Mix Examples - Probability Questions in English

(B) An event that cannot occur is called an impossible event.
The probability of an impossible event is always $0$.

(C) The probability of any event $E$,denoted by $P(E)$,always satisfies the condition $0 \le P(E) \le 1$.
Since $17/16 = 1.0625$,which is greater than $1$,it cannot be the probability of an event.
Therefore,the correct option is $C$.

(D) The probability of an event that is very unlikely to happen is closest to $0$.
Comparing the given options:
$0.1 > 0.01 > 0.001 > 0.0001$.
Among the given choices,$0.0001$ is the smallest value and is closest to $0$.

(A) The sum of the probability of an event and the probability of its complementary event is always $1$.
Let $E$ be an event and $E'$ be its complementary event.
Then,$P(E) + P(E') = 1$.
Given that the probability of the event is $p$,we have $P(E) = p$.
Substituting this into the equation,we get $p + P(E') = 1$.
Therefore,the probability of its complementary event is $P(E') = 1 - p$.

(B) The probability of any event $E$,denoted by $P(E)$,always lies in the range $0 \le P(E) \le 1$.
When expressed as a percentage,the probability range becomes $0\% \le P(E) \le 100\%$.
Therefore,the probability of a particular occurrence,when expressed as a percentage,can never be less than $0\%$.

(C) The probability of any event $A$,denoted by $P(A)$,is defined as the ratio of the number of favorable outcomes to the total number of possible outcomes in a sample space.
Since the number of favorable outcomes cannot be negative and cannot exceed the total number of possible outcomes,the value of $P(A)$ must satisfy the condition $0 \leq P(A) \leq 1$.

(D) standard deck of cards contains $52$ cards in total.
Face cards are defined as Kings,Queens,and Jacks.
There are $4$ suits in a deck: Hearts,Diamonds,Spades,and Clubs.
Hearts and Diamonds are red suits,while Spades and Clubs are black suits.
Each suit has $3$ face cards (King,Queen,Jack).
Therefore,the total number of red face cards is $3$ (from Hearts) $+ 3$ (from Diamonds) $= 6$.
The probability of selecting a red face card is given by the ratio of the number of favorable outcomes to the total number of possible outcomes.
Probability $= \frac{\text{Number of red face cards}}{\text{Total number of cards}} = \frac{6}{52}$.
Simplifying the fraction,we get $\frac{6}{52} = \frac{3}{26}$.

(A) non-leap year has $365$ days,which is equivalent to $52$ weeks and $1$ extra day.
This extra day can be any one of the $7$ days of the week: Sunday,Monday,Tuesday,Wednesday,Thursday,Friday,or Saturday.
For the year to have $53$ Sundays,this extra day must be a Sunday.
Since there is only $1$ favorable outcome (Sunday) out of $7$ possible outcomes,the probability is $\frac{1}{7}$.

(B) When a die is thrown,the total number of possible outcomes is $6$,which are $\{1, 2, 3, 4, 5, 6\}$.
The odd numbers on a die are $\{1, 3, 5\}$.
The odd numbers less than $3$ is only $\{1\}$.
Therefore,the number of favorable outcomes is $1$.
The probability of an event is given by the formula: $\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$.
Required probability $= \frac{1}{6}$.

(C) standard deck of cards contains $52$ cards in total.
There is only $1$ card that is the ace of hearts.
The event $E$ is defined as the card drawn is not an ace of hearts.
To find the number of outcomes favourable to $E$,we subtract the number of ace of hearts from the total number of cards.
Number of favourable outcomes = Total cards - Number of ace of hearts
Number of favourable outcomes = $52 - 1 = 51$.

(D) Given,total number of eggs $= 400$.
Probability of getting a bad egg $= 0.035$.
We know that,$\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$.
Therefore,$\frac{\text{Number of bad eggs}}{400} = 0.035$.
Number of bad eggs $= 0.035 \times 400$.
Number of bad eggs $= 14$.

