Theoretical Probability Questions in English

(N/A) In the experiment of tossing a coin once, the total number of possible outcomes is $2$ $-$ Head $(H)$ and Tail $(T).$
Let $E$ be the event of 'getting a head'. The number of outcomes favourable to $E$ is $1.$
Therefore, $P(E) = P(\text{head}) = \frac{\text{Number of outcomes favourable to } E}{\text{Total number of possible outcomes}} = \frac{1}{2}.$
Similarly, if $F$ is the event of 'getting a tail', the number of outcomes favourable to $F$ is $1.$
Therefore, $P(F) = P(\text{tail}) = \frac{1}{2}.$

(N/A) Kritika takes out a ball from the bag without looking into it. Since all balls are of the same size,it is equally likely that she takes out any one of them.
Let $Y$ be the event 'the ball taken out is yellow',$B$ be the event 'the ball taken out is blue',and $R$ be the event 'the ball taken out is red'.
The total number of possible outcomes $= 3$.
$(i)$ The number of outcomes favourable to the event $Y = 1$.
So,$P(Y) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} = \frac{1}{3}$.
$(ii)$ The number of outcomes favourable to the event $R = 1$.
So,$P(R) = \frac{1}{3}$.
$(iii)$ The number of outcomes favourable to the event $B = 1$.
So,$P(B) = \frac{1}{3}$.

(N/A) $(i)$ Here,let $E$ be the event 'getting a number greater than $4$'. The total number of possible outcomes is $6$ $(1, 2, 3, 4, 5, 6)$. The outcomes favourable to $E$ are $5$ and $6$. Therefore,the number of outcomes favourable to $E$ is $2$.
$P(E) = P(\text{number greater than } 4) = \frac{2}{6} = \frac{1}{3}$.
$(ii)$ Let $F$ be the event 'getting a number less than or equal to $4$'.
The total number of possible outcomes is $6$.
The outcomes favourable to the event $F$ are $1, 2, 3, 4$.
Therefore,the number of outcomes favourable to $F$ is $4$.
$P(F) = P(\text{number less than or equal to } 4) = \frac{4}{6} = \frac{2}{3}$.

(A) Well-shuffling ensures equally likely outcomes.
$(i)$ There are $4$ aces in a deck. Let $E$ be the event 'the card is an ace'.
The number of outcomes favourable to $E = 4$.
The number of possible outcomes $= 52$.
Therefore,$P(E) = \frac{4}{52} = \frac{1}{13}$.
$(ii)$ Let $F$ be the event 'card drawn is not an ace'.
The number of outcomes favourable to the event $F = 52 - 4 = 48$.
The number of possible outcomes $= 52$.
Therefore,$P(F) = \frac{48}{52} = \frac{12}{13}$.

(A) Total number of students $= 40$.
$(i)$ Number of girls $= 25$.
Probability of selecting a girl $= \frac{\text{Number of girls}}{\text{Total number of students}} = \frac{25}{40} = \frac{5}{8}$.
$(ii)$ Number of boys $= 15$.
Probability of selecting a boy $= \frac{\text{Number of boys}}{\text{Total number of students}} = \frac{15}{40} = \frac{3}{8}$.

(A) The statement that a marble is drawn at random implies that all marbles are equally likely to be drawn.
Total number of possible outcomes $= 3 + 2 + 4 = 9$.
Let $W$ denote the event 'the marble is white',$B$ denote the event 'the marble is blue',and $R$ denote the event 'the marble is red'.
$(i)$ The number of outcomes favourable to the event $W$ is $2$.
So,$P(W) = \frac{2}{9}$.
$(ii)$ The number of outcomes favourable to the event $B$ is $3$.
So,$P(B) = \frac{3}{9} = \frac{1}{3}$.
$(iii)$ The number of outcomes favourable to the event $R$ is $4$.
So,$P(R) = \frac{4}{9}$.
Note that $P(W) + P(B) + P(R) = \frac{2}{9} + \frac{3}{9} + \frac{4}{9} = \frac{9}{9} = 1$.

