Textbook - 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,$P(F) = \frac{48}{52} = \frac{12}{13}$.

(C) Let $S$ and $R$ denote the events that Sangeeta wins the match and Reshma wins the match,respectively.
The probability of Sangeeta winning is $P(S) = 0.62$ (given).
Since the events $S$ and $R$ are complementary (assuming one must win),the probability of Reshma winning is $P(R) = 1 - P(S)$.
Therefore,$P(R) = 1 - 0.62 = 0.38$.

(A) Out of the two friends,one girl,say,Savita's birthday can be any day of the year. Now,Hamida's birthday can also be any day of $365$ days in the year.
We assume that these $365$ outcomes are equally likely.
$(i)$ If Hamida's birthday is different from Savita's,the number of favourable outcomes for her birthday is $365 - 1 = 364$.
So,$P(\text{Hamida's birthday is different from Savita's birthday}) = \frac{364}{365}$.
$(ii)$ $P(\text{Savita and Hamida have the same birthday})$
$= 1 - P(\text{both have different birthdays})$
$= 1 - \frac{364}{365} \quad [\text{Using } P(\overline{E}) = 1 - P(E)]$
$= \frac{1}{365}$.

(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,$P(G) = \frac{36}{36} = 1$.

(A) $(i)$ The sum of the probabilities of an event $E$ and its complement 'not $E$' is always $1$. Thus,$P(E) + P(\text{not } E) = 1$.
$(ii)$ The probability of an event that cannot happen is $0$. Such an event is called an impossible event.
$(iii)$ The probability of an event that is certain to happen is $1$. Such an event is called a sure event or a certain event.

(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) We know that the sum of the probability of an event and the probability of its complement is $1$.
$P(E) + P(\text{not } E) = 1$
Given that $P(E) = 0.05$.
Therefore,$P(\text{not } E) = 1 - P(E)$
$P(\text{not } E) = 1 - 0.05$
$P(\text{not } E) = 0.95$
Thus,the probability of 'not $E$' is $0.95$.

(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,$P(\text{a lemon flavoured candy}) = 1$.

(C) Let $E$ be the event that $2$ students have the same birthday.
Let $\bar{E}$ be the event that $2$ students do not have the same birthday.
It is given that $P(\bar{E}) = 0.992$.
We know that the sum of the probabilities of complementary events is $1$,so $P(E) + P(\bar{E}) = 1$.
Therefore,$P(E) = 1 - P(\bar{E})$.
$P(E) = 1 - 0.992 = 0.008$.

(A) $(i)$ Total number of balls in the bag $= 3 + 5 = 8$.
The probability of drawing a red ball is given by the formula: $P(\text{red}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$.
$P(\text{red}) = \frac{3}{8}$.
$(ii)$ The probability of not drawing a red ball is the complement of drawing a red ball.
$P(\text{not red}) = 1 - P(\text{red})$.
$P(\text{not red}) = 1 - \frac{3}{8} = \frac{8 - 3}{8} = \frac{5}{8}$.

(A) Total number of marbles $= 5 + 8 + 4 = 17$.
$(i)$ Number of red marbles $= 5$.
Probability of getting a red marble $= \frac{\text{Number of favourable outcomes}}{\text{Number of total possible outcomes}} = \frac{5}{17}$.
$(ii)$ Number of white marbles $= 8$.
Probability of getting a white marble $= \frac{\text{Number of favourable outcomes}}{\text{Number of total possible outcomes}} = \frac{8}{17}$.
$(iii)$ Number of green marbles $= 4$.
Probability of getting a green marble $= \frac{4}{17}$.
Probability of not getting a green marble $= 1 - P(\text{green}) = 1 - \frac{4}{17} = \frac{13}{17}$.

(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:
$P(\text{good pen}) = \frac{11}{12}$.