(A) Given,total number of tickets sold $= 6000$.
Let the number of tickets she bought be $x$.
The probability of winning the first prize is given by the ratio of the number of tickets she bought to the total number of tickets sold.
Therefore,$\frac{x}{6000} = 0.08$.
Multiplying both sides by $6000$,we get $x = 0.08 \times 6000$.
$x = 480$.
Hence,she bought $480$ tickets.

(B) The total number of tickets in the bag is $40$,so the total number of possible outcomes is $40$.
The tickets numbered from $1$ to $40$ that are multiples of $5$ are: $5, 10, 15, 20, 25, 30, 35, 40$.
Counting these,we find there are $8$ such tickets.
Therefore,the number of favorable outcomes is $8$.
The probability of an event is given by the formula: $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$.
Substituting the values,we get: $P(E) = \frac{8}{40} = \frac{1}{5}$.

(C) The total number of possible outcomes when selecting a number from $1$ to $100$ is $100$.
The prime numbers between $1$ and $100$ are: $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97$.
Counting these,we find there are $25$ prime numbers in total.
The probability of an event is given by the formula: $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$.
Therefore,the required probability is $\frac{25}{100} = \frac{1}{4}$.

(D) Total number of students $= 23$.
Number of students in houses $A, B$,and $C = 4 + 8 + 5 = 17$.
Number of students not in houses $A, B$,or $C$ (i.e.,from houses $D$ and $E$) $= 23 - 17 = 6$.
Alternatively,students from house $D = 2$,and students from house $E = 23 - (4 + 8 + 5 + 2) = 23 - 19 = 4$. Total $= 2 + 4 = 6$.
The probability that the selected student is not from $A, B$,or $C = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{6}{23}$.

(B) The statement is False.
Probability of an outcome is $\frac{1}{2}$ only if the two outcomes are equally likely.
For example,if you toss a biased coin,the probability of getting heads or tails is not necessarily $\frac{1}{2}$ even though there are only two possible outcomes.

(B) No,the statement is incorrect because the outcomes are not equally likely.
Justification:
For a family with three children,the sample space $S$ consists of $2^3 = 8$ equally likely outcomes: $S = \{BBB, BBG, BGB, GBB, BGG, GBG, GGB, GGG\}$,where $B$ represents a boy and $G$ represents a girl.
Let $X$ be the number of girls. The possible values for $X$ are $0, 1, 2, 3$.
- $P(X=0) = P(\{BBB\}) = \frac{1}{8}$
- $P(X=1) = P(\{BBG, BGB, GBB\}) = \frac{3}{8}$
- $P(X=2) = P(\{BGG, GBG, GGB\}) = \frac{3}{8}$
- $P(X=3) = P(\{GGG\}) = \frac{1}{8}$
Since the probabilities are $\frac{1}{8}, \frac{3}{8}, \frac{3}{8}, \text{ and } \frac{1}{8}$ respectively,they are not equal to $\frac{1}{4}$.

(N/A) No,the outcomes are not equally likely.
In the given figure,the circle is divided into three regions $1, 2,$ and $3$.
The region $3$ covers half of the circle (an area of $180^{\circ}$),while regions $1$ and $2$ each cover one-fourth of the circle (an area of $90^{\circ}$ each).
Since the areas of the regions are not equal,the probability of the arrow pointing to each region is not the same.
Therefore,the outcomes $1, 2,$ and $3$ are not equally likely.

(B) Apoorv throws two dice once.
Total number of outcomes $= 6 \times 6 = 36$.
The only way to get a product of $36$ is $(6, 6)$.
Number of favorable outcomes $= 1$.
Probability for Apoorv $= \frac{1}{36}$.
Peehu throws one die.
Total number of outcomes $= 6$.
The only way to get a square of $36$ is by rolling a $6$ $(6^2 = 36)$.
Number of favorable outcomes $= 1$.
Probability for Peehu $= \frac{1}{6} = \frac{6}{36}$.
Comparing the probabilities,$\frac{6}{36} > \frac{1}{36}$.
Therefore,Peehu has a better chance of getting the number $36$.