(C) We write $H$ for 'head' and $T$ for 'tail'. When two coins are tossed simultaneously,the possible outcomes are $(H, H), (H, T), (T, H), (T, T)$,which are all equally likely.
Here $(H, H)$ means head up on the first coin (say on $Rs. 1$) and head up on the second coin (say on $Rs. 2$). Similarly,$(H, T)$ means head up on the first coin and tail up on the second coin,and so on.
The outcomes favourable to the event $E$,'at least one head',are $(H, H), (H, T)$,and $(T, H)$.
So,the number of outcomes favourable to $E$ is $3$.
The total number of possible outcomes is $4$.
Therefore,$P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} = \frac{3}{4}$.
i.e.,the probability that Harpreet gets at least one head is $3/4$.

(1/4) Here,the possible outcomes are all the numbers between $0$ and $2$. This is the portion of the number line from $0$ to $2$.
Let $E$ be the event that 'the music is stopped within the first half-minute'.
The outcomes favourable to $E$ are points on the number line from $0$ to $\frac{1}{2}$.
The total distance from $0$ to $2$ is $2$,while the distance favourable to $E$ from $0$ to $\frac{1}{2}$ is $\frac{1}{2}$.
Since all the outcomes are equally likely,we can argue that,of the total distance of $2$,the distance favourable to the event $E$ is $\frac{1}{2}$.
So,$P(E) = \frac{\text{Distance favourable to the event } E}{\text{Total distance in which outcomes can lie}} = \frac{\frac{1}{2}}{2} = \frac{1}{4}$.

(D) The helicopter is equally likely to crash anywhere in the region.
Total area of the rectangular region $= (9 \times 4.5) \, km^2 = 40.5 \, km^2$.
The lake is a rectangle with length $= (9 - 6) \, km = 3 \, km$ and width $= (4.5 - 2) \, km = 2.5 \, km$.
Area of the lake $= (3 \times 2.5) \, km^2 = 7.5 \, km^2$.
The probability that the helicopter crashed inside the lake is given by the ratio of the area of the lake to the total area.
$P(\text{helicopter crashed in the lake}) = \frac{\text{Area of the lake}}{\text{Total area}} = \frac{7.5}{40.5} = \frac{75}{405} = \frac{5}{27}$.

(A) Total number of shirts in the carton $= 100$.
$(i)$ Jimmy accepts only good shirts. The number of good shirts $= 88$.
Therefore,the probability that the shirt is acceptable to Jimmy $= \frac{88}{100} = 0.88$.
$(ii)$ Sujatha rejects only shirts with major defects. This means she accepts shirts that are good and shirts that have minor defects.
Number of shirts acceptable to Sujatha $= 88 + 8 = 96$.
Therefore,the probability that the shirt is acceptable to Sujatha $= \frac{96}{100} = 0.96$.

(N/A) When the blue die shows $1$,the grey die could show any one of the numbers $1, 2, 3, 4, 5, 6$. The same is true when the blue die shows $2, 3, 4, 5,$ or $6$. The possible outcomes of the experiment are listed in the table below; the first number in each ordered pair is the number appearing on the blue die and the second number is that on the grey die.
Note that the pair $(1, 4)$ is different from $(4, 1)$.
So,the total number of possible outcomes $= 6 \times 6 = 36$.
$(i)$ The outcomes favourable to the event "the sum of the two numbers is $8$" denoted by $E$,are: $(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)$.
i.e.,the number of outcomes favourable to $E = 5$.
Hence,$P(E) = \frac{5}{36}$.
$(ii)$ As you can see,there is no outcome favourable to the event $F$,"the sum of two numbers is $13$".
So,$P(F) = \frac{0}{36} = 0$.
$(iii)$ As you can see,all the outcomes are favourable to the event $G$,"sum of two numbers $\leq 12$".
So,$P(G) = \frac{36}{36} = 1$.

(N/A) $(i)$ The sum of the probabilities of all elementary events of an experiment is $1$.
$(ii)$ The probability of an event $E$ is such that $0 \le P(E) \le 1$. Therefore,the probability of an event is greater than or equal to $0$ and less than or equal to $1$.

(III, IV) $(i)$ This is not an equally likely event because the outcome depends on the mechanical condition of the car. The probability of the car starting is not necessarily equal to the probability of it not starting.
$(ii)$ This is not an equally likely event because the outcome depends on the player's skill and practice,which are not uniform for every attempt.
$(iii)$ This is an equally likely event because there are only two possible outcomes (right or wrong),and assuming no prior knowledge,each has an equal chance of occurring.
$(iv)$ This is an equally likely event because the biological probability of a baby being a boy or a girl is considered to be $50\%$ each,assuming no other external factors.

