frac{n\left({\mathrm{E}}_3\right)}{n\left(\mathrm{S}\right)}=\frac{24}{40}=\frac{3}{5}$

Answer:

The number of integers between 0 and 100 is n(S) = 99

(i) Let E₁ = Event of choosing an integer which is divisible by 7
= Event of choosing and integer which is multiple of 7
= {7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98}
∴ n(E₁) = 14
$\therefore \mathrm{P}\left({\mathrm{E}}_1\right) = \frac{n\left({\mathrm{E}}_1\right)}{n\left(\mathrm{S}\right)}=\frac{14}{99}$

(ii) Let Event of choosing an integer which is not divisible by 7
∴ n(E₂) = n(S) $-$n(E₁)
             = 99 $-$14
             = 85
$\therefore \mathrm{P}\left({\mathrm{E}}_2\right) = \frac{n\left({\mathrm{E}}_2\right)}{n\left(\mathrm{S}\right)}=\frac{85}{99}$

Answer:

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

(i) Let E₁ = Event of selecting a card which is an even number = {2, 4, 6, …100}

In an AP, l = a + (n$-$1)d, here l = 100, a = 2 and d = 2
⇒ 100 = 2 + (n $-$ 1)2
⇒ (n $-$ 1) = 49
⇒ n = 50
n(E₁) = 50
∴ Required probability = $\frac{n\left({\mathrm{E}}_1\right)}{n\left(\mathrm{S}\right)}=\frac{50}{100}=\frac{1}{2}$


(ii) Let E₂ = Event of selecting a card which is a square number
                 = {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}
                 = n(E₂) = 9
Hence, required probability = $\frac{n\left({\mathrm{E}}_2\right)}{n\left(\mathrm{S}\right)}=\frac{9}{100}$

Answer:

We know that, in English alphabets, there are (5 vowels + 21 consonants) = 26 letters. So, total number of outcomes in English alphabets is n(S) = 26
Let E = Event of choosing an English alphabet, which is a consonant
          = {b, c, d, f, g, h, j, k, l, m, n, p, q, r, s t, v, w, x, y, z}

∴ n(E) = 21
Hence, required probability $=\frac{n\left(\mathrm{E}\right)}{n\left(\mathrm{S}\right)}=\frac{21}{26}$

Answer:

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,
n(E) = 1000 $-$ 310 = 690
P(E) = $\frac{n\left(\mathrm{E}\right)}{n\left(\mathrm{S}\right)}=\frac{690}{1000}=\frac{69}{100}$

Answer:

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 Rs 1.
∴ Required probability = $\frac{\mathrm{No}. \mathrm{of} \mathrm{slips} \mathrm{other} \mathrm{than} \mathrm{Re}. 1}{\mathrm{Total} \mathrm{no}. \mathrm{of} \mathrm{slips}}=\frac{11}{75}$

Answer:

Total number of bulbs, n(S) = 24

Let E₁ = Event of selecting not defective bulb = Event of selecting good bulbs
n(E₁) = 18

$\therefore \mathrm{P}\left(\frac{{\mathrm{E}}_1}{\mathrm{S}}\right)=\frac{18}{24}=\frac{3}{4}$

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.
∴ P(selecting second defective bulb) = $\frac{5}{23}$
 

Answer:

Total number of figures


n(S) = 8 triangles + 10 squares = 18


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


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


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


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

Answer:

Total possible outcomes of tossing a coin 3 times,

S = {(HHH), (TTT), (HTT), (THT), (TTH), (THH), (HTH), (HHT)}
∴ n(S) = 8
(i) Let E₁ = Event that Sweta losses the entry fee
= She tosses tail on three times
n(E₁) = {(T T T)}
$\mathrm{P}\left({\mathrm{E}}_1\right)=\frac{n\left({\mathrm{E}}_1\right)}{n\left(\mathrm{S}\right)}=\frac{1}{8}$