(C) $(i)$ Total number of bulbs $= 20$.
Total number of defective bulbs $= 4$.
$P(\text{defective bulb}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{4}{20} = \frac{1}{5}$.
$(ii)$ After drawing one non-defective bulb,the remaining total number of bulbs $= 20 - 1 = 19$.
The number of non-defective bulbs remaining $= (20 - 4) - 1 = 15$.
$P(\text{non-defective bulb}) = \frac{\text{Number of non-defective bulbs}}{\text{Remaining total bulbs}} = \frac{15}{19}$.

(N/A) Total number of discs $= 90$.
$(i)$ The two-digit numbers from $1$ to $90$ are $10, 11, \dots, 90$. The total count is $90 - 9 = 81$.
$P(\text{two-digit number}) = \frac{81}{90} = \frac{9}{10}$.
$(ii)$ The perfect square numbers between $1$ and $90$ are $1^2=1, 2^2=4, 3^2=9, 4^2=16, 5^2=25, 6^2=36, 7^2=49, 8^2=64, 9^2=81$. Total count is $9$.
$P(\text{perfect square}) = \frac{9}{90} = \frac{1}{10}$.
$(iii)$ The numbers between $1$ and $90$ divisible by $5$ are $5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90$. Total count is $18$.
$P(\text{divisible by } 5) = \frac{18}{90} = \frac{1}{5}$.

(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 $7$.

(C) The possible outcomes when tossing a coin $3$ times are:
$(HHH, HHT, HTH, HTT, THH, THT, TTH, TTT)$
Total number of possible outcomes $= 8$.
Hanif wins if all tosses give the same result,which are $(HHH)$ or $(TTT)$.
Number of favourable outcomes for winning $= 2$.
Probability of winning $P(\text{Win}) = \frac{2}{8} = \frac{1}{4}$.
Since the sum of the probability of an event and its complement is $1$,the probability of losing is:
$P(\text{Lose}) = 1 - P(\text{Win}) = 1 - \frac{1}{4} = \frac{3}{4}$.

(A) Total number of outcomes $= 6 \times 6 = 36$.
$(i)$ Let $E$ be the event that $5$ comes up at least once. The outcomes where $5$ appears are $(5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (1,5), (2,5), (3,5), (4,5), (6,5)$.
Number of favourable outcomes for $E = 11$.
Probability of $5$ coming up at least once,$P(E) = \frac{11}{36}$.
The probability that $5$ will not come up either time is $P(\text{not } E) = 1 - P(E) = 1 - \frac{11}{36} = \frac{25}{36}$.
$(ii)$ The probability that $5$ will come up at least once is $P(E) = \frac{11}{36}$.

(A) $(i)$ $Incorrect$
When two coins are tossed,the possible outcomes are $(H, H), (H, T), (T, H),$ and $(T, T)$.
It can be observed that there can be one of each in two possible ways: $(H, T)$ and $(T, H)$.
Therefore,the probability of getting two heads is $\frac{1}{4}$,the probability of getting two tails is $\frac{1}{4},$ and the probability of getting one of each is $\frac{2}{4} = \frac{1}{2}$.
Thus,the probability for each of these outcomes is not $\frac{1}{3}$.
$(ii)$ $Correct$
When a die is thrown,the possible outcomes are $1, 2, 3, 4, 5,$ and $6$. Out of these,$1, 3, 5$ are $odd$ and $2, 4, 6$ are $even$ numbers.
Since there are $3$ odd numbers out of $6$ total outcomes,the probability of getting an odd number is $\frac{3}{6} = \frac{1}{2}$.

(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 $10$.

(3) Total number of balls $= 12$.
Total number of black balls $= x$.
The probability of drawing a black ball is $P_1 = \frac{x}{12}$.
If $6$ more black balls are added,the new total number of balls $= 12 + 6 = 18$.
The new number of black balls $= x + 6$.
The new probability of drawing a black ball is $P_2 = \frac{x+6}{18}$.
According to the problem,$P_2 = 2 \times P_1$.
Therefore,$\frac{x+6}{18} = 2 \times \frac{x}{12}$.
$\frac{x+6}{18} = \frac{x}{6}$.
Multiplying both sides by $18$,we get $x + 6 = 3x$.
$2x = 6$,which gives $x = 3$.

(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 $= 24 - 16 = 8$.