(A) The total number of possible outcomes when tossing a coin is $2$ (Head and Tail).
Since the coin is fair,both outcomes are equally likely.
The probability of an event $E$ is given by the formula: $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$.
For Head: $P(H) = \frac{1}{2}$.
For Tail: $P(T) = \frac{1}{2}$.
Thus,the probability of each outcome is $\frac{1}{2}$ because they are equally likely events.

(B) No,this is not correct.
When a fair die is thrown,the total number of possible outcomes is $6$,which are ${1, 2, 3, 4, 5, 6}$.
The probability of getting $1$ is $P(1) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{1}{6}$.
The probability of getting 'not $1$' is $P(\text{not } 1) = 1 - P(1) = 1 - \frac{1}{6} = \frac{5}{6}$.
Since $\frac{1}{6} \neq \frac{5}{6}$,the student's claim that each probability is $\frac{1}{2}$ is incorrect.

$(D)$ When tossing three coins together,the total number of possible outcomes is $2^3 = 8$.
These outcomes are: $(HHH), (HHT), (HTH), (THH), (HTT), (THT), (TTH), (TTT)$.
Each of these $8$ outcomes is equally likely.
The event of getting 'no heads' corresponds to only one outcome: $(TTT)$.
Therefore,the probability of getting no heads is $\frac{1}{8}$.
The conclusion that the probability is $\frac{1}{4}$ is incorrect because it assumes that the four outcomes (no heads,$1$ head,$2$ heads,$3$ heads) are equally likely,which is not true.

(N/A) No,we cannot say that the probability of getting a head is $1$.
When tossing a fair coin,the two possible outcomes (head and tail) are equally likely.
The probability of getting a head in a single toss is $P(\text{Head}) = \frac{1}{2}$.
Even if a head appears $6$ times in a row,each toss is an independent event.
The probability of getting a head remains $\frac{1}{2}$ for every individual toss,regardless of the previous outcomes.

(N/A) The outcome of the next toss may or may not be a tail. This is because tossing a coin is a random experiment where the occurrence of a head or a tail are equally likely events. Each toss is an independent event,meaning the result of the previous tosses does not influence the outcome of the next toss. Therefore,the probability of getting a tail in the next toss remains $1/2$.

(N/A) No,the outcome of each coin toss is an independent event. When we toss a fair coin,the probability of getting a head or a tail is equally likely,i.e.,$P(\text{Head}) = \frac{1}{2}$ and $P(\text{Tail}) = \frac{1}{2}$. Since the coin has no memory of previous outcomes,the probability of getting a tail in the $4^{th}$ toss remains $\frac{1}{2}$,which is the same as any other toss. Therefore,there is no reason to expect a higher chance for a tail.

(A) We know that,between $1$ to $100$,half the numbers are even and half are odd. Specifically,$50$ numbers $(2, 4, 6, 8, \dots, 96, 98, 100)$ are even and $50$ numbers $(1, 3, 5, 7, \dots, 97, 99)$ are odd.
Since the total number of slips is $100$,the number of favorable outcomes for an even number is $50$,and for an odd number is $50$.
Probability of getting an even number $= \frac{\text{Number of even slips}}{\text{Total number of slips}} = \frac{50}{100} = \frac{1}{2}$.
Probability of getting an odd number $= \frac{\text{Number of odd slips}}{\text{Total number of slips}} = \frac{50}{100} = \frac{1}{2}$.
Since both events have an equal number of favorable outcomes,they are equally likely,and the probability of each is indeed $\frac{1}{2}$.

(A) When two dice are thrown at the same time,the total number of possible outcomes is $6 \times 6 = 36$.
$(i)$ Let $E_1$ be the event of getting the same number on both dice.
The favorable outcomes are $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)$.
Number of favorable outcomes $= 6$.
Probability $P(E_1) = \frac{6}{36} = \frac{1}{6}$.
$(ii)$ Let $E_2$ be the event of getting different numbers on both dice.
This is the complement of event $E_1$.
Number of favorable outcomes $= 36 - 6 = 30$.
Probability $P(E_2) = \frac{30}{36} = \frac{5}{6}$.