(B) When we toss a coin,the possible outcomes are only two,head or tail,which are equally likely outcomes.
Since the probability of getting a head is $1/2$ and the probability of getting a tail is $1/2$,the result of an individual toss is completely unpredictable and unbiased.
Therefore,tossing a coin is considered a fair way to decide which team gets the ball.

(D) The probability of an event $E$,denoted by $P(E)$,always lies in the range $0 \le P(E) \le 1$.
This means that the probability of an event can never be negative and cannot exceed $1$.
Comparing the given options:
$(A)$ $2/3 \approx 0.66$,which is between $0$ and $1$.
$(B)$ $0.7$ is between $0$ and $1$.
$(C)$ $15 \% = 0.15$,which is between $0$ and $1$.
$(D)$ $-1.5$ is less than $0$.
Therefore,$-1.5$ cannot be the probability of an event.

(A) $(i)$ The bag contains lemon flavoured candies only. It does not contain any orange flavoured candies. This implies that every time,she will take out only lemon flavoured candies. Therefore,the event that Malini will take out an orange flavoured candy is an impossible event.
Hence,$P(\text{an orange flavoured candy}) = 0$.
$(ii)$ As the bag has lemon flavoured candies,Malini will take out only lemon flavoured candies. Therefore,the event that Malini will take out a lemon flavoured candy is a sure event.
Hence,$P(\text{a lemon flavoured candy}) = 1$.

(A) Total number of coins in the piggy bank $= 100 + 50 + 20 + 10 = 180$.
$(i)$ Number of $50\, p$ coins $= 100$.
Probability of getting a $50\, p$ coin $= \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{100}{180} = \frac{5}{9}$.
$(ii)$ Number of $Rs.\, 5$ coins $= 10$.
Probability of getting a $Rs.\, 5$ coin $= \frac{10}{180} = \frac{1}{18}$.
Probability of not getting a $Rs.\, 5$ coin $= 1 - P(\text{getting a } Rs.\, 5 \text{ coin}) = 1 - \frac{1}{18} = \frac{17}{18}$.

(C) Total number of fish in the tank $=$ Number of male fish $+$ Number of female fish
$= 5 + 8 = 13$
Probability of getting a male fish $= \frac{\text{Number of favourable outcomes}}{\text{Number of total possible outcomes}}$
$= \frac{5}{13}$

(N/A) The total number of possible outcomes is $8$ (i.e.,$1, 2, 3, 4, 5, 6, 7, 8$).
$(i)$ The number of favourable outcomes for getting $8$ is $1$.
Probability $= \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} = \frac{1}{8}$.
$(ii)$ The odd numbers are $1, 3, 5, 7$. The number of favourable outcomes is $4$.
Probability $= \frac{4}{8} = \frac{1}{2}$.
$(iii)$ The numbers greater than $2$ are $3, 4, 5, 6, 7, 8$. The number of favourable outcomes is $6$.
Probability $= \frac{6}{8} = \frac{3}{4}$.
$(iv)$ All numbers $1, 2, 3, 4, 5, 6, 7, 8$ are less than $9$. The number of favourable outcomes is $8$.
Probability $= \frac{8}{8} = 1$.

(N/A) The set of all possible outcomes when a fair die is thrown is $S = \{1, 2, 3, 4, 5, 6\}$.
The total number of possible outcomes is $n(S) = 6$.
$(i)$ The prime numbers on a die are $\{2, 3, 5\}$.
The number of favorable outcomes is $3$.
Probability $P(\text{prime}) = \frac{3}{6} = \frac{1}{2}$.
$(ii)$ The numbers lying between $2$ and $6$ are $\{3, 4, 5\}$.
The number of favorable outcomes is $3$.
Probability $P(\text{between } 2 \text{ and } 6) = \frac{3}{6} = \frac{1}{2}$.
$(iii)$ The odd numbers on a die are $\{1, 3, 5\}$.
The number of favorable outcomes is $3$.
Probability $P(\text{odd}) = \frac{3}{6} = \frac{1}{2}$.