(ii) Let Event that Sweta gets double entry fee
= She tosses tail on three times = {(HHH)}
n(E₂) = 1
$\mathrm{P}\left({\mathrm{E}}_2\right)=\frac{n\left({\mathrm{E}}_2\right)}{n\left(\mathrm{S}\right)}=\frac{1}{8}$


(iii) Let event that Sweta gets her entry fee back
= Sweta gets heads one or two times
= {(HTT), (THT), (TTH), (HHT), (HTH), (THH)}
n(E₃) = 6
$\mathrm{P}\left({\mathrm{E}}_3\right)=\frac{n\left({\mathrm{E}}_3\right)}{n\left(\mathrm{S}\right)}=\frac{6}{8}=\frac{3}{4}$

Answer:

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

∴ Total sample space, n(S) = 6² = 36


(i)Number of favorable outcomes are (0, 0), (0, 1), (0, 6), (1, 0), (1, 1), (1, 6), (6, 0), (6, 1), (6, 6).
The different score which are possible are 6 scores i.e. 0, 1, 2, 6, 7 and 12.


(ii) Let E = Event of getting a 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)}
∴ n(E) = 12
P(E) = $\frac{n\left(\mathrm{E}\right)}{n\left(\mathrm{S}\right)}=\frac{12}{36}=\frac{1}{3}$

Answer:

Given, total number of mobile phones

n(S) = 48


(i) Let E₁ = Event that Varnika will buy a mobile phone
= Varnika buy only, if it is good mobile
∴ n(E₁) = 42
$\therefore P\left(E_1\right)=\frac{n\left(E_1\right)}{n\left(S\right)}=\frac{42}{48}=\frac{7}{8}$

(ii) Let E₂ = Event that trader will buy only when it has no major defects
= Trader will buy only 45 mobiles
∴ n(E₂) = 45
$\therefore P\left(E_2\right)=\frac{n\left(E_2\right)}{n\left(S\right)}=\frac{45}{48}=\frac{15}{16}$
 

Answer:

Given that,
A bag contains 24 balls of which x are red, 2x are white and 3x are blue.
Total number of balls = x + 2x + 3x =24
⇒ 6x = 24
⇒ x = 4
∴ Number of red balls = x = 4
Number of white balls = 2x = 2×4 = 8
and number of blue balls = 3x = 3×4 = 12
So, total number of outcomes for a ball is selected at random in a bag contains 24 balls.
⇒ n(S) = 24
(i) Let E₁ = Event of selecting a ball which is not red i.e.can be white or blue.
∴ n(E₁) =Number of white balls + Number of blue balls
⇒ ∴ n(E₁) = 8 + 12 = 20
$\therefore \mathrm{Required} \mathrm{Probability} = \frac{\mathrm{n}\left({\mathrm{E}}_1\right)}{\mathrm{n}\left(\mathrm{S}\right)}=\frac{20}{24}=\frac{5}{6}$

(ii) Let E₂ Event of selecting a ball which is white.
⇒ ∴ n(E₂) = 8 
So, required probability = $\frac{\mathrm{n}\left({\mathrm{E}}_2\right)}{\mathrm{n}\left(\mathrm{S}\right)}=\frac{8}{24}=\frac{1}{3}$

Answer:

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 E₁ = 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}
∴ n(E) = 9
So, required probability = $\frac{n\left(E_1\right)}{n\left(S\right)}=\frac{9}{1000}=0.009$

(ii) First, has won i.e., one card is already selected, greater than 500, has a perfect square. Since, repetition is not allowed. So, one card is removed out of 1000 cards. So, number of remaining card is 999.
∴ Total number of remaining outcomes, n(S’) = 999
Let E₂ be the event that the second player wins a prize, if the first has won.
Then, the remaining cards has a perfect square greater than 500 = 8
$\therefore n\left(E_2\right)=9-1=8$
So, required probability = $\frac{n\left(E_2\right)}{n(S\text{'})}=\frac{8}{999}$