(A-D) When two dice are thrown simultaneously,the total number of possible outcomes is $6 \times 6 = 36$.
$(i)$ Sum of the numbers appearing on the dice is $7$. The favorable outcomes are $(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)$.
Number of favorable outcomes $= 6$.
$\text{Probability} = \frac{6}{36} = \frac{1}{6}$.
$(ii)$ Sum of the numbers appearing on the dice is a prime number. The possible prime sums are $2, 3, 5, 7, 11$.
- Sum $= 2: (1, 1) \rightarrow 1$ outcome
- Sum $= 3: (1, 2), (2, 1) \rightarrow 2$ outcomes
- Sum $= 5: (1, 4), (2, 3), (3, 2), (4, 1) \rightarrow 4$ outcomes
- Sum $= 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) \rightarrow 6$ outcomes
- Sum $= 11: (5, 6), (6, 5) \rightarrow 2$ outcomes
Total favorable outcomes $= 1 + 2 + 4 + 6 + 2 = 15$.
$\text{Probability} = \frac{15}{36} = \frac{5}{12}$.
$(iii)$ Sum of the numbers appearing on the dice is $1$. Since the minimum sum of two dice is $1 + 1 = 2$,it is impossible to get a sum of $1$.
Number of favorable outcomes $= 0$.
$\text{Probability} = \frac{0}{36} = 0$.

(N/A) The total number of outcomes when two dice are thrown is $6 \times 6 = 36$.
$(i)$ For the product to be $6$,the possible outcomes are $(1, 6), (2, 3), (3, 2), (6, 1)$.
Number of favorable outcomes $= 4$.
Probability $= \frac{4}{36} = \frac{1}{9}$.
$(ii)$ For the product to be $12$,the possible outcomes are $(2, 6), (3, 4), (4, 3), (6, 2)$.
Number of favorable outcomes $= 4$.
Probability $= \frac{4}{36} = \frac{1}{9}$.
$(iii)$ For the product to be $7$,there are no possible outcomes since $7$ is a prime number and cannot be formed by the product of two numbers from $1$ to $6$ (except $1 \times 7$,but $7$ is not on a die).
Number of favorable outcomes $= 0$.
Probability $= \frac{0}{36} = 0$.

(C) When two dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$.
We need to find the outcomes where the product of the two numbers is less than $9$.
The favorable outcomes are:
For $1$ on the first die: $(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)$ (Products: $1, 2, 3, 4, 5, 6$)
For $2$ on the first die: $(2,1), (2,2), (2,3), (2,4)$ (Products: $2, 4, 6, 8$)
For $3$ on the first die: $(3,1), (3,2)$ (Products: $3, 6$)
For $4$ on the first die: $(4,1), (4,2)$ (Products: $4, 8$)
For $5$ on the first die: $(5,1)$ (Product: $5$)
For $6$ on the first die: $(6,1)$ (Product: $6$)
Counting these,we have $6 + 4 + 2 + 2 + 1 + 1 = 16$ favorable outcomes.
The probability is $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{16}{36} = \frac{4}{9}$.

(N/A) Total number of outcomes $= 6 \times 6 = 36$.
$(i)$ Event $E_1$ (sum $= 2$): Outcomes are $\{(1, 1), (1, 1)\}$. $n(E_1) = 2$. $P(E_1) = \frac{2}{36} = \frac{1}{18}$.
$(ii)$ Event $E_2$ (sum $= 3$): Outcomes are $\{(1, 2), (1, 2), (2, 1), (2, 1)\}$. $n(E_2) = 4$. $P(E_2) = \frac{4}{36} = \frac{1}{9}$.
$(iii)$ Event $E_3$ (sum $= 4$): Outcomes are $\{(1, 3), (1, 3), (2, 2), (2, 2), (3, 1), (3, 1)\}$. $n(E_3) = 6$. $P(E_3) = \frac{6}{36} = \frac{1}{6}$.
$(iv)$ Event $E_4$ (sum $= 5$): Outcomes are $\{(2, 3), (2, 3), (3, 2), (3, 2), (4, 1), (4, 1)\}$. $n(E_4) = 6$. $P(E_4) = \frac{6}{36} = \frac{1}{6}$.
$(v)$ Event $E_5$ (sum $= 6$): Outcomes are $\{(3, 3), (3, 3), (4, 2), (4, 2), (5, 1), (5, 1)\}$. $n(E_5) = 6$. $P(E_5) = \frac{6}{36} = \frac{1}{6}$.
$(vi)$ Event $E_6$ (sum $= 7$): Outcomes are $\{(4, 3), (4, 3), (5, 2), (5, 2), (6, 1), (6, 1)\}$. $n(E_6) = 6$. $P(E_6) = \frac{6}{36} = \frac{1}{6}$.
$(vii)$ Event $E_7$ (sum $= 8$): Outcomes are $\{(5, 3), (5, 3), (6, 2), (6, 2)\}$. $n(E_7) = 4$. $P(E_7) = \frac{4}{36} = \frac{1}{9}$.
$(viii)$ Event $E_8$ (sum $= 9$): Outcomes are $\{(6, 3), (6, 3)\}$. $n(E_8) = 2$. $P(E_8) = \frac{2}{36} = \frac{1}{18}$.