(A-D) Total number of cards in a well-shuffled deck $= 52$.
$(i)$ Total number of kings of red colour $= 2$ (King of Hearts and King of Diamonds).
$P(\text{king of red colour}) = \frac{2}{52} = \frac{1}{26}$.
$(ii)$ Total number of face cards (Kings,Queens,and Jacks) $= 3 \times 4 = 12$.
$P(\text{face card}) = \frac{12}{52} = \frac{3}{13}$.
$(iii)$ Total number of red face cards (Hearts and Diamonds) $= 3 + 3 = 6$.
$P(\text{red face card}) = \frac{6}{52} = \frac{3}{26}$.
$(iv)$ Total number of Jack of hearts $= 1$.
$P(\text{Jack of hearts}) = \frac{1}{52}$.
$(v)$ Total number of spade cards $= 13$.
$P(\text{spade}) = \frac{13}{52} = \frac{1}{4}$.
$(vi)$ Total number of queen of diamonds $= 1$.
$P(\text{queen of diamonds}) = \frac{1}{52}$.

(A) $(i)$ Total number of cards $= 5$.
Total number of queens $= 1$.
$P(\text{getting a queen}) = \frac{\text{Number of favourable outcomes}}{\text{Number of total possible outcomes}} = \frac{1}{5}$.
$(ii)$ When the queen is drawn and put aside,the total number of remaining cards will be $4$.
$(a)$ Total number of aces $= 1$.
$P(\text{getting an ace}) = \frac{1}{4}$.
$(b)$ As the queen is already drawn,the number of remaining queens is $0$.
$P(\text{getting a queen}) = \frac{0}{4} = 0$.

(A) Total number of pens $= 12 + 132 = 144$.
Total number of good pens $= 132$.
The probability $P$ of an event is given by the formula:
$P(\text{event}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$.
Therefore, the probability of getting a good pen is:
$P(\text{good pen}) = \frac{132}{144}$.
Dividing both the numerator and the denominator by $12$, we get:
$P(\text{good pen}) = \frac{11}{12}$.

(N/A) The total number of possible outcomes on the die is $6$ (since there are $6$ faces).
$(i)$ The number of faces having the letter $A$ is $2$.
Therefore,the probability of getting $A$ is $P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{2}{6} = \frac{1}{3}$.
(ii) The number of faces having the letter $D$ is $1$.
Therefore,the probability of getting $D$ is $P(D) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{1}{6}$.

(N/A) The area of the rectangular region is given by $Area = \text{length} \times \text{breadth} = 3 \, m \times 2 \, m = 6 \, m^2$.
The diameter of the circle is $1 \, m$, so its radius $r = \frac{1}{2} \, m$.
The area of the circle is given by $Area = \pi r^2 = \pi \left(\frac{1}{2}\right)^2 = \frac{\pi}{4} \, m^2$.
The probability that the die will land inside the circle is the ratio of the area of the circle to the total area of the rectangle:
$P(\text{landing inside the circle}) = \frac{\text{Area of circle}}{\text{Area of rectangle}} = \frac{\frac{\pi}{4}}{6} = \frac{\pi}{24}$.

(A) Total number of pens $= 144$
Total number of defective pens $= 20$
Total number of good pens $= 144 - 20 = 124$
$(i)$ Probability of getting a good pen (Nuri buys the pen) $= \frac{124}{144} = \frac{31}{36}$
$(ii)$ Probability of getting a defective pen (Nuri will not buy the pen) $= \frac{20}{144} = \frac{5}{36}$ or $1 - \frac{31}{36} = \frac{5}{36}$