(A) The possible outcomes when a coin is tossed $2$ times are:
$S = \{(H, H), (H, T), (T, H), (T, T)\}$
Therefore,the total number of outcomes $n(S) = 4$.
Let $E$ be the event of getting at most one head.
This means getting zero heads or one head.
The outcomes favorable to event $E$ are $\{(T, T), (H, T), (T, H)\}$.
Therefore,the number of favorable outcomes $n(E) = 3$.
The probability of the event $E$ is given by $P(E) = \frac{n(E)}{n(S)}$.
Thus,$P(E) = \frac{3}{4}$.

(B) The sample space $S$ for tossing a coin $3$ times is:
$S = \{(HHH), (HHT), (HTH), (HTT), (THH), (THT), (TTH), (TTT)\}$
Total number of outcomes $n(S) = 8$.
$(i)$ Let $E_1$ be the event of getting all heads.
$E_1 = \{(HHH)\}$
Number of favorable outcomes $n(E_1) = 1$.
Probability $P(E_1) = \frac{n(E_1)}{n(S)} = \frac{1}{8}$.
$(ii)$ Let $E_2$ be the event of getting at least $2$ heads.
$E_2 = \{(HHH), (HHT), (HTH), (THH)\}$
Number of favorable outcomes $n(E_2) = 4$.
Probability $P(E_2) = \frac{n(E_2)}{n(S)} = \frac{4}{8} = \frac{1}{2}$.

(C) The total number of outcomes in the sample space when two dice are thrown is $n(S) = 6 \times 6 = 36$.
Let $E$ be the event of getting numbers whose difference is $2$.
The favorable outcomes are:
$E = \{(1, 3), (2, 4), (3, 5), (4, 6), (3, 1), (4, 2), (5, 3), (6, 4)\}$.
Therefore,the number of favorable outcomes is $n(E) = 8$.
The probability of the event $E$ is given by $P(E) = \frac{n(E)}{n(S)}$.
$P(E) = \frac{8}{36} = \frac{2}{9}$.

(N/A) The total number of balls in the bag is $10 + 5 + 7 = 22$.
Thus,the total number of possible outcomes is $n(S) = 22$.
$(i)$ Let $E_1$ be the event of drawing a red ball. The number of red balls is $n(E_1) = 10$.
Therefore,the probability $P(E_1) = \frac{n(E_1)}{n(S)} = \frac{10}{22} = \frac{5}{11}$.
$(ii)$ Let $E_2$ be the event of drawing a green ball. The number of green balls is $n(E_2) = 7$.
Therefore,the probability $P(E_2) = \frac{n(E_2)}{n(S)} = \frac{7}{22}$.
$(iii)$ Let $E_3$ be the event of not drawing a blue ball. This means drawing either a red or a green ball.
The number of favorable outcomes is $n(E_3) = 10 + 7 = 17$.
Therefore,the probability $P(E_3) = \frac{n(E_3)}{n(S)} = \frac{17}{22}$.

(A) Total number of cards in a deck = $52$.
After removing the king,queen,and jack of clubs,the number of remaining cards is $n(S) = 52 - 3 = 49$.
$(i)$ Let $E_1$ be the event of getting a heart.
Since there are $13$ hearts in a deck and none were removed,$n(E_1) = 13$.
Therefore,the probability $P(E_1) = \frac{n(E_1)}{n(S)} = \frac{13}{49}$.
$(ii)$ Let $E_2$ be the event of getting a king.
There are $4$ kings in a deck. Since the king of clubs was removed,the number of remaining kings is $n(E_2) = 4 - 1 = 3$.
Therefore,the probability $P(E_2) = \frac{n(E_2)}{n(S)} = \frac{3}{49}$.

(A) $(i)$ Let $E_1$ be the event of getting a club.
Total cards remaining $= 52 - 3 = 49$.
Number of clubs remaining $= 13 - 3 = 10$.
$\therefore$ Probability $= \frac{10}{49}$.
$(ii)$ Let $E_2$ be the event of getting $10$ of hearts.
There is only one $10$ of hearts in the deck.
$\therefore$ Probability $= \frac{1}{49}$.

(N/A) Total cards in a deck = $52$.
Number of jacks = $4$,queens = $4$,kings = $4$. Total removed = $4 + 4 + 4 = 12$.
Remaining cards $n(S) = 52 - 12 = 40$.
$(i)$ Let $E_1$ be the event of getting a card with value $7$. There are $4$ such cards (one of each suit).
$P(E_1) = \frac{n(E_1)}{n(S)} = \frac{4}{40} = \frac{1}{10}$.
$(ii)$ Let $E_2$ be the event of getting a card with a value greater than $7$. The possible values are $8, 9, 10$. There are $3$ values,each with $4$ suits,so $n(E_2) = 3 \times 4 = 12$.
$P(E_2) = \frac{n(E_2)}{n(S)} = \frac{12}{40} = \frac{3}{10}$.
$(iii)$ Let $E_3$ be the event of getting a card with a value less than $7$. The possible values are $1, 2, 3, 4, 5, 6$. There are $6$ values,each with $4$ suits,so $n(E_3) = 6 \times 4 = 24$.
$P(E_3) = \frac{n(E_3)}{n(S)} = \frac{24}{40} = \frac{3}{5}$.

(D) The total number of integers between $0$ and $100$ is $99$ (i.e.,${1, 2, 3, \dots, 99}$).
$n(S) = 99$
$(i)$ Let $E_1$ be the event of choosing an integer which is divisible by $7$.
The multiples of $7$ between $0$ and $100$ are ${7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98}$.
$n(E_1) = 14$
$P(E_1) = \frac{n(E_1)}{n(S)} = \frac{14}{99}$
$(ii)$ Let $E_2$ be the event of choosing an integer which is not divisible by $7$.
$n(E_2) = n(S) - n(E_1) = 99 - 14 = 85$
$P(E_2) = \frac{n(E_2)}{n(S)} = \frac{85}{99}$

(A) Total number of outcomes with numbers $2$ to $101$ is $n(S) = 101 - 2 + 1 = 100$.
$(i)$ Let $E_1$ be the event of selecting a card which is an even number. The even numbers are $\{2, 4, 6, \dots, 100\}$.
This is an arithmetic progression where $a = 2$,$d = 2$,and $l = 100$.
Using $l = a + (n - 1)d$,we get $100 = 2 + (n - 1)2$,which implies $98 = 2(n - 1)$,so $n - 1 = 49$,and $n = 50$.
Thus,$n(E_1) = 50$.
The probability $P(E_1) = \frac{n(E_1)}{n(S)} = \frac{50}{100} = \frac{1}{2}$.
$(ii)$ Let $E_2$ be the event of selecting a card which is a square number.
The square numbers between $2$ and $101$ are $\{4, 9, 16, 25, 36, 49, 64, 81, 100\}$.
These correspond to $\{2^2, 3^2, 4^2, 5^2, 6^2, 7^2, 8^2, 9^2, 10^2\}$.
Thus,$n(E_2) = 9$.
The probability $P(E_2) = \frac{n(E_2)}{n(S)} = \frac{9}{100}$.