(N/A) $(i)$ When two dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$. The sums and their corresponding probabilities are calculated as follows:
- Sum $= 2$: $(1,1) \rightarrow \frac{1}{36}$
- Sum $= 3$: $(1,2), (2,1) \rightarrow \frac{2}{36}$
- Sum $= 4$: $(1,3), (3,1), (2,2) \rightarrow \frac{3}{36}$
- Sum $= 5$: $(1,4), (4,1), (2,3), (3,2) \rightarrow \frac{4}{36}$
- Sum $= 6$: $(1,5), (5,1), (2,4), (4,2), (3,3) \rightarrow \frac{5}{36}$
- Sum $= 7$: $(1,6), (6,1), (2,5), (5,2), (3,4), (4,3) \rightarrow \frac{6}{36}$
- Sum $= 8$: $(2,6), (6,2), (3,5), (5,3), (4,4) \rightarrow \frac{5}{36}$
- Sum $= 9$: $(3,6), (6,3), (4,5), (5,4) \rightarrow \frac{4}{36}$
- Sum $= 10$: $(4,6), (6,4), (5,5) \rightarrow \frac{3}{36}$
- Sum $= 11$: $(5,6), (6,5) \rightarrow \frac{2}{36}$
- Sum $= 12$: $(6,6) \rightarrow \frac{1}{36}$
$(ii)$ No,$I$ do not agree with the student. The outcomes $2, 3, \dots, 12$ are not equally likely because the number of ways to obtain each sum varies. For example,there is only $1$ way to get a sum of $2$,but $6$ ways to get a sum of $7$.

(A) There are a total of $5$ days (Tuesday,Wednesday,Thursday,Friday,Saturday). Shyam can visit the shop in $5$ ways and Ekta can visit the shop in $5$ ways.
Therefore,total number of outcomes $= 5 \times 5 = 25$.
$(i)$ They can reach on the same day in $5$ ways: $(T, T), (W, W), (Th, Th), (F, F), (S, S)$.
$P(\text{same day}) = \frac{5}{25} = \frac{1}{5}$.
$(ii)$ They can reach on consecutive days in $8$ ways: $(T, W), (W, Th), (Th, F), (F, S), (W, T), (Th, W), (F, Th), (S, F)$.
$P(\text{consecutive days}) = \frac{8}{25}$.
$(iii)$ $P(\text{different days}) = 1 - P(\text{same day}) = 1 - \frac{1}{5} = \frac{4}{5}$.

(N/A) The total number of possible outcomes when two dice are thrown is $6 \times 6 = 36$.
$(i)$ The outcomes where the sum is even are:
$2, 2, 4, 4, 4, 4, 4, 4, 6, 6, 6, 6, 8, 8, 8, 8, 12$ (counting all occurrences).
Total number of outcomes with an even sum $= 18$.
$P(\text{even sum}) = \frac{18}{36} = \frac{1}{2}$.
$(ii)$ The outcomes where the sum is $6$ are:
$(3, 3), (3, 3), (3, 3), (3, 3)$ (from the table).
Total number of outcomes with sum $6 = 4$.
$P(\text{sum is } 6) = \frac{4}{36} = \frac{1}{9}$.
$(iii)$ The outcomes where the sum is at least $6$ (i.e., $\ge 6$) are:
$6, 6, 6, 6, 7, 7, 8, 8, 8, 8, 9, 9, 9, 9, 12$.
Total number of outcomes with sum at least $6 = 15$.
$P(\text{sum at least } 6) = \frac{15}{36} = \frac{5}{12}$.

(C) Let the number of blue balls be $x$.
Number of red balls $= 5$.
Total number of balls $= x + 5$.
Probability of drawing a red ball,$P(\text{Red}) = \frac{5}{x + 5}$.
Probability of drawing a blue ball,$P(\text{Blue}) = \frac{x}{x + 5}$.
According to the problem,the probability of drawing a blue ball is double that of a red ball:
$P(\text{Blue}) = 2 \times P(\text{Red})$
$\frac{x}{x + 5} = 2 \times \left( \frac{5}{x + 5} \right)$
Since $x + 5 \neq 0$,we can multiply both sides by $(x + 5)$:
$x = 2 \times 5$
$x = 10$.
Therefore,the number of blue balls in the bag is $10$.

(B) Total number of marbles $= 24$.
Let the number of green marbles be $x$.
Then,the number of blue marbles $= 24 - x$.
The probability of drawing a green marble is given by $P(\text{Green}) = \frac{\text{Number of green marbles}}{\text{Total number of marbles}} = \frac{x}{24}$.
According to the problem,$P(\text{Green}) = \frac{2}{3}$.
So,$\frac{x}{24} = \frac{2}{3}$.
Multiplying both sides by $24$,we get $x = \frac{2}{3} \times 24 = 16$.
Thus,the number of green marbles is $16$.
The number of blue marbles $= 24 - 16 = 8$.

(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 $0$.

(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 $= \frac{1}{6}$.

(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.

$(D)$ 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.