(B) We know that in the English alphabet,there are $5$ vowels and $21$ consonants,making a total of $26$ letters.
The total number of possible outcomes is $n(S) = 26$.
Let $E$ be the event of choosing a consonant from the English alphabet.
The set of consonants is ${b, c, d, f, g, h, j, k, l, m, n, p, q, r, s, t, v, w, x, y, z}$.
Therefore,the number of favorable outcomes is $n(E) = 21$.
The probability of an event is given by $P(E) = \frac{n(E)}{n(S)}$.
Thus,the required probability is $\frac{21}{26}$.

(C) Total number of sealed envelopes in a box,$n(S) = 1000$.
Number of envelopes containing a cash prize $= 10 + 100 + 200 = 310$.
Number of envelopes containing no cash prize,$n(E) = 1000 - 310 = 690$.
The probability $P(E)$ that the envelope contains no cash prize is given by the ratio of the number of favorable outcomes to the total number of outcomes.
$P(E) = \frac{n(E)}{n(S)} = \frac{690}{1000} = 0.69$.

(D) Total number of slips in the third box,$n(S) = 25 + 50 = 75$.
From the given information:
Number of slips marked $Rs. 1$ in Box $A = 19$.
Number of slips marked $Rs. 5$ in Box $A = 25 - 19 = 6$.
Number of slips marked $Rs. 1$ in Box $B = 45$.
Number of slips marked $Rs. 13$ in Box $B = 50 - 45 = 5$.
Total number of slips marked other than $Rs. 1 = 6 + 5 = 11$.
Therefore,the required probability $= \frac{\text{Number of slips marked other than } Rs. 1}{\text{Total number of slips}} = \frac{11}{75}$.

(N/A) Total number of bulbs, $n(S) = 24$.
Number of defective bulbs $= 6$.
Number of non-defective (good) bulbs $= 24 - 6 = 18$.
Part $1$: Probability that the bulb is not defective:
$P(\text{not defective}) = \frac{\text{Number of good bulbs}}{\text{Total number of bulbs}} = \frac{18}{24} = \frac{3}{4}$.
Part $2$: If the first bulb selected is defective and not replaced:
Remaining total number of bulbs $= 24 - 1 = 23$.
Remaining number of defective bulbs $= 6 - 1 = 5$.
Probability that the second bulb is defective $= \frac{\text{Remaining defective bulbs}}{\text{Remaining total bulbs}} = \frac{5}{23}$.

(N/A) Total number of pieces $= 8 \text{ triangles} + 10 \text{ squares} = 18$.
$(i)$ Probability that the lost piece is a triangle $= \frac{8}{18} = \frac{4}{9}$.
$(ii)$ Probability that the lost piece is a square $= \frac{10}{18} = \frac{5}{9}$.
$(iii)$ Number of blue squares $= 6$. Probability that the lost piece is a blue square $= \frac{6}{18} = \frac{1}{3}$.
$(iv)$ Number of red triangles $= 8 - 3 = 5$. Probability that the lost piece is a red triangle $= \frac{5}{18}$.

(A) Total possible outcomes of tossing a coin $3$ times:
$S = \{(HHH), (TTT), (HTT), (THT), (TTH), (THH), (HTH), (HHT)\}$
$\therefore n(S) = 8$
$(i)$ Let $E_1$ be the event that Sweta loses the entry fee.
This happens if she gets zero heads (i.e.,$TTT$).
$E_1 = \{(TTT)\}$,so $n(E_1) = 1$.
$P(E_1) = \frac{n(E_1)}{n(S)} = \frac{1}{8}$.
$(ii)$ Let $E_2$ be the event that Sweta gets double the entry fee.
This happens if she gets $3$ heads (i.e.,$HHH$).
$E_2 = \{(HHH)\}$,so $n(E_2) = 1$.
$P(E_2) = \frac{n(E_2)}{n(S)} = \frac{1}{8}$.
$(iii)$ Let $E_3$ be the event that Sweta gets her entry fee back.
This happens if she gets one or two heads.
$E_3 = \{(HTT), (THT), (TTH), (HHT), (HTH), (THH)\}$,so $n(E_3) = 6$.
$P(E_3) = \frac{n(E_3)}{n(S)} = \frac{6}{8} = \frac{3}{4}$.

(C) Given,a die has its six faces marked ${0, 1, 1, 1, 6, 6}$.
Total sample space,$n(S) = 6^2 = 36$.
$(i)$ The possible sums are obtained by adding the values on the two dice:
$0+0=0, 0+1=1, 0+6=6, 1+1=2, 1+6=7, 6+6=12$.
Thus,the different scores possible are ${0, 1, 2, 6, 7, 12}$. There are $6$ possible scores.
$(ii)$ Let $E$ be the event of getting a sum of $7$.
The pairs $(d_1, d_2)$ that result in a sum of $7$ are:
$(1, 6)$ occurs $3 \times 2 = 6$ times.
$(6, 1)$ occurs $2 \times 3 = 6$ times.
Total favorable outcomes $n(E) = 6 + 6 = 12$.
$P(E) = \frac{n(E)}{n(S)} = \frac{12}{36} = \frac{1}{3}$.

(N/A) Given,total number of mobile phones $n(S) = 48$.
$(i)$ Let $E_{1}$ be the event that Varnika will buy a mobile phone.
Varnika buys only if it is a good mobile.
Therefore,$n(E_{1}) = 42$.
$P(E_{1}) = \frac{n(E_{1})}{n(S)} = \frac{42}{48} = \frac{7}{8}$.
$(ii)$ Let $E_{2}$ be the event that the trader will buy a mobile phone.
The trader will buy only if it has no major defects.
This means the trader will buy the $42$ good phones and the $3$ phones with minor defects.
Therefore,$n(E_{2}) = 42 + 3 = 45$.
$P(E_{2}) = \frac{n(E_{2})}{n(S)} = \frac{45}{48} = \frac{15}{16}$.

(A) Given that,the total number of balls in the bag $= 24$.
Number of red balls $= x$,number of white balls $= 2x$,and number of blue balls $= 3x$.
According to the condition,$x + 2x + 3x = 24$.
$6x = 24$,which gives $x = 4$.
Therefore,the number of red balls $= 4$,the number of white balls $= 2 \times 4 = 8$,and the number of blue balls $= 3 \times 4 = 12$.
The total number of outcomes $n(S) = 24$.
$(i)$ Let $E_1$ be the event of selecting a ball which is not red. This means the ball can be white or blue.
$n(E_1) = \text{Number of white balls} + \text{Number of blue balls} = 8 + 12 = 20$.
Required probability $P(E_1) = \frac{n(E_1)}{n(S)} = \frac{20}{24} = \frac{5}{6}$.
$(ii)$ Let $E_2$ be the event of selecting a ball which is white.
$n(E_2) = \text{Number of white balls} = 8$.
Required probability $P(E_2) = \frac{n(E_2)}{n(S)} = \frac{8}{24} = \frac{1}{3}$.

(A) Total number of cards in the box is $n(S) = 1000$.
$(i)$ Let $E_1$ be the event that the first player wins a prize. This happens if the player selects a perfect square greater than $500$.
The perfect squares between $1$ and $1000$ are $1^2, 2^2, \dots, 31^2 = 961$.
The perfect squares greater than $500$ are $23^2=529, 24^2=576, 25^2=625, 26^2=676, 27^2=729, 28^2=784, 29^2=841, 30^2=900, 31^2=961$.
There are $9$ such cards.
Thus, $n(E_1) = 9$.
The probability that the first player wins is $P(E_1) = \frac{n(E_1)}{n(S)} = \frac{9}{1000} = 0.009$.
$(ii)$ If the first player has won, one card (which is a perfect square $> 500$) is removed from the box.
The remaining number of cards is $n(S') = 1000 - 1 = 999$.
The number of remaining perfect squares greater than $500$ is $n(E_2) = 9 - 1 = 8$.
The probability that the second player wins given the first has won is $P(E_2|E_1) = \frac{n(E_2)}{n(S')} = \frac{8}{999}$.